Inheritance: Biology unit 12 notes (FBISE class 11)

Heredity and variation

Children look like their parents but are never exactly the same as them or as each other. This chapter explains how characteristics are passed on and why they also differ.

Heredity: The passing of similar characteristics from parents to their offspring.
Variations: The differences between offspring and between offspring and their parents. The resemblance is not complete.
Genetics: The science that deals with the mechanism of heredity and variation. Since genes control heredity and variations, genetics is also called the study of genes.

Both similarity and differences are parts of inheritance, and both play a significant role in the formation of new species.

GeneOften examined

What is a gene?

The gene is the basic unit of heredity. The basic unit of DNA is a nucleotide. Nucleotides are joined together to form long chains called polynucleotides. A gene is actually a small segment of DNA, made up of a sequence of nucleotides.

Each gene consists of two polynucleotide chains, which become unzipped. One of the chains determines the type of protein to be made. This chain is called the template. Three bases code for one amino acid.

From gene to polypeptide: transcription copies the template into mRNA in the nucleus, and translation at the ribosome builds the polypeptide.

From gene to protein

A cell cannot use DNA directly to make protein. The message in the template first has to be copied into messenger RNA (mRNA). This copying is called transcription and it takes place in the nucleus.
The bases in the mRNA make up codons (three bases each). The mRNA carries the message to the cytoplasm.
In the cytoplasm the ribosome helps to translate the message in the mRNA into a protein molecule. This is called translation.

Gene: A sequence of nucleotides as part of DNA, which codes for the formation of a polypeptide.

12.1Mendelian inheritance

The science of genetics originated in the year 1900 with the rediscovery of an article originally published in 1866 by an Augustinian monk named Gregor Johann Mendel. Mendel was the first who successfully explained the mechanism of inheritance, during his research work on pea plants.

12.1.1Association of inheritance with laws of MendelOften examined

In the 1860s Mendel discovered the mechanism of inheritance through his experimental work with pea plants. His work had two steps.

Mendel's two steps

Develop true (pure) breeding varieties. A plant that produces all the offspring of its own phenotype upon self-fertilization is called a true or pure breeding plant. For example, if a round-seeded plant is self-fertilized and all its offspring are also round-seeded, the parent and offspring are both true breeding.
A pure breeding plant can be developed by repeated self-fertilization through successive generations until the pure breed is achieved. Mendel studied seven pairs of contrasting traits, and for each contrasting phenotype he developed a pure breeding plant.
Hybridization. In the second step he performed hybridization, a cross-fertilization between two individuals having contrasting phenotypes.

True (pure) breeding plant: A plant that produces all the offspring of its own phenotype upon self-fertilization.
Hybridization: A cross-fertilization between two individuals having contrasting phenotypes.
Monohybrid cross: A cross between two plants that differ in a single trait, for example a round-seeded plant crossed with a wrinkled-seeded plant.
Dihybrid cross: A cross between two individuals that differ in two traits, for example round yellow seeds with wrinkled green seeds.

The study of the inheritance of a single trait, and then the combination of two or three traits one by one, was the secret of Mendel's success. From these experiments Mendel proposed two generalizations, now known as Mendel's laws of inheritance: the law of segregation (monohybrid cross) and the law of independent assortment (dihybrid cross).

Inheritance of single trait (monohybrid cross)Often examined

Mendel studied the inheritance of a single trait in a monohybrid cross, in which two plants that differ in one trait are crossed.

Procedure and observations

Mendel crossed a pure breeding round-seeded plant with a pure breeding wrinkled-seeded plant (the parents, P). This is a hybridization experiment.
The **F $_{1}$ (first filial generation)** was made up entirely of individuals showing only one of the parental phenotypes: all had round seeds.
When the F $_{1}$ plants were interbred (self-fertilized), their offspring, the **F $_{2}$ (second filial generation), showed round and wrinkled seeds in a 3 : 1** ratio.
Mendel obtained the same 3 : 1 ratio in monohybrid crosses for all the seven contrasting pairs of traits.
He then self-fertilized the F $_{2}$ plants to get the F $_{3}$ generation. One third of the F $_{2}$ round plants produced only round seeds (true breeding, like the round P $_{1}$ parent). Two thirds of the F $_{2}$ round plants produced both round and wrinkled seeds in a 3 : 1 ratio (not pure breeding, like the F $_{1}$ round). The F $_{2}$ wrinkled plants produced only wrinkled seeds (pure breeding).

Interpretation of the results

From these observations Mendel concluded that each contrasting form of a trait, such as roundness or wrinkledness of seed, was determined by a hereditary factor which he called an elementen (now called a gene). The factors carrying hereditary information are transmitted from parents to offspring through gametes.

Alleles: The alternative forms of a gene on the same locus. Each pea plant has a pair of them, one from the male parent and one from the female parent.
Genotype: The pair of factors (alleles) that controls the expression of the trait.
Phenotype: The expression of the trait, what can be seen, for example round or wrinkled seed.
Dominant: The factor that was expressed in the F $_{1}$ generation. It is written as a capital letter (R for round).
Recessive: The factor that failed to express in the F $_{1}$ generation. It is written as a small letter (r for wrinkled).
Homozygous: Having both alleles of a gene pair the same ( $R R$ or $r r$ ). The individual is called a homozygote.
Heterozygous: Having the two alleles of a gene pair different ( $R r$ ). The individual is called a heterozygote.
Monohybrid: An offspring heterozygous for the single trait being studied, for example the F $_{1}$ plant $R r$ .

How the factors explain the results

The true breeding round parent (P $_{1}$ ) carried the alleles $R R$ and the true breeding wrinkled parent carried $r r$ .
The two alleles of a pair separated from each other during gamete formation, so each gamete got only one factor for each trait. All gametes of the round parent carry $R$ , all gametes of the wrinkled parent carry $r$ .
Fertilization was random. A male gamete carrying $R$ fertilized a female gamete carrying $r$ , so the complete set of two factors for the trait ( $R r$ ) was restored in the zygote.
The zygote developed into the F $_{1}$ offspring $R r$ . It is heterozygous because its two alleles are different. It was round in phenotype but heterozygous in genotype, and its alleles also segregated during gamete formation (half the gametes carry $R$ , half carry $r$ ).

Figure 12.1: Monohybrid cross showing the inheritance of a single trait (seed shape).

Punnett square of the F $_{2}$

The Punnett square shows that, of the F $_{2}$ progeny, $\frac{1}{4}$ would be $R R$ (homozygous round), $\frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ would be $R r$ (heterozygous round) and $\frac{1}{4}$ would be $r r$ (homozygous wrinkled). Mendel actually observed a 3 : 1 phenotypic ratio. His F $_{3}$ data can also be explained by the 1 : 2 : 1 genotypic ratio of the F $_{2}$ .

Punnett square of the cross $R r \times R r$ . Genotype ratio 1 $R R$ : 2 $R r$ : 1 $r r$ ; phenotype ratio 3 round : 1 wrinkled.
F $_{1}$ gametes	$R$	$r$
$R$	$R R$ round	$R r$ round
$r$	$R r$ round	$r r$ wrinkled

Law of segregation

Mendel compared the results of all the seven separately studied characters and found them strikingly similar, so he was able to formulate the law of segregation.

Law of segregation: The two coexisting alleles for each trait in an individual segregate (separate) from each other at meiosis, so that each gamete receives only one of the two alleles. Alleles unite at random at fertilization of gametes when the zygote is formed.

Inheritance of two traits (dihybrid cross)Often examined

The inheritance of two traits at the same time can be studied in a dihybrid cross, a cross between two individuals that differ in two traits. Of the seven characters Mendel studied, two were seed shape (round dominant, wrinkled recessive) and seed colour (yellow dominant, green recessive).

Procedure and observations

Mendel crossed a homozygous round yellow plant ( $R R Y Y$ ) with a homozygous wrinkled green plant ( $r r y y$ ). All the F $_{1}$ offspring showed both dominant phenotypes: round yellow.
To analyse the genotype of the F $_{1}$ plants he self-fertilized them to produce the F $_{2}$ generation. He expected dominant and recessive combinations in a 3 : 1 ratio, as in the monohybrid cross.
Instead the offspring appeared in four phenotypic combinations: round yellow, round green, wrinkled yellow and wrinkled green, in the ratio 9 : 3 : 3 : 1. The new combinations, round green and wrinkled yellow, were surprising. They are called recombinant phenotypes.

Interpretation of the results

The F $_{1}$ offspring (round yellow) were dihybrid: heterozygous ( $R r Y y$ ) for both traits.
The key event was the self-pollination of the F $_{1}$ plants. An F $_{1}$ plant can produce four classes of gametes, $R Y$ , $r Y$ , $R y$ and $r y$ , in equal quantities.
If the sperm of the four classes are crossed with eggs of the four types, there are 16 ( $4 \times 4$ ) equally probable ways in which the alleles can combine in the F $_{2}$ generation (the Punnett square below).
These combinations make up four phenotypic categories in the ratio 9 : 3 : 3 : 1.

Figure 12.2: Dihybrid cross showing the inheritance of two traits (the 16 squares of the Punnett square are given in the table below).

Punnett square of $R r Y y \times R r Y y$ (F $_{2}$ generation, 16 equally probable combinations).
Gametes	$R Y$	$R y$	$r Y$	$r y$
$R Y$	$R R Y Y$	$R R Y y$	$R r Y Y$	$R r Y y$
$R y$	$R R Y y$	$R R y y$	$R r Y y$	$R r y y$
$r Y$	$R r Y Y$	$R r Y y$	$r r Y Y$	$r r Y y$
$r y$	$R r Y y$	$R r y y$	$r r Y y$	$r r y y$

Sorting the 16 squares into phenotypes gives 9 : 3 : 3 : 1.
Phenotype	Genotypes (squares)	Number of squares
Round, yellow	$R R Y Y$ (1), $R R Y y$ (2), $R r Y Y$ (2), $R r Y y$ (4)	9
Round, green	$R R y y$ (1), $R r y y$ (2)	3
Wrinkled, yellow	$r r Y Y$ (1), $r r Y y$ (2)	3
Wrinkled, green	$r r y y$ (1)	1

Mendel tested his seven pea characters in various dihybrid combinations and always observed a 9 : 3 : 3 : 1 phenotypic ratio in the F $_{2}$ generation. The results of his dihybrid experiments are the basis of what we now call the law of independent assortment.

Law of independent assortment: Each pair of alleles assorts independently of other pairs of alleles during gamete formation. In other words, the alleles of each pair of contrasting traits have equal probability to assort with the alleles of another pair.

Independent assortment of chromosomes

Independent assortment of genes occurs during meiosis in eukaryotes. During meiosis the pairs of homologous chromosomes are divided in half to form haploid cells, and this separation, or assortment, of homologous chromosomes is random.

Random orientation at metaphase I

At metaphase I each homologous pair of chromosomes is positioned on the equator independently of the other pairs.
The first meiotic division then results in each pair sorting its maternal and paternal homologues into the daughter cells independently of every other pair. This is called independent assortment.
Each daughter cell represents one outcome of all the possible combinations of maternal and paternal chromosomes.

Random orientation of homologous pairs at metaphase I: the two arrangements are equally likely, so two pairs give four kinds of gametes (

2^{2}

).

In humans the number of haploid chromosomes is 23. Thus the number of possible combinations of maternal and paternal chromosomes in the resulting gametes is $2^{23}$ , or about eight million. Each gamete that you produce contains one of eight million possible combinations of chromosomes inherited from your father and mother.

Each extra pair doubles the number of combinations.
Number of homologous pairs ( $n$ )	Possible chromosome combinations in gametes ( $2^{n}$ )
1	2
2	4
3	8
23 (human)	$2^{23}$ = about 8 million

Limitations of law of independent assortmentOften examined

Although Mendel's work forms the basis of heredity, it does not cover all situations.

Mendel's work applies to diploid organisms, and not all organisms are diploid.
Genes on the same chromosome cannot be expected to assort independently. An offspring that inherited one trait would also inherit the other, unless crossing over occurred.
If genes are located on the X chromosome, the pattern of distribution in the succeeding generations is different. Males have only one X chromosome (their other is Y, which does not carry many of the genes), so males are more likely to show recessive X-linked characteristics.

Usefulness and scope of independent assortmentOften examined

Besides mutation and crossing over, independent assortment of traits is also a major source of variations in successive generations.

It is only due to crossing over and independent assortment of the traits that the characteristics may appear in new combinations in the next generation. This is often necessary for adaptation in a varying environment.

Independent assortment leads to variation in the gametes

Meiosis results in genetically distinct gametes, so random fertilization by egg and sperm always generates different zygotes. This promotes genetic variation in the offspring.

The three main sources of genetic variation in sexual reproduction

Crossing over (in prophase I) exchanges parts of maternal and paternal chromatids.
Random assortment of chromosomes (in metaphase I) gives each gamete a different mixture of maternal and paternal chromosomes.
Random fusion of gametes from different parents at fertilization combines two unique gametes.

12.1.2Inheritance and mathematical probabilitiesOften examined

The chance of an event to occur is called probability. The probability ( $P$ ) that an event will occur is the number of favourable cases ( $a$ ) divided by the total number of possible cases ( $n$ ).

P = \frac{a}{n}

Probability

For example, when we toss a coin, the probability of head (or tail) is one out of two possibilities: $P = a / n = 1/2$ .

Mendel had a firm background of mathematics. He understood that the segregation of allele pairs during gamete formation, and their reforming into pairs at fertilization, obey the rules of probability.

The inheritance of a specific phenotype in a cross also has a certain probability. In a monohybrid cross the probability that an offspring shows the dominant phenotype in the F $_{1}$ generation is 100% and the recessive phenotype 0%. In the F $_{2}$ generation the probability of the dominant phenotype is $\frac{3}{4}$ and of the recessive phenotype $\frac{1}{4}$ . The combined probability of two or more independent events is calculated by the product rule.

Product rule

According to this rule the probability of two independent events happening together is the product of their individual probabilities. For example, the probability of a round yellow phenotype in the F $_{2}$ generation of a dihybrid cross is the product of the probabilities of round ( $\frac{3}{4}$ ) and yellow ( $\frac{3}{4}$ ).

P = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}

Round yellow in the F$_2$

This agrees with the 9 of the 9 : 3 : 3 : 1 ratio.

Genetic problems of the book (activity: solve at least four to illustrate the law of independent assortment)

Genetic problem 12.1: PKU and albinism

PKU and albinism are two autosomal recessive disorders, unlinked, in human beings. If a couple, each of them heterozygous for both traits, produce a child, what is the chance of their having a child with (a) PKU, (b) albinism, (c) both traits?

Let $P$ = normal and $p$ = PKU; $A$ = normal and $a$ = albinism. Both parents are $P p A a$ . Because the genes are unlinked, the two traits are inherited independently and probabilities can be multiplied.
(a) $P p \times P p$ gives $pp$ with probability $\frac{1}{4}$ .
$P (PKU) = \frac{1}{4}$
(b) $A a \times A a$ gives $aa$ with probability $\frac{1}{4}$ .
$P (albinism) = \frac{1}{4}$
(c) Both traits together: product rule.
$P = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$

(a) \frac{1}{4}, (b) \frac{1}{4}, (c) \frac{1}{16}

Genetic problem 12.2: CC × Cc

For any gene with a dominant allele

C

and a recessive allele

c

, what proportions of the offspring from a

C C \times C c

cross are expected to be homozygous dominant, homozygous recessive and heterozygous?

The $C C$ parent has only $C$ gametes, so each offspring gets $C$ from it.
Gametes of $C c$	$C$	$c$
$C$ (from $C C$ )	$C C$	$C c$

Half the offspring are $C C$ (homozygous dominant), half are $C c$ (heterozygous) and none are $cc$ (homozygous recessive), because every offspring receives at least one $C$ from the $C C$ parent.

\frac{1}{2} C C, \frac{1}{2} C c, 0 cc

Genetic problem 12.3: tall yellow × tall yellow

Two tall, yellow-seeded pea plants were crossed and some dwarf, green-seeded plants resulted. (a) What were the genotypes of the parent plants? (b) What possible genotypes might there be among the tall, yellow-seeded offspring?

In pea plants tall ( $T$ ) and yellow ( $Y$ ) are dominant to dwarf ( $t$ ) and green ( $y$ ). A dwarf green plant is $tt y y$ , so it must have received $t$ and $y$ from each parent.
(a) Each parent is tall and yellow, so it also carries $T$ and $Y$ . Both parents are therefore $T t Y y$ .
$T t Y y \times T t Y y$
(b) Tall, yellow offspring have at least one $T$ and at least one $Y$ . The possible genotypes are:
$T T Y Y T T Y y T t Y Y T t Y y (\frac{1}{16}) (\frac{2}{16}) (\frac{2}{16}) (\frac{4}{16})$

(a) T t Y y \times T t Y y; (b) T T Y Y, T T Y y, T t Y Y, T t Y y

Genetic problem 12.4: TtYy self-pollinates

A

T t Y y

pea plant self-pollinates and one seed is picked at random for planting. (a) What is the chance that the seed will produce a tall, green-seeded plant? (b) If it turns out to be tall and yellow-seeded, what is the chance that its genotype is

T T Y Y

?

(a) $T t \times T t$ gives tall plants with probability $\frac{3}{4}$ . $Y y \times Y y$ gives green seeds ( $y y$ ) with probability $\frac{1}{4}$ . Product rule:
$P = \frac{3}{4} \times \frac{1}{4} = \frac{3}{16}$
(b) Tall yellow plants are $\frac{9}{16}$ of all progeny. $T T Y Y$ is $\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$ of all progeny. Among tall yellow plants only:
$P = \frac{1/16}{9/16} = \frac{1}{9}$

(a) \frac{3}{16}, (b) \frac{1}{9}

Worked example: the product rule in a dihybrid cross

A tall plant that is heterozygous for seed colour (

T t Y y

) is crossed with a dwarf plant that is also heterozygous for seed colour (

tt Y y

). Without drawing a 16-box square, find the probability of each phenotype among the offspring, and show that the four values add to 1.

Treat each gene separately. For height, $T t \times tt$ gives half tall and half dwarf.
$P (tall) = \frac{1}{2}, P (dwarf) = \frac{1}{2}$
For seed colour, $Y y \times Y y$ gives three quarters yellow and one quarter green.
$P (yellow) = \frac{3}{4}, P (green) = \frac{1}{4}$
The genes are on different chromosomes, so the two traits are independent. Multiply the probabilities for each combination (product rule). The result is set out in the table.
Check: add the four probabilities.
$\frac{3}{8} + \frac{1}{8} + \frac{3}{8} + \frac{1}{8} = 1$

Phenotype ratio 3 : 1 : 3 : 1.
Phenotype	Working	Probability
Tall, yellow	$\frac{1}{2} \times \frac{3}{4}$	$\frac{3}{8}$
Tall, green	$\frac{1}{2} \times \frac{1}{4}$	$\frac{1}{8}$
Dwarf, yellow	$\frac{1}{2} \times \frac{3}{4}$	$\frac{3}{8}$
Dwarf, green	$\frac{1}{2} \times \frac{1}{4}$	$\frac{1}{8}$

P = \frac{3}{8}, \frac{1}{8}, \frac{3}{8}, \frac{1}{8}

12.2Exceptions to Mendelian inheritance

Since Mendel's time our knowledge of the mechanisms of genetic inheritance has grown immensely. It is now understood in how many different ways alleles interact with their contrasting partner alleles at the same locus. These relationships between the contrasting alleles at the same locus in the heterozygous state are called dominance relations.

Mendel observed only one form of dominance relation, complete dominance. Later, many geneticists explained several exceptions to Mendelian inheritance that could not be explained on the basis of complete dominance. These exceptions are said to have non-Mendelian inheritance patterns.

Incomplete dominance
Co-dominance
Multiple alleles

12.2.1Incomplete dominanceOften examined

Studies of the inheritance of many traits have shown that one member of a pair of alleles may not be completely dominant to the other. For example, red and white are common flower colours in the Japanese four o'clock (Mirabilis jalapa). Each colour phenotype produces the same phenotype when the plants are self-pollinated.

Without knowing which is dominant, we might predict that all would have red flowers or all would have white flowers. In 1899 this cross was first made by the German botanist Carl Correns, who found that all F_1 offspring have pink flowers.

When two of these pink-flowered plants are crossed, red-flowered, pink-flowered and white-flowered offspring appear in a ratio of 1 : 2 : 1. The pink-flowered plants are clearly the heterozygous individuals, and neither the red allele nor the white allele is completely dominant. When the heterozygote has a phenotype that is intermediate between those of its two parents, the dominance relation between the parental genes is said to be incomplete dominance.

Incomplete dominance: A dominance relation in which neither allele of a pair is completely dominant, so the heterozygote shows a phenotype intermediate between those of its two homozygous parents.

Figure 12.3: Incomplete dominance in four o'clock plants (Mirabilis jalapa).

Punnett square of the F_2: phenotype ratio 1 red : 2 pink : 1 white, which is the same as the genotype ratio 1 : 2 : 1.
F_1 gametes	R_1	R_2
R_1	R_1R_1 red	R_1R_2 pink
R_2	R_1R_2 pink	R_2R_2 white

Genetic problem 12.5: pink × red four o'clock

When a pink-flowered four o'clock plant is crossed with a red-flowered plant, what is the probability of (a) a red-flowered plant, (b) a pink-flowered plant, (c) the ratio of pink to red?

Pink is R_1R_2 and red is R_1R_1. Gametes: pink gives R_1 or R_2 (half each); red gives only R_1.

Gametes	R_1 (red parent)
R_1 (pink parent)	R_1R_1 red
R_2 (pink parent)	R_1R_2 pink

(a) Red: \tfrac12. (b) Pink: \tfrac12. (c) Pink : red = 1 : 1.

\text{(a) }\tfrac12,\ \text{(b) }\tfrac12,\ \text{(c) }1:1

Worked example: pink × white four o'clock

A pink-flowered four o'clock plant is crossed with a white-flowered plant. Give the genotypes and the phenotypes of the offspring, and say how this cross shows incomplete dominance.

Choose symbols. Red is R_1R_1, white is R_2R_2, pink is R_1R_2. The cross is R_1R_2\times R_2R_2.
R_1R_2\times R_2R_2
Gametes: the pink parent makes R_1 and R_2 in equal numbers; the white parent makes only R_2.
Fill the Punnett square.

Half the offspring are pink and half are white (1 : 1).
Gametes	R_2 (white parent)
R_1 (pink parent)	R_1R_2 pink
R_2 (pink parent)	R_2R_2 white

The heterozygous offspring are pink, a colour between red and white, and neither allele hides the other. If red were completely dominant they would be red, so the 1 : 1 pink : white result is possible only because dominance is incomplete.

\tfrac12\ \text{pink }(R_1R_2),\ \tfrac12\ \text{white }(R_2R_2)

12.2.2Co-dominanceOften examined

Another variation of dominance relationships between alleles is called co-dominance, in which both contrasting alleles at the same locus express independently without influencing each other, so the phenotypes of both alleles become apparent.

Co-dominance: A dominance relation in which both contrasting alleles at the same locus express themselves independently in the heterozygote, so the phenotype of both alleles is seen at the same time.

For example, the human MN blood group is determined by two co-dominant alleles (L^M and L^N) for two specific molecules located on the surface of the red blood cells, the M and N molecules. A single gene locus at which two allelic variations are possible determines the phenotypes of this blood group.

Individuals homozygous for the L^M allele (L^ML^M) have red blood cells with only M molecules.
Individuals homozygous for the L^N allele (L^NL^N) have red blood cells with only N molecules.
Both M and N molecules are present on the red blood cells of individuals heterozygous for the M and N alleles (L^ML^N).

Table 12.1: MN blood group system as an example of co-dominance.
Blood group phenotype	Antigen	Genotype
M blood	M antigen	L^ML^M
N blood	N antigen	L^NL^N
MN blood	Both antigens	L^ML^N

Incomplete dominance and co-dominance as variations of Mendel's research

Many patterns of inheritance which cannot be explained on the basis of Mendel's laws alone were discovered in plants and animals. Such patterns are described as non-Mendelian inheritance.

Incomplete dominance is a type of interaction where both the alleles of a given trait express as a blend (mixture), as against a normal Mendelian pattern where one allele is dominant over the other. As a result of this blending an intermediate character is expressed.
Co-dominance represents a situation where two allelic genes when present together in an individual express their traits independently instead of showing a typical dominant-recessive relationship.
In both cases the heterozygous progeny of the F_2 generation shows a phenotype that is different from both homozygous parents.

Table 12.2: Difference between incomplete dominance and co-dominance.
Incomplete dominance	Co-dominance
In the heterozygous state both genes blend their phenotypic effects.	In the heterozygous state both genes independently express their phenotypic effects.
The heterozygotes show an intermediate phenotype between the two parental phenotypes.	The heterozygotes show both parental phenotypes at a time.
Example: flower colour of the 4 o'clock plant.	Example: human MN and AB blood groups.

Gene and allele

Alleles are the alternative forms of a gene present at the same locus on a homologous chromosome. Many genes have more than two allelic forms, which is known as multiple alleles.

A diploid cell contains any two of the alleles, one each on the homologous chromosomes, and a haploid cell contains only one allele.

Alleles occupy the same locus on the two homologous chromosomes. A diploid cell carries two alleles of each gene.

Difference between genes and alleles.
Gene	Allele
The basic unit of heredity: a segment of DNA, a sequence of nucleotides that codes for a polypeptide.	An alternative form of a gene.
Occupies a particular position (locus) on a chromosome.	Alleles of a gene occupy the same locus on homologous chromosomes.
A diploid cell has two copies of each gene.	The two copies may be the same allele (homozygous) or different alleles (heterozygous).

Multiple allelesOften examined

Multiple alleles: Three or more alternative forms (alleles) of a gene, which all occupy the same locus. They are produced by gene mutation.

Some traits may be controlled by as many as 100 or more multiple alleles, but each individual carries only two of them, because each locus is represented twice in a diploid individual.

For example, if a trait is controlled by three multiple alleles, A_1, A_2 and A_3, every individual carries any two of them. The ABO blood groups in humans are one example of multiple alleles.

Three multiple alleles give six genotypes, but any one person carries only two alleles.
Alleles in the population	Possible genotypes of an individual
A_1, A_2, A_3	A_1A_1, A_1A_2, A_1A_3, A_2A_2, A_2A_3, A_3A_3 (six genotypes)

12.3Blood group systems

There are a number of different blood group systems found in humans. The International Society of Blood Transfusion has recognized up to 30 major blood group systems. These systems are characterized by the presence or absence of specific molecules, called antigens, that are situated on the surface of the red blood cells. Most antigens are glycoprotein molecules.

The two main blood group systems are the ABO system and the Rh (Rhesus) system. These two are more significant because incompatibility between donor and recipient blood with respect to them may become dangerous to life.

There are more than two hundred minor blood groups (those of the blood group systems other than ABO and Rh) that usually do not complicate blood transfusions. These are known as rare blood types. One example of a blood group system other than ABO and Rh is the MN blood group system, already given under co-dominance.

Antigens (of blood groups): Specific molecules, mostly glycoproteins, situated on the surface of the red blood cells. Blood group systems are characterized by their presence or absence.

12.3.1ABO blood group systemOften examined

A well-known example of multiple alleles is the ABO blood group system in humans. It was discovered by Karl Landsteiner in 1901. ABO blood groups are also found in many other primates such as the chimpanzees, bonobos and gorillas.

Antigens of the ABO system

There are several surface markers on the red blood cell (RBC) membrane, generally called antigens. The ABO system is based upon two such antigens, the A antigen and the B antigen, which are glycoproteins in nature.

If the A antigen is present on the surface of the RBC, the blood group is type A.
If the B antigen is present, the blood group is type B.
If both antigens are present, the blood group is type AB.
If none of these is present, the blood group is type O.

Genetic basis of the ABO system

The ABO blood group is controlled by an autosomal single polymorphic gene I (which stands for isohaemagglutinogen). The four blood types result from various combinations of three different alleles, symbolized as I^A (for the ability to make substance A), I^B (for B) and i (for neither A nor B).

I^A and I^B are completely dominant over the allele i.
I^A and I^B are co-dominant to each other, because each expresses equally in the heterozygous (I^AI^B) state to produce the AB phenotype.
Therefore the genotypes I^AI^A or I^Ai produce phenotype A; I^BI^B or I^Bi produce phenotype B; the homozygous ii produces phenotype O.

The blood group alleles start their expression at an early embryonic stage and keep on expressing themselves till death. Therefore the blood group phenotype of a person never changes throughout life.

The ABO blood groups: antigens on the red blood cells and antibodies in the plasma.

Table 12.3: Multiple alleles and ABO blood groups. In the book the last column is headed "Transfusions"; it lists the groups from which a person can safely receive blood.
Blood group (phenotype)	Antigen	Genotypes	Antibodies	Transfusions (can receive from)
A	A	I^AI^A or I^Ai	anti-B	A and O
B	B	I^BI^B or I^Bi	anti-A	B and O
AB	Both	I^AI^B	None	Any
O	None	ii	Both (anti-A and anti-B)	Only O

Antibodies of the ABO system

If a wrong transfusion is carried out, the recipient's blood starts agglutination (an antigen-antibody reaction) and clumping occurs. This is due to the presence of antibodies against the wrong antigen. The antibodies are produced in the absence of their corresponding antigens.

Type A people have anti-B antibody in their plasma.
Type B people have anti-A antibody.
Type AB people have no anti-A or anti-B antibodies.
Type O individuals have no antigens but have both anti-A and anti-B antibodies in the plasma.

The production of these ABO antibodies does not require any stimulus; it begins from embryonic life and remains throughout life.

Various human blood group systems

At present 33 blood group systems representing over 300 antigens have been listed by the International Society of Blood Transfusion. Some of the important groups, with their symbols, are given here: ABO (ABO), MNS (MNS), Rhesus (Rh), Lutheran (LU), Kell (KELL), Lewis (LE), Duffy (FY) and Kidd (JK).

Some important human blood group systems.
System	Points given in the book
ABO	Among the 33 systems, ABO remains the most important in transfusion and transplantation, since any person above the age of 6 months possesses clinically significant anti-A and/or anti-B antibodies in the serum.
Rhesus	The second most important blood group system after ABO. Prophylaxis with anti-D immunoglobulin is given to pregnant Rh-negative mothers who have given birth to an Rh-positive child.
MNS	Controlled by an autosomal locus on chromosome 4 and by a pair of co-dominant alleles L^M and L^N. Anti-M and anti-N antibodies are usually IgM types and are rarely associated with transfusion reactions.
Lutheran	Made up of four pairs of allelic antigens representing a single amino acid substitution in the Lutheran glycoprotein at chromosome 19. Antibodies against this group are rare and generally not considered clinically significant.
Kell	Erythrocyte antigens defined by an immune antibody, anti-K. First noticed in the serum of Mrs Kellacher, who reacted to the erythrocytes of her newborn infant, resulting in haemolytic reactions. Since then 25 Kell antigens have been discovered. Anti-K causes severe haemolytic disease of the fetus and newborn (HDFN) and haemolytic transfusion reactions (HTR).
Duffy	First isolated in a patient called Duffy who had haemophilia. Also known as Fy glycoprotein, present on the surface of RBCs. The antigens Fy^a and Fy^b can result in four possible phenotypes. The antibodies are IgG subtypes and can cause HTR.
Kidd	The Kidd antigen (known as Jk antigen) is a glycoprotein present on the membrane of RBCs; it acts as a urea transporter in RBCs and renal endothelial cells. Kidd antibodies are rare but can cause severe transfusion reactions.

12.3.2Transfusion principleOften examined

When a transfusion is carried out between two incompatible (different) blood groups, the antigens of the donor react with the antibodies (also called agglutinins) of the recipient, and the red blood cells clump with one another. The clumping of red blood cells is known as agglutination. Therefore transfusions are carried out on the basis of the donor's antigens and the recipient's antibodies.

Agglutinins: The antibodies of the recipient that react with the antigens of an incompatible donor.
Agglutination: The clumping of red blood cells when the donor's antigens react with the recipient's antibodies.

Figure 12.4: Blood transfusion model.

Transfusion rules from the donor's antigens and the recipient's antibodies.
Recipient group	Antibodies in the recipient	Can safely receive blood from	Why
A	anti-B	A and O	Blood carrying the B antigen would be attacked by anti-B.
B	anti-A	B and O	Blood carrying the A antigen would be attacked by anti-A.
AB	none	Any group	It has no antibody to react with the donor's blood, so AB is the universal recipient.
O (negative)	anti-A and anti-B	Only O	It has both antibodies, which react with any antigenic blood (A, B or AB). But it can donate to anyone because it has no antigen to interfere with the recipient's blood, so O negative is the universal donor.

Persons with type A blood can receive blood from type A or from type O because they have anti-B antibody, so they cannot be given blood carrying the B antigen. Similarly, persons with type B blood can receive blood from type B or from type O. Persons with type AB can receive any blood, hence they are called universal recipients. Persons with type O negative blood can receive blood only from type O, but can donate to anyone, hence they are called universal donors.

Genetic problem 12.6: type A husband, type B wife

A person has type A blood group while his wife has type B. They have four children, each with a different blood group of the ABO system. Explain how this is possible.

The four ABO groups are A, B, AB and O. A child of group O is ii, so each parent must carry an i allele. Hence the husband (group A) is I^Ai and the wife (group B) is I^Bi.
I^Ai\times I^Bi
The gametes are I^A or i from the father and I^B or i from the mother. The four combinations are all different.

Each of the four groups has a probability of \tfrac14.
Gametes	I^B	i
I^A	I^AI^B (AB)	I^Ai (A)
i	I^Bi (B)	ii (O)

I^Ai\times I^Bi\ \text{gives AB, A, B and O}

Genetic problem 12.7: woman of type B with a type O child

A woman with blood type B has a child with blood type O. What are the genotypes of the mother and child? Which genotypes could the father not have?

The child is type O, so its genotype is ii. It received one i from each parent.
ii
The mother is type B and gave the child an i, so she is heterozygous: I^Bi.
I^Bi
The father must also have given an i, so he must carry at least one i. He could be I^Ai, I^Bi or ii. He could not be I^AI^A, I^BI^B or I^AI^B (so he is not group AB, and not homozygous A or B).

\text{mother } I^Bi;\ \text{child } ii;\ \text{father not } I^AI^A,\ I^BI^B,\ I^AI^B

Worked example: the children of an AB father and an O mother

A man of blood group AB marries a woman of group O. Give the possible blood groups of their children and say whether a child of group AB or group O is possible.

Write the genotypes. Group AB is always I^AI^B and group O is always ii.
I^AI^B\times ii
Gametes: the father gives I^A or I^B; the mother gives only i.
Fill the Punnett square.

Half the children are group A and half are group B.
Gametes	i (mother)
I^A (father)	I^Ai (group A)
I^B (father)	I^Bi (group B)

A child of group AB is impossible because the mother has no I^A or I^B allele to give. A child of group O is impossible because the father has no i allele to give.

\tfrac12\ \text{group A},\ \tfrac12\ \text{group B}

Alternatives to blood transfusion

Alternatives to blood transfusion (box in the book, "Derive an idea to get alternatives of blood transfusion"). Some approaches decrease the need for a transfusion: (1) volume expanders (normal saline, and saline with some chemicals added such as Ringer's solution) prevent or treat the shock associated with loss of body fluids; (2) haematopoietic growth factors, made in the laboratory, encourage the bone marrow to make more red blood cells in people with low blood cell counts; (3) erythropoietin, a hormone naturally produced by the kidneys, stimulates the body to produce more red blood cells and is widely used as a transfusion alternative to treat anaemia; (4) aprotinin, a drug given before heart surgery to reduce the risk of bleeding and the need for transfusion.

12.3.3Rh blood group systemOften examined

The Rh blood group system is clinically the most important blood group system after ABO. The name of the system (Rh) is derived from the Rhesus monkey, because its antigen was first discovered in it by Landsteiner in the 1930s.

Antigens of the Rh blood group system

The Rh system is characterized by the presence or absence of the D antigen, also called the Rh factor. The persons having this antigen are called Rh positive and those in which it is absent are called Rh negative. Incompatibility of the D antigen between donor and recipient can cause problems not only during blood transfusion, but it is also a relevant cause of the haemolytic disease of the newborn or erythroblastosis foetalis.

Genetic basis of the Rh blood group system

The Rh system is mainly controlled by the D gene, which determines the formation of the D antigen or Rh factor, while its alternative allele d inhibits the formation of the Rh factor. D is completely dominant over d. Therefore persons with genotypes DD or Dd have the D antigen on their RBCs and are Rh positive. Persons with genotype dd do not have the Rh factor and are Rh negative.

Table 12.4: Rh blood group system.
Blood group (phenotype)	Rh antigen (factor)	Genotypes	Anti-Rh antibody	Transfusions (can receive from)
Rh positive	Present	DD or Dd	Not produced	Rh positive, Rh negative
Rh negative	Absent	dd	Produced (if stimulated)	Rh negative

The blood group O negative is the actual universal donor because it has no antigen of the ABO system and of the Rh system that can interfere with the recipient's blood, whereas AB positive is the actual universal recipient because it has neither anti-A and anti-B nor anti-Rh antibodies, therefore it cannot resist any donor's blood.

Anti-Rh antibody and transfusion principle

The Rh system also has a mechanism of antibody production, the anti-Rh antibody, which is produced in Rh-negative blood. It differs from ABO antibody production: the production of anti-Rh antibody always requires a stimulus in the form of exposure to the Rh factor.

Rh incompatibility in transfusion

An Rh-negative person does not produce anti-Rh antibodies unless he is exposed to the Rh antigen.
An Rh-positive donor is totally incompatible for an Rh-negative recipient. If an Rh-negative person receives Rh antigen through a wrong Rh-positive transfusion, he will begin to produce anti-Rh antibodies against the Rh antigens.
Once the mechanism of anti-Rh antibody production is stimulated, it continues for the whole life.
An Rh-negative donor, clear of any anti-Rh antibody because he has never been exposed to the Rh antigen, can be transfused to an Rh-positive recipient.

Antibody production in the two main systems (from the book's text).
ABO antibodies	Anti-Rh antibody
Produced without any stimulus, from embryonic life.	Produced only after exposure to the Rh antigen (needs a stimulus).
Present in the plasma throughout life against the antigen the person lacks.	Once stimulated, production continues for the whole life.

12.3.4Erythroblastosis foetalisOften examined

Erythroblastosis foetalis develops in a foetus when anti-Rh antibodies produced by the mother pass through the placenta.

Problems and complications in the foetus

The mother's anti-Rh antibodies seep through the placenta into the blood circulation of the foetus.
They agglutinate the foetal Rh-positive red cells, which start haemolysis (breakdown of blood).
As this destruction continues, the foetus becomes anaemic. The anaemic foetus starts to release many immature erythroblasts into his blood stream; therefore the disease is called erythroblastosis foetalis. The anaemia may lead to abortion or still birth.
Even if the pregnancy continues, the liver and spleen of the foetus swell as they rapidly produce RBCs. The breakdown product of RBCs, bilirubin, also accumulates in the foetus. Bilirubin damages brain cells and turns the skin and whites of the eye yellow, which is called jaundice.
The baby, if born alive, suffers from severe haemolytic anaemia and jaundice. Such a baby's blood should be immediately replaced by Rh-negative blood free of anti-Rh antibodies.

Figure 12.5: Maternal-foetal Rh incompatibility. Maternal anti-Rh antibodies cross the placenta, and agglutination of the foetal Rh-positive red cells leads to haemolysis.

Causes and risk factors

Erythroblastosis foetalis most commonly happens when a woman with Rh-negative blood marries a man with Rh-positive blood and conceives a baby with Rh-positive blood (maternal-foetal Rh incompatibility). If the man's genotype is DD, all of their offspring will have the Dd genotype and will be Rh positive. If the man's genotype is Dd, half of their offspring will have the Dd genotype and will be Rh positive. There is always a chance of erythroblastosis foetalis whenever an Rh-positive foetus is conceived by an Rh-negative mother.

Chance of an Rh-positive foetus in an Rh-negative mother.
Father	Mother (dd)	Children
DD (Rh positive)	dd (Rh negative)	All Dd, all Rh positive (all at risk)
Dd (Rh positive)	dd (Rh negative)	Half Dd (Rh positive, at risk), half dd (Rh negative)

Prevention of erythroblastosis foetalis

Erythroblastosis foetalis is preventable. A medication called Rh immunoglobulin (Rhig), also known as RhoGAM, can help to prevent Rh sensitization. This medication prevents the pregnant woman from developing Rh-positive antibodies. The injection contains anti-Rh antibodies that attack any of the baby's red blood cells in the mother's blood before these cells stimulate her immune system to produce her own antibodies. However, this will not help women who have already undergone Rh sensitization.

Women at risk for Rh sensitization should receive RhoGAM doses at specific times during their pregnancy and after delivery:

at 28 weeks of gestation;
within 72 hours following delivery, if the newborn is Rh positive;
within 72 hours of a miscarriage, abortion or ectopic pregnancy (when the fertilized egg implants and grows outside the main cavity);
following an invasive prenatal test, such as an amniocentesis or CVS;
after any vaginal bleeding.

Genetic problem 12.8: Rh-negative woman and Rh-positive man

An Rh-negative woman is married to an Rh-positive man, whose father was also Rh negative. What are the possible genotypes of each person in the family, and what are the chances that their child will be affected with erythroblastosis foetalis?

The woman is Rh negative, so she is dd.
dd
The man's father was Rh negative (dd), so he passed a d to his son. The son is Rh positive, so his other allele must be D: the man is Dd.
Dd
Cross dd\times Dd: half the children are Dd (Rh positive) and half are dd (Rh negative).

The mother makes only d gametes.
Gametes	D (father)	d (father)
d (mother)	Dd Rh positive	dd Rh negative

Only an Rh-positive child can be affected, since the mother's anti-Rh antibodies attack the D antigen. So the chance of an affected child is \tfrac12 (with the mother sensitized by an earlier exposure to Rh-positive cells). The Rh-negative children (dd) are not affected.

\text{mother } dd;\ \text{father } Dd;\ \text{chance of an Rh-positive (at risk) child}=\tfrac12

Worked example: Rh-positive parents with an Rh-negative child

Two Rh-positive parents have an Rh-negative child. Explain how this is possible and find the probability that their next child is Rh negative.

An Rh-negative child is dd, so it received a d allele from each parent.
Each parent is Rh positive, so each has the D antigen (DD or Dd). Since each also carries a d, both parents must be heterozygous: Dd\times Dd.
Dd\times Dd
Fill the Punnett square.

Genotype ratio 1 DD : 2 Dd : 1 dd; phenotype ratio 3 Rh positive : 1 Rh negative.
Gametes	D	d
D	DD Rh+	Dd Rh+
d	Dd Rh+	dd Rh−

Each pregnancy is an independent event, so the probability that the next child is Rh negative is again \tfrac14, whatever happened before.

P(\text{Rh negative})=\dfrac14

12.4Polygenic inheritance and epistasisOften examined

Some traits have a large number of alternative phenotypes with small and less striking differences between them, so they show continuous variations. Examples are height, weight, intelligence and skin colour in humans, and grain colour in wheat.

Such traits cannot be encoded by a single gene with two alleles. Even a few multiple alleles of a single gene cannot make a large number of phenotypes. Such traits are encoded by the alleles of two or more different gene pairs found at different loci, all influencing the same trait in an additive way: the intensity of the phenotype depends upon the number of particular effect-causing alleles.

Polygenic traits: Quantitative traits that are encoded by two or more gene pairs at different loci, all influencing the same trait in an additive way. They show continuous variation.
Polygenes: All the genes that control a quantitative trait. Each has a small positive or negative effect on the character.

Polygenes supplement each other, and the sum of the positive and negative effects of all the individual polygenes produces the quantitative phenotype of a continuously varying trait.

Comparison drawn from the book's description of polygenic traits.
Single gene (Mendelian) trait	Polygenic trait
One gene pair, two alleles.	Two or more gene pairs at different loci.
A few distinct phenotypes (for example round or wrinkled).	Many phenotypes with small differences between them (continuous variation).
Phenotype depends on which allele is dominant.	Phenotype depends on the number of effect-causing alleles (additive effect).

12.4.1Wheat grain colourOften examined

Wheat grain colour is a good example of polygenic (multiple gene) inheritance. Wheat grains show a continuous variation in colour from white to dark red. Approximately seven different phenotypes (Table 12.5) are found in the wheat population all over the world.

When a homozygous dark red grain plant was crossed with a homozygous white grain plant, all the F_1 grains had light red colour, intermediate between the two parental shades. It seemed as if it were a case of incomplete dominance. But when the F_1 grains were grown to mature plants and crossed with each other, the F_2 grains had exactly seven shades of colour in the ratio given in Table 12.5.

Table 12.5: Range of phenotypes in wheat grain colours.
Pigment alleles	6	5	4	3	2	1	0
Range of phenotypes	Dark red	Moderately dark red	Red	Light red	Pink	Light pink	White
Ratio	1	6	15	20	15	6	1

Figure 12.6: Inheritance of wheat grain colour (P, F_1 and the eight gamete types of the F_1 plant; the 64 squares are summarised in the next table).

The 64 squares of Fig 12.6, each shown as the number of pigment-producing (capital) alleles in the zygote. Counting them gives 1 + 6 + 15 + 20 + 15 + 6 + 1.
Gametes	ABC	ABc	AbC	Abc	aBC	aBc	abC	abc
ABC	6	5	5	4	5	4	4	3
ABc	5	4	4	3	4	3	3	2
AbC	5	4	4	3	4	3	3	2
Abc	4	3	3	2	3	2	2	1
aBC	5	4	4	3	4	3	3	2
aBc	4	3	3	2	3	2	2	1
abC	4	3	3	2	3	2	2	1
abc	3	2	2	1	2	1	1	0

F_2 grain colours of wheat

Number of F_2 plants out of 64 for each number of pigment alleles: a bell-shaped (continuous) distribution.

Finally it was identified that actually three different gene pairs, Aa, Bb and Cc, at three different loci contribute to the wheat grain colour. Each individual contains six alleles for the trait. The alleles A, B and C code for an equal amount (dose) of red pigment, which is a positive effect. None of a, b and c encodes red pigment, which is a negative effect.

The colour depends on the number of pigment-producing alleles, not on which of A, B, C they are.
Number of pigment alleles	Examples of genotypes	Grain colour
6	AABBCC	Dark red
5	AABBCc, AaBBCC, AABbCC	Moderately dark red
4	AABBcc, aaBBCC, AAbbCC	Red
3	AaBbCc, AABbcc, AAbbCc	Light red
2	AaBbcc, aaBbCc, AabbCc	Pink
1	Aabbcc, aaBbcc, aabbCc	Light pink
0	aabbcc	White

Thus the colour phenotypes of grains depend upon the number of pigment-producing alleles (A, B and C). Environmental factors like light, water and nutrients also influence the amount of grain colour.

12.4.2Inheritance of human skin colourOften examined

Human skin colour is a good example of polygenic inheritance. Skin colour is largely determined by the amount of melanin the skin produces. Dark-skinned individuals produce more melanin than light-skinned individuals. At least three genes A, B and C regulate the amount of melanin produced.

Each gene has two forms. The dark-skin alleles are represented by capital letters (A, B and C) and the light-skin alleles by small letters (a, b and c). No allele is completely dominant (incomplete dominance), and heterozygotes exhibit an intermediate phenotype, because A, B and C act as dark-skin alleles: they add pigment by increasing melanin production. On the other hand, a, b and c act as light-skin alleles in the genotype because they inhibit melanin production.

There are seven different shades of skin colour ranging from very light (aabbcc) to very dark (AABBCC). Most individuals have the intermediate skin colour (AaBbCc). This AaBbCc genotype would be characteristic of a mulatto (an offspring of a black and a white parent).

Figure 12.7 (part 1): Inheritance of human skin colour, P and F_1 generations.

Skin colour in the F_2 generation

Figure 12.7 (part 2): the proportion of the F_2 population in each skin-colour class is 1/64, 6/64, 15/64, 20/64, 15/64, 6/64 and 1/64, a bell-shaped curve with the intermediate shade the most common.

Examples of polygenic traits in humans

The examples of polygenic traits in humans are many, such as:

height
skin colour
the risk for diseases and resistance
intelligence
blood pressure
autism
longevity
cancer
heart disease
diabetes
bipolar disorder (formerly called manic-depressive illness or manic depression)

12.4.3EpistasisOften examined

An example of non-allelic interaction is epistasis.

Epistasis: The phenomenon in which the effect caused by the genes at one locus interferes with or hides the effect caused by another gene at another locus.
Epistatic gene (inhibiting gene): The gene that suppresses or masks the effect of action of a gene at another locus.
Hypostatic gene: The gene whose effect is suppressed.

Epistasis must not be confused with dominance. Dominance is the relationship between alleles of the same gene occupying the same locus, but epistasis is the interaction between different genes occupying different loci.

Table 12.6: Difference between dominance and epistasis.
Dominance	Epistasis
It involves a single pair of alleles.	It involves two pairs of alleles (two genes).
A gene suppresses the expression of its own allele.	A gene suppresses the expression of an allele of another gene.
Only the recessive allele is suppressed.	It suppresses the effect of both dominant and recessive alleles of another gene.
The effect is only due to the dominant allele.	Both a dominant and a recessive allele can become epistatic.

Epistasis and the ABO blood group: the Bombay phenotype

The expression of the ABO genotypes (locus on chromosome 9) also depends upon another gene H (locus on chromosome 19) that encodes a particular H substance. The H substance is a precursor to the A and B antigens. For instance, the I^B allele must be present to produce the B enzyme that modifies the H substance to become the B antigen. It is the same for the I^A allele. However, if only recessive alleles for the H substance are inherited (hh), the H substance will not be produced. Subsequently the A and B antigens will also not be produced. The result is an O phenotype by default, since a lack of A and B antigens is the O type. This seemingly impossible phenotype result has been referred to as a Bombay phenotype because it was first described in that Indian city.

The H gene is epistatic to the ABO gene: without H substance the I^A and I^B alleles cannot show their effect.

Relationship of epistasis with polygenic inheritance

Both epistasis and polygenic inheritance involve the interaction of more than one gene. The distinction between the two phenomena is that in epistasis the traits show discontinuous variations, whereas in polygenic inheritance the traits show continuous variations in the population.

Relationship between epistasis and polygenic inheritance.
Epistasis	Polygenic inheritance
Interaction of two or more genes at different loci.	Interaction of two or more genes at different loci.
One gene hides the effect of another.	The genes add their effects together.
Traits show discontinuous variation (distinct classes).	Traits show continuous variation in the population.

Coat colour in the Labrador retrieverOften examined

Coat colour in the Labrador retriever (a highly popular type of dog) is an excellent example of epistasis in animals. They have coats of three basic colours: yellow, black and chocolate.

The allele for black coat colour (B) is dominant to the chocolate colour (b). A puppy will be chocolate only if each parent contributes the chocolate allele (bb). If one (Bb) or both (BB) parents contribute the black allele, the puppy will be black (BB or Bb).
The gene that determines yellow coat colour is at a different locus in the DNA from the black versus chocolate gene. In order to be yellow, a Labrador must have two recessive copies of the yellow gene (ee).
In this case the yellow colour genes become epistatic and completely inactivate the black or brown genes, and the puppy is yellow. This means both parents contributed a yellow gene (e).
If only one (Ee) or no (EE) yellow genes are contributed, the puppy will be either black or chocolate, as determined by what is on the black/chocolate gene.
Two yellow Labradors (ee) can only have yellow puppies (ee), since both have two copies of the yellow gene and that is all they can contribute.

How the two genes decide the coat colour of a Labrador retriever.

If a black Labrador homozygous for both gene pairs (BBEE) is crossbred with a yellow (bbee) partner, all the offspring will be black (BbEe). If such black Labradors, heterozygous for both gene pairs (BbEe), are interbred, all three coat colours (black, chocolate and yellow) are expected in the offspring in a 9 : 3 : 4 ratio.

Figure 12.9: Inheritance of coat colour of Labrador retriever, BbEe\times BbEe. Each gamete type has probability \tfrac14.
Gametes	BE	Be	bE	be
BE	BBEE black	BBEe black	BbEE black	BbEe black
Be	BBEe black	BBee yellow	BbEe black	Bbee yellow
bE	BbEE black	BbEe black	bbEE chocolate	bbEe chocolate
be	BbEe black	Bbee yellow	bbEe chocolate	bbee yellow

The ratio 9 : 3 : 4 is a modified form of the Mendelian 9 : 3 : 3 : 1, because the two yellow classes (B\_ee and bbee) look the same.
Phenotype	Genotypes	Squares of 16
Black	B\_E\_ (BBEE, BBEe, BbEE, BbEe)	9
Chocolate	bbE\_ (bbEE, bbEe)	3
Yellow	B\_ee (BBee, Bbee) and bbee	3 + 1 = 4

Genetic problem 12.9: two chocolate Labradors

When two chocolate coloured Labradors were crossed, a yellow puppy was born. What is the probability of a yellow coat coloured puppy if the parents are again crossed?

A chocolate dog is bb and has at least one E (bbE\_). A yellow puppy is ee, so each parent must have given an e.
Therefore both parents are bbEe. The cross is bbEe\times bbEe.
bbEe\times bbEe
Ee\times Ee gives ee with probability \tfrac14. Every puppy is bb, and ee hides this, so these puppies are yellow.
P(\text{yellow})=\dfrac14

P(\text{yellow})=\tfrac14

Worked example: a test cross in Labradors

A black Labrador of genotype BbEe is crossed with a yellow Labrador of genotype bbee. Give the genotypes and the ratio of the phenotypes in the litter.

The black dog makes four kinds of gametes, BE, Be, bE and be, in equal numbers. The yellow dog makes only be gametes.
Combine each gamete of the black dog with be and read the phenotype using the epistasis rule: any dog with ee is yellow; otherwise B\_ is black and bb is chocolate.

Each offspring class has probability \tfrac14.
Gamete from BbEe	Offspring genotype	Phenotype
BE	BbEe	Black
Be	Bbee	Yellow
bE	bbEe	Chocolate
be	bbee	Yellow

Two of the four classes are yellow, so the ratio is 1 black : 1 chocolate : 2 yellow.

1\ \text{black} : 1\ \text{chocolate} : 2\ \text{yellow}

Epistasis in foxglove flower

Dominant epistasis: Epistasis in which the presence of a dominant allele at one gene locus masks the effects of alleles at another locus. It is in essence the opposite of recessive epistasis.

In dominant epistasis the phenotypic ratio is 12 : 3 : 1. Example: foxglove (Digitalis purpurea).

In foxglove two genes interact to determine petal coloration, but the two genes are unlinked.

In foxglove flower colour, the wild-type allele d produces light red pigment. The mutant allele D produces dark red pigment. The D or d works on a precursor to produce a different pigment.
The wild-type allele w distributes colour throughout the flower, and the mutant allele W restricts pigment to the throat of the flower.
In effect, W asserts an epistatic effect by restricting flower colour.

Foxglove flower colours.
Flower	Genotype	Explanation
White	DW (that is D\_W\_)	Dark red pigment (D) only deposited to throat spots.
White	dW (that is ddW\_)	Light red pigment (d) only deposited to throat spots.
Red	Dw (that is D\_ww)	Dark red pigment not restricted by W.
Light red	dw (that is ddww)	Light red pigment not restricted in the flower (no dominant epistatic allele W).

Figure 12.10 (a): Petal colour of the foxglove flower exhibits epistasis (DdWw\times DdWw).

Punnett square of DdWw\times DdWw: 12 white : 3 red : 1 light red.
Gametes	DW	Dw	dW	dw
DW	DDWW white	DDWw white	DdWW white	DdWw white
Dw	DDWw white	DDww red	DdWw white	Ddww red
dW	DdWW white	DdWw white	ddWW white	ddWw white
dw	DdWw white	Ddww red	ddWw white	ddww light red

12.5.1Gene linkageOften examined

Mendel did not know anything about the physical nature of genes or that genes are part of chromosomes, because the nature of chromosomes was not discovered until long after his experiments were concluded.

The number of genes in a cell is far greater than the number of chromosomes. In fact each chromosome has hundreds and thousands of genes. Genes located on the same chromosome that tend to be inherited together in genetic crosses are said to be linked genes.

Gene linkage: The phenomenon of staying together of more than one gene on the same chromosome.
Autosomal linkage: Linkage of genes on an autosome.
Sex linkage: Linkage of genes on a sex chromosome.
Linkage group: All the linked genes found on the same homologous pair of chromosomes. The number of linkage groups in an organism is equal to the number of homologous pairs of chromosomes in that organism.

The linked genes tend to be inherited together in the offspring, so usually they do not show recombination and do not assort independently. So the ideal Mendelian ratio of independent assortment is deviated.

Genes on different chromosomes assort independently; linked genes on the same chromosome are inherited together.

Detection of gene linkage

Gene linkage can easily be detected by performing a test cross between two gene pairs (a dihybrid test cross). In this type of test cross, a heterozygous individual for two traits (F_1) is back crossed with its recessive parent (P_1).

How a test cross shows linkage.
Result of the dihybrid test cross	Conclusion
All four phenotypic combinations (parental and recombinants) in equal 1 : 1 : 1 : 1 ratio.	No linkage between the genes (independent assortment).
The ratio is deviated: more parental types and fewer recombinant types.	Incomplete linkage.
Only parental types are produced.	Complete linkage.

In a typical dihybrid cross, complete (tight) linkage inhibits the outcome of recombinant types and disturbs the 9 : 3 : 3 : 1 ratio of independent assortment. As a result only parental combinations are produced, in a 3 : 1 ratio.

Morgan's experimentOften examined

T. H. Morgan studied about 85 pairs of contrasting traits in the fruit fly Drosophila melanogaster. Two of them were wing length and width of the abdomen.

The allele for long wings (Vg) is dominant over short or vestigial wing (vg).
The allele for broad abdomen (A) is dominant over narrow abdomen (a).

The experiment

Morgan crossed a fly with long wings and a broad abdomen with one having vestigial wings and a narrow abdomen.
The F_1 offspring all had long wings and broad abdomens.
Two of these F_1 flies were mated. In the F_2 generation about \tfrac34 of the offspring had long wings and a broad abdomen, and nearly all the remaining flies, about \tfrac14 of the total, had vestigial wings and a narrow abdomen.

Gene linkage encounters independent assortment

Morgan's results were very different from the results he expected based on the law of independent assortment, that is, 9 : 3 : 3 : 1. What had happened?

From his data Morgan concluded that the genes for abdomen width and wing length were located on the same chromosome, so they did not assort independently during meiosis.
Instead they were inherited together. Therefore no recombinant types were produced.
The standard ratio of independent assortment, 9 : 3 : 3 : 1, is modified to 3 : 1.

Figure 12.11: Linkage in fruit fly. A filled chromosome carries Vg and A together; an open chromosome carries vg and a together.

12.5.2Crossing-over modifies the progenyOften examined

Subsequent experiments demonstrated that the process which is responsible for the recombination of linked genes is crossing over.

Crossing over: An exchange of maternal and paternal chromatid parts that occurs while homologous chromosomes are paired during prophase of meiosis I.

Homologous chromosomes pair up in prophase of meiosis I.
Maternal and paternal chromatids exchange parts. The recombinant chromatids resulting from crossing over may bring alleles together in new combinations.
When these chromatids are distributed to different gametes, a wide variety of gametes is produced.
So crossing over leads to recombination of genes and is responsible for genetic variability in sexually reproducing organisms.

Crossing over between non-sister chromatids (drawn from the book's description): the outer chromatids remain parental and the two inner chromatids become recombinant.

Crossovers result in recombination and the exchange of genetic material between the maternal and paternal chromosomes. As a result the offspring can have different combinations of genes than their parents. Thus crossing over modifies the progeny.

12.6Sex determination

Although the anatomical and physical differences between women and men are numerous, the chromosomal basis for determining sex is rather simple. A clear picture of the genetic basis of sex determination emerged after the discovery of the sex chromosomes.

12.6.1Genetic identification of sex phenotypesOften examined

Most animals and plants have a genetic difference between male and female individuals, which is reflected in their particular array (arrangement) of chromosomes, that is, their karyotype. Let us see this difference and identify the male and female individuals in Drosophila and humans.

Chromosomes in Drosophila

There are four pairs, that is, 8 chromosomes, in Drosophila. Three pairs of chromosomes are identical in male and female; these are called autosomes. The fourth pair of chromosomes is different in male and female and determines the gender; these are called sex chromosomes.

In females both the sex chromosomes are identical, so these are called XX chromosomes.
In males one is rod shaped and is like the sex chromosome of the female, so it is also called the X chromosome, while the other chromosome is hooked in shape and is called the Y chromosome.

Chromosomes in man

In humans there are 23 pairs of chromosomes, 22 pairs of which are autosomes, while one pair is the sex chromosomes. In a female it is XX and in a male it is XY. Human females have two copies of the X chromosome.

Figures 12.12 and 12.13: Chromosomes in Drosophila and in human. Only the sex pair differs between the sexes.

How the sex of a child is decided

All the egg cells produced by a woman contain one X chromosome.
In males half of the sperms contain an X chromosome and half contain a Y chromosome. The sex chromosome carried by the sperm therefore determines the gender of the child.
If a sperm carrying an X chromosome fertilizes the egg, the child will be a girl (XX). If a sperm carrying a Y chromosome fertilizes the egg, the child will be a boy (XY).

12.6.2Patterns of sex determinationOften examined

There is a wide variety of sex-determining mechanisms, but three patterns are more common. The book describes the XY-XX type.

XY-XX type

This pattern of sex determination is found in Drosophila, man and many other organisms. The male is XY and the female is XX.

The male is heterogametic: it produces two types of sex-determining sperms. Half of the sperms carry the X chromosome and the other half carry the Y chromosome. The chances for both types of sperm are equal.
The female is homogametic: it produces only one type of eggs, each with an X chromosome.
The sex of the offspring is determined by the type of sperm. If an X-carrying sperm fertilizes the egg, the zygote will be XX and a female offspring is produced. If a Y-carrying sperm fertilizes the egg, the zygote will be XY and a male offspring is produced.
The sex ratio between male and female offspring is 1 : 1. The sex ratio indicates the chances for the sex of the offspring. The chances for a son or daughter in human birth are equal.

Figure 12.14: Sex determination in animals (the X-Y system).

Punnett square of XX\times XY: 50% girls, 50% boys.
Gametes	X (egg)	X (egg)
X (sperm)	XX female	XX female
Y (sperm)	XY male	XY male

Heterogametic: Producing two types of gametes with respect to the sex chromosomes (the XY male).
Homogametic: Producing only one type of gamete with respect to the sex chromosomes (the XX female).

Genetic problem 12.10: why is the chance of a boy or a girl 50 : 50?

Why are the chances for a husband and wife having either a boy or a girl 50 : 50?

The wife is XX, so every egg carries an X.
The husband is XY, so half his sperms carry X and half carry Y.
The sex of the child depends only on which sperm fertilizes the egg. An X sperm gives a girl (XX) and a Y sperm gives a boy (XY). The two types are equally likely, so the chances are 50 : 50.

\tfrac12\ \text{boy},\ \tfrac12\ \text{girl}

Genetic problem 12.11: four daughters, then a fifth child

A woman produced four daughters in successive pregnancies. What are the prospects of her getting a son in the fifth pregnancy?

Each pregnancy is an independent event. The sex of the fifth child depends only on whether an X or a Y sperm fertilizes the egg, and the earlier daughters do not change this. The probability of a son is again \tfrac12.

P(\text{son})=\tfrac12

Genetic problem 12.12: A positive parents with an O negative child

If both parents are A positive and have an O negative child, calculate the probability of an A positive daughter and an A positive son in the next pregnancy.

An O negative child is iidd. Each A positive parent must have given i and d, so both parents are I^Ai\,Dd.
I^Ai\,Dd\times I^Ai\,Dd
Blood group A (I^AI^A or I^Ai) has probability \tfrac34. Rh positive (DD or Dd) has probability \tfrac34. The genes are independent, so A positive has probability:
\dfrac34\times\dfrac34=\dfrac9{16}
The sex of the child is independent of its blood group. The probability of a girl is \tfrac12 and of a boy \tfrac12.
P(\text{A positive daughter})=\dfrac9{16}\times\dfrac12=\dfrac9{32}

P(\text{A}^+\text{ daughter})=\tfrac9{32},\quad P(\text{A}^+\text{ son})=\tfrac9{32}

12.7Sex linkage

Traits controlled by alleles on the sex chromosomes are said to be sex-linked, and the phenomenon is called sex linkage.

X-linked: An allele that is only on the X chromosome.
Y-linked: An allele that is only on the Y chromosome.
X-Y linked (pseudo-autosomal): Traits controlled by genes which have alleles found on both X and Y. Unlike other sex-linked traits they have just an autosomal mode of inheritance.

Very few alleles have been found on the Y chromosome.

Types of sex-linked traits.
Type	Where the allele is	Mode of inheritance
X-linked	Only on the X chromosome	Sex-linked pattern
Y-linked	Only on the Y chromosome	Sex-linked pattern (males only)
X-Y linked (pseudo-autosomal)	On both X and Y	Like autosomal inheritance

12.7.1Sex linkage in DrosophilaOften examined

Consider the inheritance of eye colour in Drosophila, where females are XX and males are XY. Organisms having normal characteristics are called wild types. Organisms having characteristics other than the normal are called mutants. The wild type Drosophila has bright red eyes. T. H. Morgan raised cultures of the fruit fly Drosophila to study different traits.

Wild type: An organism having the normal characteristics.
Mutant: An organism not having the normal characteristics.

Morgan's experiments

Morgan crossed a white-eyed male with a red-eyed female. About 1237 flies were produced in the F_1 generation and all were red-eyed.
Would the white-eye trait reappear in the F_2 progeny as Mendel had predicted? In the F_2 there were 3470 red-eyed flies and 782 white-eyed flies. Morgan observed two deviations from what Mendel had explained: a slightly deviated phenotypic ratio from the standard 3 : 1 of a monohybrid cross, and **all the white-eyed flies in the F_2 generation were male**.
Morgan wondered if the white-eyed trait somehow killed female flies preferentially. A straightforward test was a test cross: crossing red-eyed F_2 females back to the original white-eyed male. He obtained 129 red-eyed females, 132 red-eyed males, 88 white-eyed females and 86 white-eyed males. This was a rather poor fit to the expected 1 : 1 : 1 : 1 ratio, due to a deficiency in recessives. The important thing was that there were fully 88 white-eyed female flies. This observation finally discarded his doubt that the white-eyed trait can kill female flies.
The appearance of a white-eyed female provided an opportunity for a further confirmatory test. Morgan mated a white-eyed female with a red-eyed male. All the female offspring had red eyes and all the male offspring had white eyes.

Figure 12.15: Morgan's experiments on eye colour in Drosophila. Letter R represents the dominant allele for red eye and letter r the recessive allele for white eye. The alleles are carried on the X chromosome, so they are shown as superscripts to the letter X.

The Y chromosome does not have a gene locus for eye colour; the phenotype of the male results entirely from the single X-linked gene.
Genotype	Phenotype
X^RX^R and X^RX^r (female)	Red eyes
X^rX^r (female)	White eyes
X^RY (male)	Red eyes
X^rY (male)	White eyes

Morgan's conclusion

A simple explanation of his results: Morgan concluded that the white-eye trait gene resided on the X chromosome. He had only to assume that the Y chromosome did not have this gene (it was later shown to carry almost no functional genes). Knowing from his previous crosses that white-eye is a recessive trait, the results he obtained could be seen as a natural consequence of Mendelian segregation.

12.7.2Sex linkage in humansOften examined

Humans have many X-linked traits, of which some, like haemophilia and colour blindness, are recessive, while others, like hypophosphatemic or vitamin D resistant rickets, are dominant. An X-linked dominant trait is determined by an X-linked dominant gene, while an X-linked recessive trait is determined by an X-linked recessive gene. Their patterns of inheritance are very different from each other.

Experimental mating is not practically possible in humans. The mode of inheritance of human traits can be traced through pedigrees.

Symbols used in pedigrees (science tit-bit in the book).

X-linked recessive inheritance

Females possessing one X-linked recessive mutation are considered carriers and will generally not show clinical symptoms of the disorder.
All males possessing an X-linked recessive mutation will be affected (males have a single X chromosome and therefore only one copy of X-linked genes).
All offspring of a carrier female have a 50% chance of inheriting the mutation.
All female children of an affected father will be carriers (daughters get their father's X chromosome).
No male children of an affected father will be affected (sons do not inherit their father's X chromosome).

Examples: haemophilia A and B, colour blindness, and testicular feminization.

Figure 12.17: Pedigree of recessive X-linked inheritance (drawn from the book's rules): the trait skips a generation through carrier daughters.

X-linked dominant inheritance

A male or female child of an affected mother has a 50% chance of inheriting the mutation and thus being affected with the disorder.
All female children of an affected father will be affected (daughters possess their father's X chromosome).
No male children of an affected father will be affected (sons do not inherit their father's X chromosome).

Example: hypophosphatemic rickets.

Figure 12.18: Pedigree of dominant X-linked inheritance: every daughter of an affected father is affected and no son is.

Y-linked inheritance

Y-linked traits are called holandric traits. The genes located on the Y chromosome, whose alleles are absent on the X chromosome, are Y-linked genes or holandric genes (also hemizygous). They are located only on the non-homologous region of the Y chromosome. Their phenotypic expression is observed only in males. They are inherited from father to son, as a son only receives the Y chromosome from the father.

Examples of Y-linked characters in man: hypertrichosis (growth of hair on the rim of the pinna), porcupine man (straight hair on the body) and webbing of toes.
Recently two more genes have been discovered: the testis determining factor (TDF) and the minor histocompatibility gene (H-Y).

Figure 12.19: Pedigree of a Y-linked trait: it passes only from father to son.

Summary of the patterns described in the book. The recessive row for an affected mother follows from the male having a single X.
Pattern	Affected father	Affected mother	Typical sex affected
X-linked recessive	All daughters carriers; no sons affected	All sons affected; daughters carriers (if the father is normal)	Males more often
X-linked dominant	All daughters affected; no sons affected	50% of sons and 50% of daughters	Both sexes (females more often)
Y-linked (holandric)	All sons affected	Not applicable	Males only

Genetics of haemophiliaOften examined

Haemophilia is a rare X-linked recessive trait. The blood of a haemophiliac fails to clot properly after an injury, because it has either a reduction or malfunction or complete absence of blood clotting factors. It is a serious hereditary disease because a haemophiliac may bleed to death even from minor cuts.

Haemophilia is of three types: A, B and C. Haemophilia A and B are X-linked recessive, but haemophilia C is an autosomal recessive trait.

Table 12.6 (second table with this number in the book): Haemophilia.
S. No	Types of haemophilia	Cause	Percentage	Inheritance
1	Haemophilia A	Abnormality of blood clotting factor VIII	80%	Recessive X-linked
2	Haemophilia B	Disturbance in blood clotting factor IX	20%	Recessive X-linked
3	Haemophilia C	Reduction in blood clotting factor XI	Less than 1%	Recessive autosomal

Being X-linked recessives, haemophilia A and B affect men more than women, but haemophilia C affects both sexes equally because it is autosomal. A woman can suffer from haemophilia A or B only when she is homozygous for the recessive allele, but a man with just one recessive allele will display the trait (the book says the chance for a man is double that of a woman). Haemophilia A and B zigzag from the maternal grandfather through a carrier daughter to a grandson. They never pass directly from father to son. The gene for normal is H and the gene for haemophilia A is h.

Genotypes and phenotypes for haemophilia A.
Genotype	Phenotype
X^HX^H	Normal female
X^HX^h	Carrier female (normal)
X^hX^h	Haemophiliac female
X^HY	Normal male
X^hY	Haemophiliac male

Blood clotting

The cascade of clotting (box in the book)

Within seconds of cutting a blood vessel, the damaged tissue causes the platelets to become sticky and clump together around the cut.
These activated platelets and the damaged tissue release chemicals. These chemicals react with other chemicals and proteins in the plasma, called clotting factors.
Next to the cut, a complex series of reactions involving these clotting factors happens quickly. Each reaction triggers the next reaction. This is called a cascade.
The final step of this cascade is to convert factor I (fibrinogen, a soluble protein) into thin strands of a solid protein called fibrin.
The strands of fibrin form a meshwork and trap blood cells, which form into a solid clot.

The blood-clotting cascade. In haemophilia one clotting factor of the cascade is faulty or missing, so the cascade breaks down.

Genetic problem 12.13: can two normal parents have a haemophiliac child?

Can two normal parents have a haemophiliac child? Work out the probability.

Yes, if the mother is a carrier. She is normal (X^HX^h) and the father is normal (X^HY).
X^HX^h\times X^HY
The mother makes X^H and X^h eggs; the father makes X^H and Y sperms.

One of four children, \tfrac14, is a haemophiliac boy.
Gametes	X^H (father)	Y (father)
X^H (mother)	X^HX^H normal female	X^HY normal male
X^h (mother)	X^HX^h carrier female	X^hY haemophiliac male

So the probability of a haemophiliac child is \tfrac14 (25%). It is half of the sons, and none of the daughters is affected, though half of the daughters are carriers.

P(\text{haemophiliac child})=\tfrac14

Genetics of colour-blindnessOften examined

Normal trichromatic colour vision is based on three different kinds of cone cells in the retina, each sensitive to only one of the three primary colours: red, green or blue. Each type of cone cell has specific light-absorbing proteins called opsins.

The genes for the red and green opsins are on the X chromosome, while the gene for the blue opsin is present on autosome 7. Mutations in the opsin genes cause colour blindness like dichromacy and monochromacy.

Kinds of colour blindness (book).
Type	What the person sees	Examples
Dichromacy	A dichromate can perceive two primary colours but is unable to perceive the one whose opsin is missing due to mutation.	Protanopia is red blindness; deuteranopia is green blindness; tritanopia is blue blindness.
Monochromacy	A monochromate can perceive only one colour. Monochromacy is true colour blindness.	Blue cone monochromacy is an X-linked recessive trait in which both red and green cone cells are absent. That is why it is also called red-green colour blindness.

Red-green colour blindness is a common hereditary disease. Like any sex-linked recessive trait it zigzags from the maternal grandfather through a carrier daughter to a grandson. It never passes directly from father to son. This type of colour blindness is more common in men than in women, because the chances for a male to be affected by it are double those of a female.

Genotypes for red-green colour blindness (symbols C and c chosen here; the book gives no symbols).
Genotype	Phenotype
X^CX^C	Normal female
X^CX^c	Carrier female (normal vision)
X^cX^c	Colour-blind female
X^CY	Normal male
X^cY	Colour-blind male

Genetic problem 12.14: carrier woman × normal man

When a carrier woman gets married to a normal man, what will be the probability of a colour-blind son in their children?

The carrier woman is X^CX^c and the normal man is X^CY.
X^CX^c\times X^CY

Gametes	X^C (father)	Y (father)
X^C (mother)	X^CX^C normal girl	X^CY normal boy
X^c (mother)	X^CX^c carrier girl	X^cY colour-blind boy

One of the four equally likely children is a colour-blind son: the probability is \tfrac14 of all children, which is half of the sons.

P(\text{colour-blind son})=\tfrac14

Worked example: colour-blind father and carrier mother

A colour-blind man marries a woman who is a carrier for red-green colour blindness. List the possible children and the chance that a daughter is colour-blind.

The father is X^cY and the mother is X^CX^c.
X^CX^c\times X^cY
The mother makes X^C and X^c eggs; the father makes X^c and Y sperms. Fill the Punnett square.

Ratio 1 : 1 : 1 : 1.
Gametes	X^c (father)	Y (father)
X^C (mother)	X^CX^c carrier girl	X^CY normal boy
X^c (mother)	X^cX^c colour-blind girl	X^cY colour-blind boy

Half of the daughters (\tfrac12) are colour blind, and half of the sons are colour blind. A girl is colour blind only if she gets an X^c from both parents, which needs an affected father and a carrier or affected mother.

\text{daughters: }\tfrac12\ \text{colour blind},\ \tfrac12\ \text{carriers}

Linkage can be observed if the number of progeny is too largeOften examined

Morgan suggested that (a) linkage is directly related to the distance between the linked genes in a chromosome and (b) the parental combination of linked genes remains unchanged during the process of inheritance. Morgan predicted that the frequency of crossing over is governed largely by the distance between genes.

It is not possible to measure such distances in the light microscope, so scientists use an arbitrary unit of measure, the map unit, to describe the distances between linked genes.

Map unit: A unit of distance between linked genes equal to 1 per cent crossing over (recombination).

The map provides the exact location and combination of genes in a linkage group. Maps indicate the strength of linkage between two genes and the chances of crossing over.

Why a large number of progeny is needed

A large sample size provides more statistical power.
With more offspring you increase the chances of observing a wider range of genetic combinations, making it easier to detect any patterns or deviations from independent assortment that might indicate linkage.
This helps ensure the reliability of the linkage analysis.

Gene linkage becomes more apparent when observing a large number of progeny. As an example (given in the book) consider two traits in humans, hair colour and the ability to roll the tongue. Assume there are two versions for each trait: brown hair (B) or blonde hair (b), and the ability to roll the tongue (T) or inability to roll (t). Assume that these genes are located on the same chromosome and are linked. If there is no crossing over during the formation of reproductive cells, certain combinations of these traits will be inherited together more frequently.

When observing a small group of people, some mixing of traits might be noticed. With a larger number of individuals, consistent patterns are more visible: individuals with brown hair (B) might be more likely to have the ability to roll their tongue (T), while those with blonde hair (b) might be more likely to lack this ability (t). In a larger sample size it is observed that these traits tend to be inherited together more often than if they were assorted independently.

Techniques employed for embryonic screeningOften examined

Embryo screening checks an embryo to see if it is carrying any alleles for a genetic disease. A number of techniques are now available to test for chromosomal and gene mutations in human foetuses.

Amniocentesis

Before the start of the procedure, a local anaesthetic can be given to the mother.
A needle is usually inserted through the mother's abdominal wall, then through the wall of the uterus, and finally into the amniotic sac.
With the aid of ultrasound guidance, a physician punctures the sac in an area away from the foetus and extracts approximately 20 ml of amniotic fluid.
After the amniotic fluid is extracted, the foetal cells are separated from the sample. The cells are grown in a culture medium, then fixed and stained.
Under a microscope the chromosomes are examined for abnormalities, for example Down syndrome (trisomy 21).

Figure 12.21: Amniocentesis.

Chorionic villus sampling (CVS)

CVS is a prenatal test that involves taking a sample of tissue from the placenta to test for chromosomal abnormalities and certain other genetic problems. The chorionic villi are tiny projections of placental tissue that look like fingers and contain the same genetic material as the foetus.

The sample can be taken through the cervix (transcervical) or through the abdominal wall (transabdominal).
In this technique a physician inserts a narrow tube through the cervix into the uterus and suctions out a tiny sample of tissue from the placenta.
The cells of the chorionic villi, the portion of placenta sampled, are derived from the foetus and have the same genotype as the new individual.
These cells are proliferating rapidly enough to allow karyotyping to be carried out immediately.

Chorionic villus sampling is a safe and established option for early diagnostic prenatal genetic testing. Complications are minimal with a skilled provider and the advancement of ultrasound. The results of a chorionic villus sampling can help families make informed decisions about their pregnancy.

Figure 12.22: Chorionic villus sampling, transcervical and transabdominal.

Comparison of the two techniques as described in the book.
	Amniocentesis	Chorionic villus sampling
Sample taken	About 20 ml of amniotic fluid containing foetal cells	Tissue of the chorionic villi of the placenta
Route	Needle through the abdominal wall and uterus into the amniotic sac	Through the cervix (transcervical) or the abdominal wall (transabdominal)
Guidance	Ultrasound	Ultrasound
Cells used	Foetal cells separated from the fluid, grown in culture, fixed and stained	Cells of the chorionic villi, derived from the foetus, which divide fast enough for immediate karyotyping
Example of what is looked for	Chromosomes examined for abnormalities, e.g. Down syndrome (trisomy 21)	Chromosomal abnormalities and certain other genetic problems

Formula sheet

Probability

P=\dfrac{a}{n}

a is the number of favourable cases and n is the total number of possible cases.

Units: A number from 0 to 1 (or a percentage).

Product rule

P(A\ \text{and}\ B)=P(A)\times P(B)

P(A) and P(B) are the probabilities of two independent events; for example round yellow in the F_2 is \tfrac34\times\tfrac34=\tfrac9{16}.

Unit: A number from 0 to 1.

Chromosome combinations in gametes

N=2^{n}

n is the haploid number of chromosomes (homologous pairs); for humans n=23, so N=2^{23}, about eight million.

Unit: A number of combinations.

Key terms

Heredity: The passing of similar characteristics from parents to their offspring.
Variations: The differences between offspring and between offspring and their parents.
Genetics: The science that deals with the mechanism of heredity and variation; the study of genes.
Gene: A sequence of nucleotides as part of DNA, which codes for the formation of a polypeptide; the basic unit of heredity.
True (pure) breeding plant: A plant that produces all the offspring of its own phenotype upon self-fertilization.
Hybridization: A cross-fertilization between two individuals having contrasting phenotypes.
Monohybrid cross: A cross between two individuals that differ in a single trait.
Dihybrid cross: A cross between two individuals that differ in two traits.
F_1 and F_2 generations: The first filial generation (F_1) is the offspring of the parental (P) cross; the second filial generation (F_2) is produced by interbreeding (self-fertilizing) the F_1.
Allele: An alternative form of a gene present at the same locus on a homologous chromosome.
Genotype: The pair of alleles that controls the expression of a trait.
Phenotype: The expression of a trait; what can be seen.
Dominant: The allele (factor) that is expressed in the F_1 generation.
Recessive: The allele (factor) that fails to express in the F_1 generation.
Homozygous: Having both alleles of a gene pair the same; such an individual is a homozygote.
Heterozygous: Having the two alleles of a gene pair different; such an individual is a heterozygote.
Law of segregation: The two coexisting alleles for each trait in an individual separate at meiosis, so each gamete receives only one of the two alleles.
Law of independent assortment: Each pair of alleles assorts independently of other pairs of alleles during gamete formation.
Probability: The chance of an event to occur: the number of favourable cases divided by the total number of possible cases.
Product rule: The combined probability of two or more independent events is the product of their individual probabilities.
Dominance relations: The relationships between the contrasting alleles at the same locus in the heterozygous state.
Incomplete dominance: A dominance relation in which the heterozygote has a phenotype intermediate between those of its two homozygous parents.
Co-dominance: A dominance relation in which both contrasting alleles express independently in the heterozygote.
Multiple alleles: Three or more alternative forms of a gene that occupy the same locus; produced by gene mutation.
Antigen: A specific molecule, mostly a glycoprotein, on the surface of red blood cells that characterizes a blood group.
Rare blood types: The more than two hundred minor blood groups of systems other than ABO and Rh.
ABO blood group system: A system based on the A and B antigens, controlled by the three alleles I^A, I^B and i of the gene I.
Agglutination and agglutinins: Agglutination is the clumping of red blood cells when the donor's antigens react with the recipient's antibodies (agglutinins).
Universal donor: A person of blood group O negative, who has no ABO or Rh antigen to interfere with the recipient's blood.
Universal recipient: A person of blood group AB positive, who has no anti-A, anti-B or anti-Rh antibody and so cannot resist any donor's blood.
Rh factor (D antigen): The antigen whose presence makes a person Rh positive and whose absence makes a person Rh negative.
Erythroblastosis foetalis: The haemolytic disease of the newborn, which develops in a foetus when anti-Rh antibodies produced by the mother pass through the placenta.
Bilirubin: The breakdown product of red blood cells that accumulates in the foetus in erythroblastosis foetalis, damages brain cells and causes jaundice.
Rh immunoglobulin (RhoGAM): A medication containing anti-Rh antibodies that prevents an Rh-negative mother from developing her own anti-Rh antibodies.
Polygenic traits: Quantitative traits encoded by two or more gene pairs at different loci that influence the trait in an additive way.
Polygenes: All the genes that control a quantitative trait, each with a small positive or negative effect.
Epistasis: The phenomenon in which the effect of the genes at one locus interferes with or hides the effect of another gene at another locus.
Epistatic and hypostatic genes: The epistatic (inhibiting) gene suppresses or masks the effect of a gene at another locus; the gene whose effect is suppressed is the hypostatic gene.
Bombay phenotype: The O phenotype produced in people who lack the H substance (hh), so that neither A nor B antigen is formed.
Dominant epistasis: Epistasis in which a dominant allele at one locus masks the effects of alleles at another locus (ratio 12 : 3 : 1).
Linked genes and gene linkage: Linked genes are genes on the same chromosome that tend to be inherited together; gene linkage is this phenomenon of staying together.
Linkage group: All the linked genes found on the same homologous pair of chromosomes.
Test cross: A cross of a heterozygous individual with its recessive parent, used to detect linkage.
Map unit: A unit of distance between linked genes, equal to 1 per cent crossing over.
Crossing over: The exchange of maternal and paternal chromatid parts while homologous chromosomes are paired in prophase of meiosis I.
Karyotype: The particular array (arrangement) of the chromosomes of an organism.
Autosomes: The chromosomes that are identical in male and female.
Sex chromosomes: The pair of chromosomes that is different in male and female and determines the gender.
Heterogametic and homogametic: Heterogametic: producing two types of gametes for the sex chromosomes (XY male). Homogametic: producing only one type (XX female).
Sex linkage: The inheritance of traits controlled by alleles on the sex chromosomes.
X-linked: An allele that is only on the X chromosome.
Y-linked (holandric): A gene located on the non-homologous region of the Y chromosome; its trait appears only in males and passes from father to son.
Wild type and mutant: A wild type organism has the normal characteristics; a mutant does not have the normal characteristics.
Carrier: A female who has one X-linked recessive allele and does not show the disorder, but may pass it to her children.
Pedigree: A family tree used to trace the inheritance of a trait in humans.
Haemophilia: A rare X-linked recessive disorder in which the blood fails to clot properly because clotting factors are reduced, faulty or absent.
Opsins: The specific light-absorbing proteins of the cone cells of the retina.
Colour blindness: A defect of colour vision caused by mutation of the opsin genes; it may be dichromacy or monochromacy.
Amniocentesis: A prenatal test in which about 20 ml of amniotic fluid is withdrawn with a needle and the foetal cells are examined for chromosome abnormalities.
Chorionic villus sampling (CVS): A prenatal test in which a sample of placental tissue (chorionic villi) is taken through the cervix or abdominal wall to test for chromosomal abnormalities and genetic problems.

Multiple choice questionsp. 341–342

1

Mendel's law of independent assortment can be demonstrated by

Atest cross
Bback cross
Cmonohybrid cross
Ddihybrid crossAnswer

Why

Independent assortment concerns the segregation of two gene pairs at once, so it is shown by a dihybrid cross: the 9:3:3:1 F2 ratio is the evidence (book p.314).

A monohybrid cross follows one gene pair only, which shows segregation (law of segregation). A test cross or back cross reveals genotype and does not by itself show assortment of two pairs.

2

Investigators do a dihybrid cross between two heterozygous and get about a 3:1 ratio among the offspring. The reason must be due to

Apolygenes
Bpleiotropic genes
Clinked genesAnswer
Depistatic gene

Why

Two unlinked genes give 9:3:3:1, and epistasis modifies that ratio (for example 9:3:4 or 12:3:1) but does not give 3:1. Complete linkage of the two genes on one chromosome lets only the parental combinations appear, in 3:1 (book p.331).

Pleiotropy (one gene with several effects) and polygenes (continuous variation) do not produce a 3:1 result from two genes in a dihybrid cross.

Note. The stem asks for a dihybrid ratio of about 3:1. The reason is complete linkage in the coupling arrangement, where the two parental types are produced in 3:1 and recombinants are missing.

3

Whether an allele is dominant or recessive depends on

Ahow common the allele is relative to other alleles
Bwhether it is inherited from the mother or the father
Cwhich chromosome it is on
Dwhether it shows expression in heterozygous state or notAnswer

Why

Dominance is a relation between alleles of the same gene in the heterozygote: a dominant allele is expressed in the heterozygous state and masks the recessive one (book p.312, p.317). Frequency, parent of origin or chromosome number do not decide which allele is dominant.

A frequent allele can be recessive and a rare one can be dominant, so option A is wrong.

4

All the offspring of a white hen and a black rooster are grey. The simplest explanation for this pattern of inheritance is

Alinkage
Bsex linkage
Cindependent assortment
Dincomplete dominanceAnswer

Why

Each offspring is intermediate between the two parents, a blend of white and black. This is incomplete dominance, where the heterozygote shows an intermediate phenotype (book p.317, Table 12.2).

Sex linkage would make the sexes differ, co-dominance would show both colours as separate patches rather than a grey blend, and linkage and independent assortment concern two traits, not one colour.

5

A man who has type B blood and a woman who has type A blood could have children of which of the following phenotypes?

AA or B only
BAB only
CAB or O only
DA, B, AB or OAnswer

Why

The man can be I^B I^B or I^B i and the woman I^A I^A or I^A i. If both are heterozygous (I^B i \times I^A i), the children are I^A I^B (AB), I^A i (A), I^B i (B) and ii (O), so all four groups are possible (book p.319).

Options A, B and C each leave out groups that some parental genotypes can produce, so they are wrong.

6

When mother's anti-Rh negative antibodies seep through placenta into blood circulation of foetus they start ______ of RBC of the foetus

Aplasmolysis
Bcrenation
ChaemolysisAnswer
Ddeplasmolysis

Why

The maternal anti-Rh antibodies bind the D antigen on foetal red blood cells, which causes agglutination and then haemolysis, the breakdown of blood (book p.324).

Plasmolysis, crenation and deplasmolysis are changes in cell volume caused by the osmotic movement of water, not immune destruction of cells.

Note. The stem says 'anti-Rh negative antibodies'. The antibodies are anti-Rh (anti-D) antibodies produced by an Rh-negative mother, so the wording is loose but the answer is unaffected.

7

All chromosomes other than sex-chromosomes are called

AautosomeAnswer
Bmesosome
Cpolysome
Dlysosome

Why

Autosomes are the chromosomes that are not sex chromosomes. They occur as identical pairs in males and females (book p.333).

A mesosome is an infolding of the bacterial plasma membrane, a polysome is a chain of ribosomes and a lysosome is a cell organelle with digestive enzymes. None of them is a chromosome.

8

If a man of M blood group marries a woman of N blood group all their children will have

AM blood group
BN blood group
CO blood group
DMN blood groupAnswer

Why

The man is L^M L^M and the woman L^N L^N. Each child receives one L^M from the father and one L^N from the mother, so every child is L^M L^N, the MN blood group, in which both M and N antigens are expressed (book Table 12.1).

A child cannot be pure M or pure N, because each child has one allele from each parent.

9

Rh blood group system is encoded by three genes C, D and E which occupy ------ tightly linked loci

Afour
BthreeAnswer
Cfive
Dtwo

Why

Three genes at three tightly linked loci give three loci, so option B is correct. The unit's text names only the D gene, so this fact is taken from the question itself.

Four, five and two do not match the three genes named in the stem.

Note. The unit text describes only the D gene of the Rh system (book p.323). The three-locus statement (C, D and E) is assumed by the question.

10

Albinism is a recessive gene. A woman with albino father marries an albino man. The proportion of her progeny is

A2 normal : 1 albino
Ball normal
Call albino
D1 normal : 1 albinoAnswer

Why

Albinism is recessive (aa). The woman's father is aa, so she received an a from him. The question does not state her phenotype; if she is normal she is Aa. The man is aa.

Aa \times aa gives Aa (normal) and aa (albino) in a 1 : 1 ratio, so option D is correct. Option A (2 : 1) is not produced by a single-gene cross, and B and C would require the woman to be AA or aa.

Note. The woman's own phenotype is not given. The answer assumes she is normal, which makes her Aa.

11

Phenomena of an allele of one gene suppressing the activity of allele of another gene is called

Adominance
BepistasisAnswer
Csuppression
Dinactivation

Why

Epistasis is the interaction of different genes at different loci, in which the epistatic gene masks the effect of a gene at another locus (book p.327).

Dominance acts between alleles of the same gene at the same locus, so option A is excluded by the stem, which speaks of two different genes.

12

Blood group in human beings are controlled by

A4 alleles in which A is dominant
B3 alleles in which A and B are co-dominant and i is recessiveAnswer
C3 alleles in which none is dominant
D3 alleles in which A is recessive

Why

The ABO system is controlled by three alleles, I^A, I^B and i. I^A and I^B are co-dominant to each other and both are dominant over i (book p.319).

A has the wrong number of alleles, C says none is dominant (but I^A and I^B are dominant over i), and D makes A recessive, which is not the case.

13

Genes located on same locus but having different expressions are

Amultiple allelesAnswer
Boncogenes
Cpolygenes
Dcodominants

Why

Alleles are the alternative forms of a gene that occupy the same locus, and multiple alleles are three or more such forms of one gene (book p.318). The different alleles express differently, which is what the stem describes.

Oncogenes are cancer-causing genes and polygenes act at several loci. Co-dominant alleles are also at one locus, so D is partly right, but A is the general term the unit uses for genes at the same locus.

Note. Co-dominant alleles also sit at the same locus and express differently, so D is a narrower match. The key follows the unit's wording, where all multiple alleles occupy the same locus (book p.318).

Short answer questionsp. 342–344

13 marks

In monohybrid cross why did only about one fourth of Mendel's F2 plants exhibit the recessive trait?

Solution

The F1 plants are Rr (round, heterozygous). The dominant allele R masks r, so only plants with two recessive alleles, rr, show the wrinkled trait.

The F1 plants produce gametes R and r in equal numbers.
Random fertilisation of these gametes in the F2 gives RR, Rr, Rr and rr (Punnett square), so the genotypic ratio is 1 : 2 : 1.

Only the rr class (one in four) is wrinkled, so the phenotypic ratio is 3 round : 1 wrinkled (book p.313).

23 marks

Hypothesize that in a dihybrid inheritance pattern of colour and texture of pea seed, the two traits are not interdependent.

Solution

Hypothesis: seed colour and seed shape are inherited independently, so each pair of alleles segregates without regard to the other. The prediction is an F2 ratio of 9 : 3 : 3 : 1.

The genes for seed colour and seed shape lie on different pairs of homologous chromosomes (book p.316).
At metaphase I each homologous pair is positioned independently, so the four gamete types (RY, Ry, rY, ry) are formed in equal numbers.
Test: the F2 of a dihybrid cross shows four phenotypes in 9 : 3 : 3 : 1, including round green and wrinkled yellow seeds, which supports the hypothesis.

33 marks

What are the exceptions to the Mendel's laws of inheritance?

Solution

Mendel's laws describe simple complete dominance at one locus, and several patterns depart from them:

Group	Exceptions (book section)	Example
Dominance relations between alleles	Incomplete dominance (12.2.1), co-dominance (12.2.2), multiple alleles (12.2)	4 O'clock flower colour, MN blood group, ABO blood group
Interaction of genes	Polygenic inheritance (12.4), epistasis (12.4.3)	Wheat grain colour, Labrador coat colour
Genes on the same chromosome	Gene linkage (12.5), sex linkage (12.7)	Morgan's fruit fly, haemophilia, colour blindness

43 marks

What is probability? How does probability relate to genetics?

Solution

Probability is the chance that an event will occur. It is the number of favourable cases a divided by the total number of possible cases n: P = a/n (book p.316). A coin toss has P = 1/2.

Fertilisation and the separation of alleles at meiosis are random events, so genetics follows the rules of probability. Each gamete carries one allele of a pair with probability 1/2 and the product rule gives the chance of combined events, for example two independent traits 3/4 \times 3/4 = 9/16 for round yellow seed in F2.

Mendel's ratios are probabilities: 1/4 RR, 1/2 Rr and 1/4 rr in F2, so the 3 : 1 ratio is the expected result, and the offspring counts only approximate it.

53 marks

Write the differences between incomplete dominance and co-dominance.

Solution

Book Table 12.2.
Feature	Incomplete dominance	Co-dominance
Heterozygote	Both genes blend their phenotypic effects	Both genes express their phenotypic effects independently
Phenotype of heterozygote	An intermediate phenotype between the two parents	Both parental phenotypes at the same time
Example	Flower colour of the 4 O'clock plant	Human MN blood group and AB blood group

63 marks

Describe the multiple alleles and state the alleles responsible for the traits of ABO blood groups.

Solution

Multiple alleles are three or more alternative forms of one gene, all occupying the same locus. Each diploid individual still carries only two of them, because each locus is present twice (book p.318).

The ABO system is controlled by three alleles: I^A (makes antigen A), I^B (makes antigen B) and i (makes neither).
I^A and I^B are co-dominant to each other, and both are dominant over i.
The genotypes are I^A I^A or I^A i (group A), I^B I^B or I^B i (group B), I^A I^B (group AB) and ii (group O).

73 marks

Why can multiple alleles provide many different phenotypes for a trait?

Solution

A trait controlled by many alleles has many possible pairs of alleles, so many genotypes. With three alleles A_1, A_2 and A_3 there are six genotypes: A_1A_1, A_1A_2, A_1A_3, A_2A_2, A_2A_3 and A_3A_3 (book p.318).

Some traits have 100 or more alleles in a population, so the number of possible genotypes is very large.
Each allele can be dominant, co-dominant or recessive to the others, so different pairs give different phenotypes.
Each individual carries only two alleles, but the population as a whole carries the whole series, so the variety of phenotypes is large.

83 marks

Name various human blood group systems.

Solution

The International Society of Blood Transfusion recognises 33 blood group systems with over 300 antigens. Some important ones, with their symbols, are:

ABO (ABO)
MNS (MNS)
Rhesus (Rh)
Lutheran (LU)
Kell (KELL)
Lewis (LE)
Duffy (FY)
Kidd (JK)

93 marks

Describe the antigens of ABO blood group system.

Solution

The ABO system is based on antigens, which are glycoprotein markers on the surface of red blood cells (book p.319).

Antigen A is present on the red cells of group A and antigen B on those of group B.
Group AB red cells carry both antigens, and group O red cells carry neither.
The H substance is the precursor of both A and B antigens and is present on all red blood cells; its absence produces the Bombay phenotype (book p.328).

103 marks

Describe the antibodies of ABO blood group system.

Solution

The antibodies of the ABO system are present in the plasma and are directed against the antigen that the person lacks (book p.320).

Blood group	Antigen on red cells	Antibodies in plasma
A	A	anti-B
B	B	anti-A
AB	A and B	none
O	none	anti-A and anti-B

These antibodies appear in early life without any stimulus and remain throughout life.

113 marks

Why Rh-incompatibility could be a danger to the developing foetus and mother?

Solution

An Rh-negative mother carrying an Rh-positive foetus can be sensitised when foetal blood enters her circulation at birth or through bleeding. She then makes anti-Rh antibodies.
In a later pregnancy these antibodies cross the placenta and agglutinate the Rh-positive red cells of the foetus, which are then broken down (haemolysis). The result is anaemia, a swollen liver and spleen, and bilirubin that causes jaundice and damages the brain (erythroblastosis foetalis).
The same antibodies can cause a severe transfusion reaction if an Rh-positive person is given Rh-positive blood after sensitisation, and the condition can end in miscarriage or stillbirth.

122025 Annual I, Set D, Q.2 (x)3 marks

Investigate the reasons for O- (negative) individuals as the universal donor and AB+ (positive) as the universal recipient.

Solution

Group O- red cells have neither A, B nor D (Rh) antigens. A recipient's antibodies therefore have nothing to agglutinate, so O- blood can be given to almost anyone (book p.322, p.323).
Group AB+ plasma contains neither anti-A, anti-B nor anti-Rh antibodies, so it cannot react with the antigens on donor red cells, and AB+ people can receive blood of any group.
Both are defined by what is absent: a donor needs no antigen that can be attacked, and a recipient needs no antibody that can attack.

133 marks

Why a recessive group allele "i" is more frequent in population?

Solution

Allele i is recessive, so the people who carry it are hidden. A person with I^A i or I^B i is group A or B and carries i without showing it (book p.322).

Group O occurs only in ii people. Two carriers produce ii children: I^A i \times I^A i or I^B i \times I^B i, or I^A i \times I^B i gives 1/4 group O children.
Carriers of i are common because they cannot be recognised by their blood group, so the allele is maintained in the population.
The book quotes frequencies of about 37 to 47 percent for several countries, such as Iceland and Ireland at 47 percent (book p.322), which shows how common the allele is.

Note. The book's frequency figures (Australia 40%, Canada 39%, Iceland 47%, Ireland 47%, UK 37%, USA 37%) are quoted for group O; the printed text does not state the blood group explicitly.

143 marks

List at least five polygenic traits discovered in humans.

Solution

Polygenic traits in humans are quantitative, vary continuously, and are influenced by several genes (book p.326). Five of them are:

Height
Skin colour
Intelligence
Blood pressure
Longevity

153 marks

Write the differences between dominance and epistasis.

Solution

Book Table 12.6, summarised.
Feature	Dominance	Epistasis
Loci involved	Single pair of alleles at one locus	Two pairs of alleles at two loci
What is suppressed	A gene suppresses the expression of its own allele	A gene suppresses the expression of alleles of another gene, both dominant and recessive
Source of the effect	Only the dominant allele is effective	Either a dominant or a recessive allele can be epistatic

163 marks

What is gene linkage?

Solution

Gene linkage is the phenomenon in which genes located on the same chromosome tend to be inherited together. The group of linked genes on one homologous pair is a linkage group, so the number of linkage groups equals the number of homologous pairs (book p.331).

Linked genes do not assort independently at meiosis, so they do not give the 9 : 3 : 3 : 1 ratio.
Recombination between linked genes is rare, and the closer two genes are on the chromosome, the less often they are separated.
Linkage can be complete (only parental types) or incomplete (some recombinants, from crossing over).

173 marks

Explain crossing over.

Solution

Crossing over is the exchange of corresponding segments between maternal and paternal non-sister chromatids of homologous chromosomes.

It takes place during prophase I of meiosis, when the homologous chromosomes are paired and form chiasmata.
The exchanged segments carry alleles into new combinations, producing recombinant chromatids (book p.332).
When these chromatids are distributed to gametes, the gametes carry new combinations of alleles, which increases genetic variation.

182025 Annual I, Set D, Q.2 (v) OR3 marks

Describe XX-XY type pattern of sex determination.

Solution

In the XX-XY pattern, females are homogametic (XX) and males are heterogametic (XY). This pattern occurs in Drosophila, in man and in many other organisms (book p.334).

All eggs of a woman carry one X chromosome.
Half of the sperms carry an X chromosome and half carry a Y chromosome, so the male determines the sex of the child.
A sperm with X gives an XX zygote (a girl) and a sperm with Y gives an XY zygote (a boy), so the sex ratio is 1 : 1.

193 marks

Why is human male referred as heterogametic?

Solution

A male is referred to as heterogametic because he produces two different kinds of gametes, while the female produces only one (book p.333, p.334).

Males have one X and one Y chromosome, which are different in size and shape.
At meiosis the X and Y separate, so half the sperms carry X and half carry Y.
Females have two X chromosomes and produce only X-carrying eggs, which is why they are homogametic.

203 marks

What is sex linkage?

Solution

Sex linkage is the inheritance of traits controlled by genes on the sex chromosomes (book p.334). Most are on the X chromosome, and a few are on the Y chromosome.

An allele only on the X chromosome is called X-linked; one only on the Y is Y-linked.
Because males have one X, a recessive X-linked allele is expressed in males at once, and the pattern of inheritance differs from autosomal traits.
Examples are eye colour in Drosophila and haemophilia and colour blindness in man.

212025 Annual I, Set B, Q.2 (ix)3 marks

Describe Y-linked inheritance in man.

Solution

Y-linked traits are called holandric. The genes are on the non-homologous region of the Y chromosome, so they have no allele on the X (book p.336).

They appear only in males, because only males carry a Y chromosome.
A son always receives the Y chromosome from his father, so the trait passes from father to all sons and never to daughters.
Examples are hypertrichosis (hair on the rim of the pinna), webbing of the toes and porcupine man (straight hair on the body).

223 marks

Name some of the sex-linked disorders of man and Drosophila.

Solution

In man: haemophilia A and B (X-linked recessive), red-green colour blindness (X-linked recessive), testicular feminisation, and hypophosphatemic rickets (X-linked dominant).
In Drosophila: white eye colour, which is recessive and X-linked (book p.335).

233 marks

How gene linkage encounters independent assortment?

Solution

Independent assortment requires the genes to be on different chromosomes. Linkage places two genes on one chromosome, so the two genes tend to be inherited together and the expected 9 : 3 : 3 : 1 ratio is modified.

Linked alleles travel together during meiosis, so parental combinations are in excess and recombinant types are fewer.
Crossing over separates linked genes only occasionally, so the closer the genes, the stronger the linkage and the closer the ratio is to 3 : 1.
Only if genes are far apart on a chromosome, or on different chromosomes, do they assort independently.

243 marks

Under what circumstances is it possible for father and son to suffer from haemophilia?

Solution

Haemophilia A and B are X-linked recessive. The father is affected, with genotype X^h Y, and the son is affected with X^h Y. The son receives his Y from his father, so his X chromosome must come from his mother (book p.336, p.337).

The mother must carry the haemophilia allele: she is a carrier (X^H X^h) or affected (X^h X^h), so she can give X^h to her son.
The father's X^h passes only to his daughters, who become carriers, and never to his sons.
Haemophilia C is autosomal recessive, so father and son can both be affected when the mother also carries the allele.

253 marks

Explain dichromacy.

Solution

Dichromacy is a colour-vision defect in which a person can perceive two of the three primary colours, because one of the three cone types is missing (book p.338).

The missing cone type results from a mutation in the opsin gene for that colour.
Protanopia is the absence of red perception (red-blindness), deuteranopia the absence of green perception (green-blindness) and tritanopia the absence of blue perception (blue-blindness).
The red and green opsin genes lie on the X chromosome, while the blue opsin gene lies on autosome 7.

263 marks

Explain monochromacy.

Solution

Monochromacy is true colour blindness, in which a person can perceive only one colour (book p.338).

Only one type of cone works, or none, so the person sees the world in shades of a single colour.
Blue cone monochromacy is an X-linked recessive trait in which both red and green cone cells are absent.
Because the red and green cones are both missing, it is also called red-green colour blindness in its most complete form.

273 marks

What is a sex limited trait?

Solution

A sex-limited trait is controlled by a gene that is present in both sexes but is expressed only in one sex, because of the anatomy or physiology of that sex.

The gene is usually autosomal, so males and females both carry it.
Because the trait depends on the sex-specific body structure, it cannot be expressed in the other sex even when the genotype is present.
An example is a trait of the reproductive organs or of milk production.

Note. Sex-limited and sex-influenced traits are not explained in the unit's text. The answer uses the general definition needed to answer the question.

283 marks

What is sex influenced trait?

Solution

A sex-influenced trait is an autosomal trait whose expression differs between males and females, because of the sex hormones present in each.

The gene is on an autosome, so both sexes carry it.
In males, a dominant allele is expressed in the heterozygote, while in females the same trait appears only in the homozygote.
Male pattern baldness is the common example: a heterozygous man is bald, but a heterozygous woman is not.

Note. Sex-influenced traits are not explained in the unit's text. The answer uses the general definition needed to answer the question.

293 marks

How do multiple alleles and polygenic traits differ?

Solution

Feature	Multiple alleles	Polygenic traits
Genes involved	One gene with many alleles at one locus	Several genes at different loci
Phenotypes	Usually discrete classes (for example A, B, AB, O)	Continuous variation (for example height, skin colour)
Effect of each gene	Each allele has a specific dominance relation	Each polygene adds a small positive or negative effect

303 marks

Solve the following.

Solution

(a) Problem: Show how the "test cross" works, using round-seeded pea of unknown generation.

A round pea is RR or Rr. It is crossed with a wrinkled pea, which must be rr (the recessive homozygote) so that its own alleles show in the offspring.

Unknown genotype	Cross	Offspring	Conclusion
RR	RR \times rr	All Rr, all round	Pure breeding round
Rr	Rr \times rr	1 Rr (round) : 1 rr (wrinkled)	Heterozygous round

If any wrinkled seeds appear in the offspring, the unknown pea is Rr; if all are round, it is RR.

(b) Problem: if a four o'clock plant having red flowers is crossed with a white-flowered four o'clock, the F1 plants are all pink flowered. From this evidence alone, what ratios would you expect in these crosses: red crossed with red? red X pink? pink X white? white X white?

Red is R_1R_1, white is R_2R_2 and pink is R_1R_2, with incomplete dominance (book p.317).

Cross	Genotypes	Expected offspring ratio
Red \times red	R_1R_1 \times R_1R_1	All red
Red \times pink	R_1R_1 \times R_1R_2	1 red : 1 pink
Pink \times white	R_1R_2 \times R_2R_2	1 pink : 1 white
White \times white	R_2R_2 \times R_2R_2	All white

(c) Problem: Suppose a man with blood type A marries a woman who has type AB blood. What blood types would you expect to find among their children? In this problem there is no way of knowing whether the man is homozygous or heterozygous for I^A allele. What would tell you which of the genotype he has?

The woman is I^A I^B, so she gives I^A or I^B. The man is I^A I^A or I^A i.

If the man is I^A I^A, all children are group A (I^A I^A) or AB (I^A I^B), in equal numbers.
If the man is I^A i, the children are group A (I^A I^A or I^A i), AB (I^A I^B) and group B (I^B i).

A group B child can only come from the I^B of the mother and the i of the father, so the appearance of group B children shows the father is I^A i. A large family with only A and AB children points to I^A I^A.

(d) Problem: Two white sheep produce a black offspring. What is the parent's genotype for the colour? What is the probability that their next offspring will be black?

Let white be dominant (W) and black recessive (w). Black offspring (ww) need one w from each parent, so both parents are Ww.

Ww \times Ww gives 1\,WW : 2\,Ww : 1\,ww. The chance of a black offspring is P = 1/4 (25 percent).

(e) Problem: Albinism (lack of pigment) in man is caused by a recessive gene. If normal parents have an albino child, what is the probability that their next child will be normal for colour?

Albinism is recessive (aa). Normal parents with an albino child must both be Aa, since each gave an a.

Aa \times Aa gives 1/4 aa (albino) and 3/4 normal (AA or Aa). The probability that the next child is normal is 3/4 (75 percent).

313 marks

Write the differences between:

Solution

(a) Genes and alleles

Feature	Gene	Allele
Definition	A segment of DNA that codes for a polypeptide	An alternative form of a gene
Position	Occupies a locus on a chromosome	Occupies the same locus as its alternative
Example	The gene for seed shape	R (round) and r (wrinkled)

(b) Alleles and multiple alleles

Feature	Alleles	Multiple alleles
Number	Two alternative forms of a gene	Three or more alternative forms of one gene in a population
In one individual	Two alleles (a pair)	Still only two alleles
Example	R and r	I^A, I^B and i in the ABO system

(c) Monohybrid and dihybrid cross

Feature	Monohybrid cross	Dihybrid cross
Traits followed	One	Two
Gene pairs	One pair of alleles	Two pairs of alleles
F2 phenotype ratio	3 : 1	9 : 3 : 3 : 1

(d) First and second filial generation

Feature	F1	F2
Produced by	Cross between two pure-breeding parents	Self-fertilisation of the F1
Phenotype	Uniform, the dominant trait	Dominant and recessive in 3 : 1
Genotype	All heterozygous	1 homozygous dominant : 2 heterozygous : 1 homozygous recessive

(e) Genotype and phenotype

Feature	Genotype	Phenotype
Meaning	The genetic constitution, the alleles present	The observable trait
Visible?	Not directly	Yes
Example	RR and Rr	Both are round

(f) Homozygous and heterozygous

Feature	Homozygous	Heterozygous
Alleles	Two identical alleles (RR or rr)	Two different alleles (Rr)
Gametes	One type	Two types, in equal numbers
Breeding	Breeds true	Segregates in the next generation

(g) Homozygous and hemizygous

Feature	Homozygous	Hemizygous
Alleles	A pair of identical alleles on homologous chromosomes	A single allele, with no homologue to pair with
Where it occurs	Autosomes and most genes	Genes on the X chromosome in males, and genes on the Y
Example	X^H X^H in a woman	X^h Y in a haemophilic man

(h) Incomplete dominance and co-dominance

Feature	Incomplete dominance	Co-dominance
Heterozygote	Blend of both phenotypes	Both phenotypes shown at once
Phenotype shown	Intermediate	Both parental phenotypes
Example	4 O'clock flower colour	MN and AB blood groups

(i) Dominance and epistasis

Feature	Dominance	Epistasis
Loci	One locus, one pair of alleles	Two loci, two pairs of alleles
Effect	An allele masks its partner allele	A gene masks alleles of another gene
Ratio	3 : 1 for a monohybrid cross	Modified ratios such as 9 : 3 : 4 or 12 : 3 : 1

(j) Sex determination in Drosophila and human

Feature	Drosophila	Human
Chromosome pairs	Four pairs: three autosome pairs and one sex-chromosome pair	23 pairs: 22 autosome pairs and one sex-chromosome pair
Pattern	XX female, XY male	XX female, XY male
Y chromosome	Hooked, with almost no functional genes	Carries the sex-determining region (SRY gene)

Both follow the XX-XY pattern in which the male is heterogametic, but the number of chromosomes and the role of the Y chromosome differ (book p.333, Science Titbits).

(k) X-linked and Y-linked inheritance

Feature	X-linked	Y-linked (holandric)
Carried by	Both sexes (female has two X, male one)	Males only
Passed on by	Mother to son, and father to daughter	Father to all sons, never to daughters
Example	Haemophilia, colour blindness	Hypertrichosis of the pinna, webbed toes

(l) Sex limited traits and sex influenced traits

Feature	Sex-limited	Sex-influenced
Expression	In one sex only	In both sexes, but differently, because of hormones
Gene	Usually autosomal, present in both sexes	Autosomal, present in both sexes
Example	Traits of the reproductive organs or milk production	Male pattern baldness

These two terms are not explained in the unit's text; the answer uses their general definitions.

(m) Pattern of baldness in human male and female

Feature	Male	Female
Dominance	The allele for baldness is dominant in males: a heterozygote is bald	The allele is recessive in females: only the homozygote is bald
Result	Bald men are common	Bald women are rare

The unit's text does not discuss baldness, so this answer applies the sex-influenced pattern in general terms.

Extensive answer questionsp. 344–345

1

Explain Mendel's law of inheritance of monohybrid cross and interpret the results.

Solution

Mendel's monohybrid cross

A monohybrid cross follows one pair of contrasting traits. Mendel crossed pure-breeding round-seeded plants (RR) with pure-breeding wrinkled-seeded plants (rr) (book p.312, p.313). The procedure has four steps:

P generation: pure-breeding parents are crossed. Each parent makes one kind of gamete.
RR \times rr \to Rr
F1 generation: all offspring are Rr and round. The dominant allele R masks r, so the F1 are monohybrids in genotype but round in phenotype.
F2 generation: F1 plants are self-fertilised. The Punnett square gives 1 RR : 2 Rr : 1 rr, so the phenotype ratio is 3 round : 1 wrinkled.
Test of the F2: one third of the round F2 plants (RR) breed true, two thirds (Rr) segregate 3 : 1 again, and all wrinkled plants breed true (rr).

Punnett square for the F2

Gamete	R (1/2)	r (1/2)
R (1/2)	RR (round)	Rr (round)
r (1/2)	Rr (round)	rr (wrinkled)

Monohybrid cross of round and wrinkled peas

The law of segregation

Mendel's observations are summarised in the law of segregation: the two alleles of a gene pair separate during gamete formation, so each gamete receives one allele. The two alleles unite again at random fertilisation when the zygote forms (book p.313).

Interpretation

Each character is controlled by a pair of factors (now called alleles), one from each parent.
A dominant allele is expressed in the heterozygote and a recessive allele shows only when it is present in two copies.
The pair separates at meiosis, so each gamete carries one factor.
Random fertilisation gives the ratio 1 : 2 : 1 of genotypes and 3 : 1 of phenotypes in F2, which is the result Mendel observed in all seven characters.

2

Describe law of independent assortment under the following heads: a. Limitations of law of independent assortment; b. Usefulness and scope of law of independent assortment.

Solution

Statement of the law

The law of independent assortment states that each pair of alleles assorts independently of other pairs of alleles during gamete formation (book p.314). The alleles for seed colour and for seed shape therefore sort into the gametes without regard to each other.

Mechanism

Independent assortment occurs at metaphase I of meiosis. Each homologous pair is positioned independently of the other pairs on the equator, so each daughter cell receives a random mixture of maternal and paternal chromosomes (book p.315).

Independent assortment at metaphase I of meiosis for two gene pairs, A/a and B/b. The two equally likely arrangements give the four gamete types, each in one quarter of the gametes.

a. Limitations of law of independent assortment

Limitations

The law applies to diploid organisms. Not all organisms are diploid, so the law does not cover them (book p.315).
It applies only to genes on different chromosomes. Genes on the same chromosome are linked and tend to be inherited together, so they do not assort independently. Only crossing over separates them, and then only occasionally.
Genes on the X chromosome show a different pattern in succeeding generations. Males have only one X, so a recessive X-linked allele is expressed in males even when it is rare, and the expected ratios do not hold.

The law is therefore a rule for unlinked genes in diploid sexually reproducing organisms.

b. Usefulness and scope of law of independent assortment

Usefulness and scope

Together with mutation and crossing over, independent assortment is a major source of variation in successive generations (book p.315).
New combinations of traits can appear in the next generation, which helps populations adapt to changing environments.
In humans the haploid number is 23, so the number of possible combinations of maternal and paternal chromosomes in gametes is 2^{23}, about eight million. Each gamete a person produces contains one of these combinations.

This shows that independent assortment provides a huge number of possible gametes, which underlies the variety seen within families and populations.

3

What is probability? How can you evaluate that the inheritance of genes and their mixing during fertilization is based on mathematical probabilities?

Solution

What probability is

Probability is the chance that an event occurs. It is the number of favourable cases a divided by the total number of possible cases n (book p.316).

P=\dfrac{a}{n}

probability

Probability in a monohybrid cross

Generation	Dominant phenotype	Recessive phenotype
F1 (Rr)	P = 1 (100 percent)	P = 0
F2 (1 RR : 2 Rr : 1 rr)	P = 3/4	P = 1/4

Mendel's ratios are therefore probabilities. Mendel had a firm background in mathematics and recognised that the separation of alleles at meiosis and their reunion at fertilisation obey the rules of probability.

Product rule and sum rule

The combined probability of two or more independent events is the product of their individual probabilities (product rule). The probability of a round yellow F2 seed in a dihybrid cross is the product of the chances for round (3/4) and yellow (3/4): P = 3/4 \times 3/4 = 9/16 (book p.316). The sum rule gives the chance of one of several alternative outcomes: round green (3/4 \times 1/4 = 3/16), wrinkled yellow (3/16) and wrinkled green (1/16) add up to 1.

F2 phenotype (dihybrid)	Calculation	Probability
Round yellow	3/4 \times 3/4	9/16
Round green	3/4 \times 1/4	3/16
Wrinkled yellow	1/4 \times 3/4	3/16
Wrinkled green	1/4 \times 1/4	1/16

Mixing of genes at fertilisation

Segregation: each gamete receives one allele of each pair with probability 1/2.
Independent assortment: the four gamete types of a dihybrid are equally likely, each with probability 1/4.
Random fertilisation: the chance that any sperm unites with any egg is the same, so the zygote combines alleles at random, with the probabilities of the Punnett square.
Humans: 2^{23} gamete types are possible from the 23 chromosome pairs, so the chance of any particular gamete is very small, and each zygote is unique.

Evaluation

The mathematical probabilities predict the expected ratios, and the observed ratios of Mendel's large experiments agreed with them, which supports the idea that inheritance is governed by probability.
Small samples deviate from the expected ratio, as Morgan's data show for linked genes, so a large number of progeny is needed to test a ratio.
Linked genes deviate from the ratio predicted for independent genes, because their alleles do not assort independently. The probabilities must therefore use the actual linkage.

42025 Annual I, Set B, Q.2 (iv) OR

Explain incomplete dominance with reference to the inheritance of flower in 4 O'clock plant.

Solution

Incomplete dominance

In incomplete dominance neither allele of a pair is completely dominant over the other. The heterozygote shows an intermediate phenotype between those of the two parents (book p.317). Red and white 4 O'clock flowers (Mirabilis jalapa) are the classic example, first crossed by Carl Correns in 1899.

The cross

Parents: pure-breeding red (R_1R_1) and pure-breeding white (R_2R_2) plants are crossed.
F1: all plants have pink flowers, the heterozygote R_1R_2, an intermediate phenotype. The red and white alleles blend, so neither is dominant.
F2: F1 plants are self-pollinated. The genotypes are R_1R_1, R_1R_2 and R_2R_2 in a 1 : 2 : 1 ratio, and the phenotypes are red, pink and white in the same 1 : 2 : 1 ratio.

Incomplete dominance in the 4 O'clock plant

Why it is not Mendelian

Mendel's monohybrid cross gives a 3 : 1 phenotype ratio because one allele completely masks the other. Here the heterozygote is pink and the ratio is 1 : 2 : 1.
The F1 does not resemble either parent, so the dominance relation between the alleles cannot be complete.
Because neither allele is completely dominant, the book writes the alleles as R_1 and R_2 with numerals, and the heterozygote is named by its intermediate colour (Science Titbits, p.317).

Conclusion

Incomplete dominance is a non-Mendelian pattern in which the heterozygote is intermediate. It explains the pink F1 and the 1 : 2 : 1 F2 of the 4 O'clock plant.

5

Give a detailed account of ABO blood group system.

Solution

Antigens and blood groups

The ABO system is based on two surface antigens of red blood cells, A and B, which are glycoproteins. Type A carries antigen A, type B carries antigen B, type AB carries both and type O carries neither (book p.319).

Genetic basis

The system is controlled by a single autosomal gene with three alleles: I^A (makes A), I^B (makes B) and i (makes neither). I^A and I^B are co-dominant, and both are dominant over i.

Genotypes and phenotypes

Book Table 12.3.
Blood group	Antigen	Genotypes	Antibodies in plasma	Transfusion
A	A	I^A I^A or I^A i	anti-B	A and O
B	B	I^B I^B or I^B i	anti-A	B and O
AB	A and B	I^A I^B	none	any
O	none	ii	anti-A and anti-B	only O

Antibodies and transfusion

The antibodies of the system are produced from early life, without any stimulus, and they remain throughout life. A donor's antigen that meets the recipient's antibody against it causes agglutination, the clumping of red blood cells. Blood is therefore given only where the donor has no antigen that the recipient's antibodies can attack.

Group O is the universal donor, because its red cells carry no A or B antigen.
Group AB is the universal recipient, because its plasma has no anti-A or anti-B antibody.
The blood group of a person does not change throughout life, and the genes start to express early in embryonic life.

a. Associate the positive and negative blood groups with the presence and absence of Rh factor

The Rh system is controlled mainly by the D gene. Dominant D produces the D antigen (Rh factor) on the red cells, while the recessive d produces no Rh factor (book p.323).

Blood group	Rh antigen (factor)	Genotypes	Anti-Rh antibody
Rh-positive	Present	DD or Dd	Not produced unless stimulated
Rh-negative	Absent	dd	Produced if the person is stimulated by Rh-positive blood

A person can therefore be A or B or O and also Rh-positive or Rh-negative. For example, O-negative is group O with genotype ii\,dd, and AB-positive is I^A I^B DD or I^A I^B Dd. The two systems are inherited independently, because the Rh and ABO genes lie on different chromosomes.

b. Explain Erythroblastosis foetalis in the light of antigen-antibody reaction

Erythroblastosis foetalis is haemolytic disease of the newborn, caused when the mother's anti-Rh antibodies cross the placenta and attack the foetal Rh-positive red cells (book p.324).

Maternal-foetal Rh incompatibility, from the text and Figure 12.5. The highlighted box is the antigen-antibody reaction that destroys the foetal red cells.

Sensitisation: during the first birth, foetal Rh-positive red cells enter the Rh-negative mother's blood, and her immune system makes anti-Rh antibodies.
Second pregnancy: the anti-Rh antibodies pass through the placenta into the foetal blood.
Antigen-antibody reaction: each antibody binds the D antigen on foetal red cells, which clump (agglutination) and are broken down (haemolysis).
Result: anaemia, bilirubin from broken-down haemoglobin (jaundice) and damage to brain cells; the foetus may be stillborn or aborted.

The reaction is specific: an antibody fits only its own antigen, so only Rh-positive foetal cells are destroyed. The first pregnancy usually does no harm, because the antibodies take time to form.

6

Suggests prevention and treatment of Erythroblastosis foetalis.

Solution

Prevention

Erythroblastosis foetalis is preventable. The medication Rh immunoglobulin (Rh-Ig, also called RhoGAM) is given to Rh-negative pregnant women. It contains anti-D antibodies that destroy any Rh-positive foetal red cells that have entered the mother's blood, before her own immune system is stimulated to make antibodies (book p.324).

When RhoGAM is given

Dosing times for Rh-negative women (book p.325).
Time	Situation
At 28 weeks of pregnancy	Routine dose during the pregnancy
Within 72 hours after delivery	If the new-born is Rh-positive
Within 72 hours of a miscarriage, abortion or ectopic pregnancy	Any pregnancy that ends early
After an invasive prenatal test	Such as amniocentesis or chorionic villus sampling (CVS)
After any vaginal bleeding	Any bleeding that may let foetal blood into the mother's circulation

Treatment

If a baby is born with the disease, the Rh-positive blood is replaced by Rh-negative blood that is free of anti-Rh antibodies. This is done immediately, because the bilirubin and anaemia threaten the baby (book p.324).
Replacement with Rh-negative blood removes the antibodies and the affected red cells from the baby's circulation, so the haemolysis stops.

RhoGAM does not help a woman who has already been sensitised, because her own antibodies are already present (book p.324).

Note. The book's list on page 325 contains an inserted parenthesis about ectopic pregnancy. The table above gives the five intended times.

7

Explain polygenic inheritance with reference to wheat grain colour.

Solution

Polygenic inheritance

Polygenic traits are quantitative traits controlled by several gene pairs at different loci. Each polygene has a small positive or negative effect, and their effects add together, so the phenotype varies continuously (book p.325).

The wheat cross

P generation: a dark red grain plant (AABBCC) is crossed with a white grain plant (aabbcc). Each parent makes one kind of gamete.
AABBCC \times aabbcc
F1 generation: all grains are light red, AaBbCc, an intermediate colour (similar to incomplete dominance at first sight).
F2 generation: F1 plants are crossed with each other. There are seven shades of colour in the ratio 1 : 6 : 15 : 20 : 15 : 6 : 1 (book Table 12.5).

Distribution of F2 grain colours in wheat, drawn from the ratio in Table 12.5 of the book. The tallest bar, light red, is the most common class.

Explanation

Three gene pairs, Aa, Bb and Cc, lie at three loci and each contributes to the red pigment. Each capital allele adds one dose of red pigment and each small-letter allele adds none, so the phenotype depends on the number of pigment alleles in the genotype (book p.326).

Pigment alleles	Phenotype	Ratio (out of 64)
6	Dark red	1
5	Moderately dark red	6
4	Red	15
3	Light red	20
2	Pink	15
1	Light pink	6
0	White	1

Why it is polygenic

Each phenotype is a class, but the classes are closely spaced, so the grain colours form a continuous series.
No single gene gives seven colours; the seven colours come from six alleles at three loci.
Environmental factors such as light, water and nutrients also change the amount of colour (book p.326).

Note. The text of the book refers to 'table 22.5' for the wheat phenotypes. The table is Table 12.5 on the same page.

8

What is epistasis? Explain relationship of epistasis with polygenic inheritance.

Solution

Epistasis

Epistasis is the phenomenon in which the effect of genes at one locus interferes with or hides the effect of genes at another locus. The gene that masks the effect is the epistatic (inhibiting) gene, and the gene whose effect is suppressed is the hypostatic gene (book p.327).

Epistasis is different from dominance. Dominance is a relation between two alleles of the same gene at one locus, while epistasis is an interaction between different genes at different loci (book Table 12.6).

Relationship with polygenic inheritance

Both epistasis and polygenic inheritance involve the interaction of more than one gene, so the same trait can be controlled by several genes. The difference lies in the result: epistasis gives discontinuous, distinct classes with modified Mendelian ratios, whereas polygenic inheritance gives continuous variation.

Feature	Epistasis	Polygenic inheritance
Genes involved	Two genes, one masking the other	Several genes, each adding a small effect
Variation	Discontinuous classes	Continuous variation
Ratio	Modified ratios such as 9 : 3 : 4 or 12 : 3 : 1	Many classes, as in the 1 : 6 : 15 : 20 : 15 : 6 : 1 of wheat
Example in the book	Coat colour in the Labrador retriever, flower colour in foxglove	Wheat grain colour and human skin colour

Conclusion

Both patterns depart from Mendel's simple ratios because more than one gene acts on a trait. In epistasis the genes act by masking one another, and in polygenic inheritance they act by adding up.

92025 Annual I, Set D, Q.5

Explain epistasis with reference to the inheritance of coat colour in Labrador retriever.

Solution

The genes

Coat colour in the Labrador retriever is controlled by two genes at different loci. The black/chocolate gene has alleles B (black, dominant) and b (chocolate). The pigment gene has E (pigment is made, dominant) and e (no pigment). A dog with ee is yellow, whatever its B or b alleles (book p.328, p.329).

Mechanism of the epistasis

The E allele allows pigment to be deposited in the coat, so black or chocolate can show.
The ee genotype blocks pigment deposition, so the black and chocolate genes cannot show, and the coat is yellow. The yellow gene is therefore epistatic to the B and b genes.

Cross of two black heterozygotes

When two dogs of genotype BbEe are mated, each gamete carries one of BE, Be, bE or be in equal numbers, so the 16 combinations give:

Dihybrid cross BbEe x BbEe.
Egg (down) / sperm (across)	BE	Be	bE	be
BE	BBEE black	BBEe black	BbEE black	BbEe black
Be	BBEe black	BBee yellow	BbEe black	Bbee yellow
bE	BbEE black	BbEe black	bbEE chocolate	bbEe chocolate
be	BbEe black	Bbee yellow	bbEe chocolate	bbee yellow

Phenotypic ratio

Phenotype	Genotypes	Out of 16
Black	B\_E\_	9
Chocolate	bbE\_	3
Yellow	\_\_ee	4

The result is a 9 : 3 : 4 ratio, a modified dihybrid ratio in which the yellow class combines four genotypes, because ee masks the effect of the B and b genes (book p.329).

Why this is epistasis

The epistatic gene is e (recessive, but masking) and the hypostatic genes are B and b.
The traits show discontinuous variation (black, chocolate and yellow), unlike the continuous variation of polygenic traits.
Mendel's 9 : 3 : 3 : 1 is modified to 9 : 3 : 4 because the yellow class merges genotypes that would otherwise show as black or chocolate.

10

Explain epistasis with reference to the inheritance of flower colour (pigment phenotype) of sweet pea, Lathyrus odoratus.

Solution

The genes

Flower colour in the sweet pea depends on two genes at different loci. Purple flowers need a dominant allele of both genes: C (colour is produced) and P (pigment is made and deposited). The homozygous recessive genotypes cc or pp prevent the purple pigment, giving white flowers.

The unit's text gives the sweet pea only in its list of learning outcomes. The gene symbols and the 9 : 7 ratio used here are standard textbook genetics and are not printed in the unit's text.

The cross

Two plants of genotype CcPp (purple) are crossed. Each gamete carries one of CP, Cp, cP or cp in equal numbers.
The 16 combinations are sorted into classes by the genotype at both loci.
Purple needs at least one C and at least one P, so the purple class is C\_P\_, with 9 out of 16 combinations.

Genotype class	Phenotype	Probability
C\_P\_	Purple	9/16
C\_pp	White	3/16
cc\_\_	White	4/16

Epistasis

The cc genotype masks the effect of the P allele, so any plant that is cc is white whatever its P alleles. The pp genotype blocks pigment at the second step, so it is white whatever its C alleles. The recessive genotype at either locus is therefore epistatic to the other gene.

The result is a 9 : 7 ratio of purple to white flowers, a modified dihybrid ratio of the same kind as the Labrador 9 : 3 : 4 and the foxglove 12 : 3 : 1.

Note. The unit text uses foxglove and Labrador for epistasis. The sweet pea genes C and P and the 9 : 7 ratio are not printed in the unit text; the answer applies standard genetics.

11

What is gene linkage? How gene linkage can be detected? Explain Morgan's experiment. Suggest why linkage could be observed or evaluated if the number of progeny is quite large.

Solution

What gene linkage is

Genes located on the same chromosome tend to be inherited together. This is gene linkage. Linked genes form a linkage group, and the number of linkage groups equals the number of homologous chromosome pairs (book p.331). Linked genes do not assort independently, so the expected Mendelian ratio is modified.

Detection by a test cross

A dihybrid test cross mates a heterozygote for two traits (F_1) with the recessive parent (P_1). The offspring are then counted in four phenotypic classes.

Result of the test cross	Conclusion
All four classes in 1 : 1 : 1 : 1	The genes are not linked; they assort independently
More parental types and fewer recombinant types	Incomplete linkage (some crossing over)
Only parental types	Complete linkage

Morgan's experiment

T. H. Morgan studied Drosophila melanogaster. Long wings (Vg) are dominant over vestigial wings (vg), and broad abdomen (A) is dominant over narrow abdomen (a). Morgan crossed a long-winged, broad-abdomened fly with a vestigial-winged, narrow-abdomened fly. All F1 flies were long and broad, and in the F2 about 3/4 were long and broad while nearly all the rest were vestigial and narrow (book p.331).

Expected under independent assortment: a 9 : 3 : 3 : 1 ratio in F2.
Observed: about 3 : 1, with almost no long-narrow or vestigial-broad flies.
Conclusion: the genes for wing length and abdomen width are on the same chromosome and are inherited together, so no recombinant types arise without crossing over.

Why linkage is seen with large progeny

A large sample gives more statistical power, so small deviations from the expected ratio are detected.
With many offspring, a wider range of genetic combinations appears, which makes patterns and deviations from independent assortment easier to see.
Small groups can show mixing of traits by chance, but in a large sample the linked traits clearly tend to be inherited together (book p.339).
The reliability of the linkage analysis depends on the number of progeny tested.

12

Explain Morgan's experiment to demonstrate the role of crossing over in gene linkage.

Solution

Morgan's observations

Morgan's first results showed almost no recombinant types, which linkage explained (book p.331). Later experiments showed that the rare recombinants that do occur come from crossing over, the process that breaks the linkage between genes on the same chromosome (book p.332).

Crossing over in prophase I

In prophase I of meiosis the homologous chromosomes pair closely.
Non-sister chromatids exchange corresponding segments at a point called the chiasma.
The exchanged chromatids now carry new allele combinations: the recombinant chromatids.
When the homologues separate, the gametes receive either parental or recombinant chromatids.

Crossing over between the genes for long wing (Vg) and broad abdomen (A) in Drosophila, drawn from Figure 12.11 and the text. The highlighted chromatids are the recombinant ones.

Morgan's test cross

Morgan crossed the F1 female Vg\,A / vg\,a with a vestigial narrow male vg\,a / vg\,a. The male passes only vg\,a gametes, so the offspring show the type of gamete the female produced.

Gametes from the F1 female	Type	Phenotype of offspring
Vg A and vg a	Parental	Long broad, vestigial narrow (most offspring)
Vg a and vg A	Recombinant	Long narrow, vestigial broad (few offspring)

Role of crossing over

The frequency of recombinants depends on the distance between linked genes: the farther apart, the more often crossing over separates them.
One map unit equals 1 per cent recombination (book p.339), so the map gives the position of genes in a linkage group.
Crossing over is therefore the reason linkage is incomplete, and it is a source of variation in the progeny (book p.332).

13

Explain genetic identification of sex phenotypes in Drosophila and man.

Solution

Karyotype

The karyotype is the array of chromosomes of a cell. Males and females differ in one pair, the sex chromosomes, and the other pairs (the autosomes) are identical in both sexes (book p.333).

Karyotypes of Drosophila and man, drawn from Figures 12.12 and 12.13. The sex chromosomes are the pair that differs between males and females.

Comparison

Feature	Drosophila	Man
Total chromosomes	8 (four pairs)	46 (23 pairs)
Autosomes	Three pairs	22 pairs
Female	XX (two identical X)	XX
Male	XY (rod-shaped X, hooked Y)	XY

Identification

Prepare the karyotype from a dividing cell and arrange the chromosomes in pairs.
Compare the sex pair: two identical X chromosomes indicate a female, while an X and a smaller Y indicate a male.
In man, the sex is set by the Y chromosome: the Y carries the sex-determining region (SRY gene). If this region is lacking, an XY individual develops as a female (book p.333, Science Titbits).

In both organisms the male is heterogametic (XY) and the female is homogametic (XX), so the sex of the offspring is set by the sperm (book p.334).

14

Explain sex linkage in Drosophila.

Solution

Eye colour in Drosophila

The wild-type Drosophila has bright red eyes. The white-eyed mutant is recessive, and its gene lies on the X chromosome, not on the Y, so the trait is sex-linked (book p.334, p.335). The alleles are written as superscripts of X: X^R (red) and X^r (white).

The cross

Morgan crossed a homozygous red-eyed female with a white-eyed male. The F1 were all red-eyed, and the F2 showed red-eyed and white-eyed flies in an unexpected pattern: all the white-eyed flies were male.

Sex-linked inheritance of eye colour in Drosophila, drawn from Figure 12.15 and the text.

Explanation

The male has one X and a Y that carries no eye-colour gene, so his one X decides the eye colour: X^r Y is white.
A daughter receives one X from each parent and so is red if she has at least one X^R. A son receives his X from his mother and so inherits only the mother's alleles.
The white-eyed trait therefore appears only in males, from carrier mothers, and the F2 ratio is 3 red : 1 white, with all whites male.

Confirmation

Morgan confirmed the sex linkage with a reciprocal cross: a white-eyed female crossed with a red-eyed male gave red-eyed daughters and white-eyed sons (book p.335). Test crosses also showed that white-eyed females exist, which disproved the idea that the trait kills females.

15

Explain sex linkage in man.

Solution

Types of sex-linked traits in man

Humans have many X-linked traits. X-linked recessive traits such as haemophilia and colour blindness show a zigzag pattern, while X-linked dominant traits such as hypophosphatemic rickets pass from affected mothers and fathers. Y-linked traits are holandric (book p.336).

Feature	X-linked recessive	X-linked dominant	Y-linked (holandric)
Gene location	X chromosome, non-homologous region	X chromosome	Y chromosome, non-homologous region
Who is affected	Males (one X) and homozygous females	Both sexes; females more often, because they have two X chromosomes	Males only
Carrier state	Females with one allele do not show it	Heterozygous females show it	No carriers
Passed by	Carrier mother to sons, and daughters become carriers	Affected father to all daughters, never to sons; affected mother to half of children	Father to all sons
Examples	Haemophilia A and B, red-green colour blindness	Hypophosphatemic rickets	Hypertrichosis of the pinna, webbed toes

Pedigree pattern

X-linked recessive: the trait skips a generation, passing from grandfather through a carrier daughter to a grandson (zigzag).
X-linked dominant: an affected father has all his daughters affected and none of his sons affected.
Y-linked: each son of an affected father is affected, and no daughter is.

Experimental mating is not possible in humans, so these patterns are read from pedigrees (book p.336).

162026 Annual I, Set B, Q.6 OR (a)

Explain genetics of haemophilia.

Solution

What haemophilia is

Haemophilia is a serious hereditary disease in which the blood fails to clot properly after an injury, because a clotting factor is reduced or absent. Even a minor cut can cause fatal bleeding (book p.337).

Book Table 12.6.
Type	Clotting factor affected	Share of cases	Inheritance
Haemophilia A	Factor VIII	80%	Recessive, X-linked
Haemophilia B	Factor IX	20%	Recessive, X-linked
Haemophilia C	Factor XI	Less than 1%	Recessive, autosomal

Genetic basis

Haemophilia A and B are X-linked recessive. The normal allele is H and the haemophilia allele is h (book p.337). A man with one X^h is affected, because he has no second X. A woman is affected only when she is homozygous X^h X^h, and a carrier woman (X^H X^h) is normal.

Cross of a carrier mother and a normal father

Mother / father	X^H (1/2)	Y (1/2)
X^H (1/2)	X^H X^H normal daughter	X^H Y normal son
X^h (1/2)	X^H X^h carrier daughter	X^h Y affected son

Half of the sons are affected, because each son gets his X from his mother and the mother gives X^h to half of her eggs.
No daughter is affected, because her father gives her a normal X^H. Half of the daughters are carriers.
Overall, one quarter of the children are affected sons (probability 1/2 \times 1/2 = 1/4).

Haemophilia in normal parents

Two normal parents can have a haemophiliac son only if the mother is a carrier, since the affected son receives his X from her. The probability of an affected son is then 1/4 of the children, as calculated above (book p.337, Genetic Problem 12.13).

Pattern of inheritance

Haemophilia A and B pass from a maternal grandfather through a carrier daughter to a grandson, and never directly from father to son. They affect men far more often than women (book p.337).

172026 Annual I, Set D, Q.6 (a)

Explain genetics of colour blindness.

Solution

Basis of colour vision

Normal trichromatic vision depends on three kinds of cone cells in the retina, each sensitive to one primary colour: red, green or blue. Each cone type has its own light-absorbing protein, the opsin (book p.338).

Gene	Location	Opsin
Red opsin	X chromosome	red-sensitive cones
Green opsin	X chromosome	green-sensitive cones
Blue opsin	Autosome 7	blue-sensitive cones

Types of colour blindness

Dichromacy: one cone type is missing, so the person sees two primary colours. Protanopia is red blindness, deuteranopia is green blindness and tritanopia is blue blindness.
Monochromacy: only one colour is perceived. Blue cone monochromacy is X-linked recessive, with red and green cones absent.

Inheritance of red-green colour blindness

Red-green colour blindness is an X-linked recessive trait. It zigzags from a maternal grandfather through a carrier daughter to a grandson, and it is more common in men, because a man with one recessive allele on his single X is affected (book p.339).

Genetic Problem 12.14 asks for the sons of a carrier woman and a normal man. Let the normal allele be X^N and the colour-blind allele X^c:

Mother / father	X^N (1/2)	Y (1/2)
X^N (1/2)	X^N X^N normal daughter	X^N Y normal son
X^c (1/2)	X^N X^c carrier daughter	X^c Y colour-blind son

Half of the sons are colour blind, since each son gets the mother's X, which is X^c in half of her eggs.
All daughters have normal vision. Half are carriers and half are homozygous normal.
The probability that a child is a colour-blind son is 1/4.

18

Explain genetics of muscular dystrophy.

Solution

Note on the unit text

Muscular dystrophy is not described in the unit's text, so the account below applies the X-linked recessive pattern that the unit teaches for haemophilia and colour blindness (book p.336).

Pattern

The common form of muscular dystrophy is an X-linked recessive disorder. A boy with one recessive allele on his single X is affected.
A carrier mother can have affected sons, and a son never passes the disorder to his own sons, because he gives them his Y.
Daughters of an affected father are carriers, and an affected daughter needs an affected father and a carrier mother.

Expected result

If the mother is a carrier and the father is normal, half of the sons are affected and half of the daughters are carriers, the same as for colour blindness in the table of Question 17.

Note. Muscular dystrophy is not covered in the unit's text. The answer applies the X-linked recessive pattern the unit teaches; the disease itself is not described in the book.

19

Describe sex related traits.

Solution

Sex-linked traits

Sex-linked traits are controlled by genes on the sex chromosomes, mainly the X chromosome (book p.334). Their pattern follows the sex chromosomes, as described for haemophilia, colour blindness and hypophosphatemic rickets.

Sex-limited traits

A sex-limited trait is controlled by a gene present in both sexes but expressed in only one sex, because of the anatomy or physiology of that sex. The unit text does not discuss these traits, so this is the general definition.

Sex-influenced traits

A sex-influenced trait is controlled by an autosomal gene, but its expression differs between the sexes because of sex hormones. Male pattern baldness is the usual example: the allele is dominant in males, so a heterozygous man is bald, while in women it is recessive, so only homozygous women are bald. The unit text does not discuss baldness.

Type	Gene location	Expression	Example
Sex-linked	Sex chromosomes	Pattern follows the X or Y chromosome	Haemophilia, colour blindness, Y-linked hypertrichosis
Sex-limited	Usually autosomal	Only in one sex	Traits of the reproductive organs
Sex-influenced	Autosomal	Differs between the sexes, because of hormones	Male pattern baldness

Note. Sex-limited and sex-influenced traits, and baldness, are not described in the unit's text; the answers use general definitions.

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 131 mark

Polygenic inheritance results in:

ASingle trait expression
BContinuous variationAnswer
CDominant traits only
DMutation only

Why

Polygenic traits are controlled by two or more gene pairs at different loci, all acting on the same trait in an additive way. The intensity of the phenotype depends on the number of effect-causing alleles, so many small steps appear between the extremes and the trait shows continuous variation (height, skin colour, wheat grain colour). Single phenotype, dominant-only and mutation-only are not what polygenic inheritance means.

2025 Annual I, Set B, MCQ 131 mark

The sex linked traits which are more common in males than female are:

AAutosomal recessive
BY- linked dominant
CX- linked dominant
DX- linked recessiveAnswer

Why

An X-linked recessive allele shows in a man with a single copy, because he has only one X, but a woman is affected only when she has two copies (X^hX^h). The trait is therefore more common in males (haemophilia, colour blindness). Autosomal traits are equally common in both sexes. A Y-linked trait occurs only in males, so it is 'only in males', not 'more common', and X-linked dominant traits are, if anything, more common in females.

2026 Annual I, Set B, MCQ 141 mark

Epistasis modifies:

ADNA
BGenetrix ratio
CPhenotypic ratioAnswer
DProtein

Why

Epistasis is an interaction between two genes at different loci in which one gene hides the effect of the other. The genotypes are formed exactly as in Mendel's dihybrid cross, but the classes that look alike merge, so the usual 9 : 3 : 3 : 1 phenotypic ratio is modified (to 9 : 3 : 4 in the Labrador retriever). The DNA itself and the proteins are not changed by the interaction, and the genotypic ratio is unchanged.

Note. Option B is printed as 'Genetrix ratio' in the paper (probably 'genotypic ratio'). Even read as genotypic ratio, it is not the answer: epistasis changes the phenotypic ratio only.

2025 Annual I, Set B, MCQ 161 mark

Identify the trait which is NOT controlled by polygenes?

AIntelligence
BHeight
CSkin color
DTongue rollingAnswer

Why

Intelligence, height and skin colour are polygenic traits with continuous variation. Tongue rolling is the book's example of a simple Mendelian trait of one gene pair (roller T dominant, non-roller t) with two clear classes.

2025 Annual I, Set B, Q.2 (iv) OR1+2 = 3 marks

What is incomplete dominance? Explain with a suitable example.

Solution

Definition (1 mark)

Incomplete dominance is a dominance relation in which neither allele of a pair is completely dominant, so the heterozygote shows a phenotype intermediate between those of its two homozygous parents.

Example: flower colour of the 4 O'clock plant (2 marks)

Red (R_1R_1) x white (R_2R_2) gives all pink (R_1R_2) F1 plants, as Carl Correns found in 1899.
F1 x F1 gives red : pink : white in the ratio 1 : 2 : 1, which is also the genotypic ratio (R_1R_1 : R_1R_2 : R_2R_2).

Incomplete dominance in the 4 O'clock plant: red x white gives all pink F1, and F1 x F1 gives red : pink : white in the ratio 1 : 2 : 1.

2025 Annual I, Set B, Q.2 (ix)1+1+1 = 3 marks

What is Y- linked inheritance? How these traits are transmitted? Also give two examples.

Solution

Y-linked inheritance (1 mark)

Y-linked (holandric) traits are controlled by genes on the non-homologous region of the Y chromosome, which have no allele on the X chromosome.

Transmission (1 mark)

They are expressed only in males, as only males have a Y chromosome.
A son always receives the Y from his father, so the trait passes from father to all his sons and never to daughters.

Examples (1 mark)

Hypertrichosis (hair on the rim of the pinna), porcupine man (straight hair on the body) and webbing of the toes.

2026 Annual I, Set B, Q.2 (x)3 marks

Explain through a cross that incomplete dominance is a deviation from Mendel's law of segregation.

Solution

Mendel's monohybrid cross gives a 3 : 1 phenotypic ratio in F2 because one allele is completely dominant and hides the other in the heterozygote. In incomplete dominance neither allele is completely dominant, so the ratio changes. The 4 O'clock plant (Mirabilis jalapa) shows this.

Cross (1 mark for parents and F1)

Parents: pure red (R_1R_1) x pure white (R_2R_2). Gametes: R_1 and R_2.
F1: all R_1R_2 with pink flowers. The heterozygote is intermediate and resembles neither parent, whereas Mendel's F1 resembled one parent.

F2 (1 mark)

F1 gametes	R_1	R_2
R_1	R_1R_1 red	R_1R_2 pink
R_2	R_1R_2 pink	R_2R_2 white

Genotypic ratio 1 : 2 : 1 and phenotypic ratio 1 : 2 : 1 (red : pink : white).

Incomplete dominance in the 4 O'clock plant: red x white gives all pink F1, and F1 x F1 gives red : pink : white in the ratio 1 : 2 : 1.

Why it is a deviation (1 mark)

The alleles still separate (segregate) into different gametes, but because dominance is incomplete the heterozygote can be told apart from both homozygotes. The F2 phenotypic ratio is therefore 1 : 2 : 1 instead of Mendel's 3 : 1, and the genotypic and phenotypic ratios become the same.

2025 Annual I, Set B, Q.2 (x) OR2+1 = 3 marks

Coat color in Labrador retriever shows epistasis. If a black(BbEe) and a chocolate (bbEe) Labradors retrievers are crossed: (a) Show genotypes and phenotypes of offspring(s). (b) Calculate the probability of black and yellow coat color.

Solution

(a) Genotypes and phenotypes (2 marks)

Parents: black BbEe x chocolate bbEe. Gametes of BbEe: BE, Be, bE, be. Gametes of bbEe: bE, be.

Eight equally likely combinations.
Gametes	bE	be
BE	BbEE black	BbEe black
Be	BbEe black	Bbee yellow
bE	bbEE chocolate	bbEe chocolate
be	bbEe chocolate	bbee yellow

Phenotype	Genotypes	Out of 8
Black	BbEE (1), BbEe (2)	3
Chocolate	bbEE (1), bbEe (2)	3
Yellow	Bbee (1), bbee (1)	2

Phenotypic ratio black : chocolate : yellow = 3 : 3 : 2. Yellow appears because ee is epistatic: it hides both B and b.

(b) Probabilities (1 mark)

Black = 3/8 (37.5%). Yellow = 2/8 = 1/4 (25%).

\text{black }3/8,\ \text{yellow }1/4

2026 Annual I, Set B, Q.6 OR (a)3 marks

Explain genetics of hemophilia with reference to pattern of inheritance.

Solution

What the disease is

Haemophilia is a hereditary disease in which blood does not clot properly after injury, because a clotting factor is reduced, faulty or absent. Haemophilia A (factor VIII, 80%) and B (factor IX, 20%) are X-linked recessive; haemophilia C is autosomal recessive.

Genotypes

Genotype	Phenotype
X^HX^H	Normal female
X^HX^h	Carrier female (normal)
X^hX^h	Haemophiliac female
X^HY	Normal male
X^hY	Haemophiliac male

Pattern of inheritance

It is carried on the X chromosome and is recessive, so a man with one X^h is affected (he has no second X to mask it), while a woman must be X^hX^h. It is therefore much more common in men.
A carrier mother passes X^h to half of her eggs, so half of her sons are affected and half of her daughters are carriers.
An affected father gives his X only to daughters (all carriers) and his Y to sons, so it never passes directly from father to son. It 'zigzags' from the maternal grandfather through a carrier daughter to a grandson.

Carrier mother x normal father.
Mother / father	X^H	Y
X^H	X^HX^H normal daughter	X^HY normal son
X^h	X^HX^h carrier daughter	X^hY haemophiliac son

2026 Annual I, Set B, Q.6 OR (b)3 marks

Can two normal parents have a hemophilic child? Work out the probability.

Solution

Yes, but only when the mother is a carrier. The haemophilic son gets his single X from his mother, and the allele is recessive, so a normal-looking mother can be X^HX^h while the father is X^HY (normal).

Mother / father	X^H (1/2)	Y (1/2)
X^H (1/2)	X^HX^H normal daughter	X^HY normal son
X^h (1/2)	X^HX^h carrier daughter	X^hY haemophiliac son

Probability that the egg carries X^h is 1/2.
Probability that the sperm carries Y (to make a son) is 1/2.
Probability of an affected son = 1/2\times 1/2 = 1/4.

So the probability of a haemophilic child is 1/4 (25%); all the affected children are sons, and half of the sons (50%) are affected. No daughter can be haemophilic, because the normal father gives every daughter X^H. If the mother were X^HX^H, no child could be affected.

Note. Matches Genetic Problem 12.13 of the book.

2026 Annual I, Set D, MCQ 131 mark

Polygenic inheritance results in:

ADiscrete variation
BMutation only
CNo variation
DContinuous variationAnswer

Why

Polygenic traits are controlled by several gene pairs whose effects add up, giving many small steps between the extremes: continuous variation. Discrete variation (clear separate classes) is what single-gene Mendelian traits and epistasis give, which is why it is the tempting wrong option.

2026 Annual I, Set D, MCQ 141 mark

In Labrador retrievers, coat colour inheritance is an example of:

AIncomplete dominance
BCodominance
CMutation
DEpistasisAnswer

Why

Two genes at different loci decide the coat: B/b (black or chocolate) and E/e. A dog that is ee is yellow whatever its B gene, so one gene hides the effect of another, which is epistasis and gives a 9 : 3 : 4 ratio. In incomplete dominance and codominance only one gene (two alleles at one locus) is involved.

2025 Annual I, Set D, MCQ 151 mark

Holandric traits show all of the following characteristics EXCEPT:

ASons of affected father are affected
BShow Y-linked inheritance
CPorcupine man is an example
DDaughters of affected father are carriersAnswer

Why

Holandric traits are Y-linked: the Y goes from father to every son, so sons of an affected father are affected, and porcupine man is a book example. A daughter receives her father's X, not his Y, so she gets no holandric gene and cannot be a carrier. Carrier daughters of an affected father occur in X-linked recessive traits, not Y-linked ones.

2025 Annual I, Set D, Q.2 (v) OR1+2 = 3 marks

Name the type of pattern of sex determination in humans. Also write mechanism of sex determination in man.

Solution

Type of pattern (1 mark)

Humans follow the XX-XY type of sex determination: the female is XX (homogametic) and the male is XY (heterogametic).

Mechanism (2 marks)

Humans have 23 pairs of chromosomes: 22 pairs of autosomes and one pair of sex chromosomes.
All eggs of a woman carry one X chromosome. Half of the sperms of a man carry X and half carry Y, with equal chance.
An X sperm fertilising the egg gives XX (a girl); a Y sperm gives XY (a boy). The sperm therefore decides the sex of the child, and the sex ratio is 1 : 1.

Gametes	X (egg)	X (egg)
X (sperm)	XX female	XX female
Y (sperm)	XY male	XY male

XX-XY sex determination in man (after Figure 12.14): the sperm that fertilises the egg decides the sex of the child.

2026 Annual I, Set D, Q.2 (x)3 marks

When Red 4'O clock flower is crossed with white flower, what is the probability of red, white and pink flowers?

Solution

Flower colour in the 4 O'clock plant shows incomplete dominance, so red is R_1R_1, white is R_2R_2 and the heterozygote is pink (R_1R_2).

Red x white (the cross asked)

Gametes	R_2	R_2
R_1	R_1R_2 pink	R_1R_2 pink
R_1	R_1R_2 pink	R_1R_2 pink

All the F1 plants are pink: probability of pink = 1 (100%), red = 0 and white = 0.

If the pink F1 plants are crossed with each other (F2)

F1 gametes	R_1	R_2
R_1	R_1R_1 red	R_1R_2 pink
R_2	R_1R_2 pink	R_2R_2 white

Red = 1/4 (25%), pink = 2/4 = 1/2 (50%), white = 1/4 (25%), a 1 : 2 : 1 ratio.

Incomplete dominance in the 4 O'clock plant: red x white gives all pink F1, and F1 x F1 gives red : pink : white in the ratio 1 : 2 : 1.

Note. The question can be read as the first cross (all pink) or as the F2 of the book's Figure 12.3; both answers are given.

2025 Annual I, Set D, Q.2 (x)3 marks

Justify that O^{-ve} persons are universal donors and AB^{+ve} are universal recipients of blood.

Solution

A transfusion works on the donor's antigens and the recipient's antibodies: if the donor's antigen meets the matching antibody, the red cells clump (agglutination).

	O negative (universal donor)	AB positive (universal recipient)
Antigens on red cells	None: no A, no B and no Rh (D) antigen	A, B and Rh (D)
Antibodies in plasma	Anti-A and anti-B	None: no anti-A, no anti-B and no anti-Rh
Result	Its red cells carry nothing that a recipient's antibodies can attack, so it can be given to anyone	Its plasma has no antibody to attack any donor's cells, so it can receive blood of any group

A donor must be free of antigens, and a recipient must be free of the matching antibodies. A person of group O negative can only receive from O negative (he has both antibodies), and AB positive can only give to AB positive (his cells carry all the antigens).

2025 Annual I, Set D, Q.56 marks

Epistasis is different from dominance. Explain with the example of coat color in Labrador retriever.

Solution

Epistasis and dominance

Dominance	Epistasis
Involves one pair of alleles at one locus.	Involves two pairs of alleles at different loci.
A gene suppresses the expression of its own allele.	A gene suppresses the expression of an allele of another gene.
Only the recessive allele is suppressed.	Both dominant and recessive alleles of the other gene can be suppressed.
The effect is due only to the dominant allele.	Either a dominant or a recessive allele can be epistatic.

Coat colour in the Labrador retriever

Gene 1 (B/b): black (B) is dominant to chocolate (b).
Gene 2 (E/e) at a different locus: it decides whether pigment is deposited. A dog that is ee is yellow, whatever its B gene, because ee is epistatic and hides B and b.

How the two genes decide the coat colour of a Labrador retriever: ee is epistatic and makes the dog yellow whatever its B gene.

Cross of BBEE (black) x bbee (yellow)

F1: all BbEe, black.
F1 x F1: each parent makes gametes BE, Be, bE, be; the 16 combinations are in the table.

Gametes	BE	Be	bE	be
BE	BBEE black	BBEe black	BbEE black	BbEe black
Be	BBEe black	BBee yellow	BbEe black	Bbee yellow
bE	BbEE black	BbEe black	bbEE chocolate	bbEe chocolate
be	BbEe black	Bbee yellow	bbEe chocolate	bbee yellow

Phenotype	Genotypes	Out of 16
Black	B\_E\_	9
Chocolate	bbE\_	3
Yellow	B\_ee and bbee	3 + 1 = 4

Why this is not dominance

Mendel's dihybrid ratio would be 9 : 3 : 3 : 1, but the F2 ratio is 9 : 3 : 4: the B\_ee and bbee classes both look yellow, because ee masks the other gene. This interaction between two different loci, not between two alleles of one locus, is epistasis.

Note. Combines the book's long question 9 with the dominance and epistasis comparison (short question 15).

2026 Annual I, Set D, Q.6 (a)3 marks

Explain genetics of colour-blindness with reference to pattern of inheritance.

Solution

Colour vision

Normal (trichromatic) vision depends on three kinds of cone cells in the retina, sensitive to red, green and blue, each with its own opsin protein. The red and green opsin genes are on the X chromosome and the blue opsin gene is on autosome 7.

Types

Dichromacy: one opsin is missing, so two primary colours are seen (protanopia red blind, deuteranopia green blind, tritanopia blue blind).
Monochromacy: only one colour is seen (true colour blindness); blue cone monochromacy is X-linked recessive.

Pattern of inheritance of red-green colour blindness

Genotype	Phenotype
X^CX^C	Normal female
X^CX^c	Carrier female (normal vision)
X^cX^c	Colour-blind female
X^CY	Normal male
X^cY	Colour-blind male

It is X-linked recessive, so it is more common in men: one X^c is enough in a man, but a woman needs two.
A carrier mother passes X^c to half of her eggs, so half of her sons are colour blind.
It never passes directly from father to son; it zigzags from the maternal grandfather through a carrier daughter to a grandson.

Carrier mother x normal father.
Mother / father	X^C	Y
X^C	X^CX^C normal daughter	X^CY normal son
X^c	X^CX^c carrier daughter	X^cY colour-blind son

2026 Annual I, Set D, Q.6 (b)3 marks

Can two normal parents have a colourblind child? Work out the probability.

Solution

Yes, if the mother is a carrier. The colour-blind son gets his only X from his mother, so a normal-sighted mother X^CX^c and a normal father X^CY can have a colour-blind son.

Mother / father	X^C (1/2)	Y (1/2)
X^C (1/2)	X^CX^C normal daughter	X^CY normal son
X^c (1/2)	X^CX^c carrier daughter	X^cY colour-blind son

Probability that the egg carries X^c is 1/2.
Probability that the sperm carries Y (to make a son) is 1/2.
Probability of an affected son = 1/2\times 1/2 = 1/4.

The probability of a colour-blind child is 1/4 (25%), always a son; half of the sons are colour blind, and no daughter is affected because the normal father gives her X^C. If the mother is X^CX^C, no child is colour blind.

Note. Matches Genetic Problem 12.14 (carrier woman x normal man).

Exam focus

Explain the genetics of haemophilia or colour blindness (X-linked recessive inheritance) and work out whether two normal parents can have an affected child. (long)
Both 2026 sets asked it as a Section C pair (Set B Q.6 OR (a) and (b) on haemophilia; Set D Q.6 (a) and (b) on colour blindness), and the book has long questions 16 and 17.
Go to answer
Epistasis with the coat colour of the Labrador retriever: genes, the 9 : 3 : 4 ratio, and how epistasis differs from dominance. (long)
A 6-mark Section C question in 2025 (Set D Q.5), a cross in 2025 Set B Q.2 (x) OR, an MCQ in 2026 Set D (MCQ 14) and 2026 Set B MCQ 14; book long question 9.
Go to answer
Incomplete dominance in the 4 O'clock plant: the 1 : 2 : 1 cross and why it departs from Mendel's 3 : 1. (short)
Asked in 2026 Set B Q.2 (x), 2026 Set D Q.2 (x) and 2025 Set B Q.2 (iv) OR; book long question 4.
Go to answer
Sex determination in man (XX-XY pattern) and why the male is heterogametic. (short)
Asked in 2025 Set D Q.2 (v) OR; book short questions 18 and 19 and SLOs 27 to 29.
Go to answer
Y-linked (holandric) inheritance: meaning, transmission from father to son and examples; X-linked recessive traits being commoner in males. (short)
2025 Set B Q.2 (ix), 2025 Set B MCQ 13 and 2025 Set D MCQ 15 all test it; book short question 21.
Go to answer
Why O negative is the universal donor and AB positive the universal recipient (antigens and antibodies, ABO and Rh). (short)
Asked in 2025 Set D Q.2 (x); book short question 12 and SLO 14.
Go to answer
Polygenic inheritance: continuous variation, additive effect of alleles and examples such as wheat grain colour and skin colour. (mcq)
MCQ 13 in both 2026 sets and MCQ 16 of 2025 Set B (which trait is not polygenic); book long question 7.
Go to answer
Dihybrid and monohybrid crosses with Punnett squares and probability: write gametes, fill the square and give the ratio. (numerical)
Every paper cross above (4 O'clock, Labrador, haemophilia, colour blindness) is a Punnett square question, so the method scores marks across the unit; practise the book's genetic problems.
Go to answer
ABO blood groups: alleles I^A, I^B, i, antigens, antibodies and genotype table. (long)
No direct paper question beyond the donor-recipient item, but it is the base for the transfusion answer and the book's long question 5 and short questions 6, 9 and 10.
Go to answer

Heredity and variation

GeneOften examined

What is a gene?

12.1Mendelian inheritance

12.1.1Association of inheritance with laws of MendelOften examined

Inheritance of single trait (monohybrid cross)Often examined

Procedure and observations

Interpretation of the results

Punnett square of the F2​

Law of segregation

Inheritance of two traits (dihybrid cross)Often examined

Procedure and observations

Interpretation of the results

Independent assortment of chromosomes

Limitations of law of independent assortmentOften examined

Usefulness and scope of independent assortmentOften examined

Independent assortment leads to variation in the gametes

12.1.2Inheritance and mathematical probabilitiesOften examined

Product rule

Genetic problems of the book (activity: solve at least four to illustrate the law of independent assortment)

12.2Exceptions to Mendelian inheritance

12.2.1Incomplete dominanceOften examined

12.2.2Co-dominanceOften examined

Incomplete dominance and co-dominance as variations of Mendel's research

Gene and allele

Multiple allelesOften examined

12.3Blood group systems

12.3.1ABO blood group systemOften examined

Antigens of the ABO system

Genetic basis of the ABO system

Antibodies of the ABO system

Various human blood group systems

12.3.2Transfusion principleOften examined

12.3.3Rh blood group systemOften examined

Antigens of the Rh blood group system

Genetic basis of the Rh blood group system

Anti-Rh antibody and transfusion principle

12.3.4Erythroblastosis foetalisOften examined

Problems and complications in the foetus

Causes and risk factors

Prevention of erythroblastosis foetalis

12.4Polygenic inheritance and epistasisOften examined

12.4.1Wheat grain colourOften examined

12.4.2Inheritance of human skin colourOften examined

Examples of polygenic traits in humans

12.4.3EpistasisOften examined

Epistasis and the ABO blood group: the Bombay phenotype

Relationship of epistasis with polygenic inheritance

Coat colour in the Labrador retrieverOften examined

Epistasis in foxglove flower

12.5.1Gene linkageOften examined

Detection of gene linkage

Morgan's experimentOften examined

Gene linkage encounters independent assortment

12.5.2Crossing-over modifies the progenyOften examined

12.6Sex determination

12.6.1Genetic identification of sex phenotypesOften examined

Chromosomes in Drosophila

Chromosomes in man

12.6.2Patterns of sex determinationOften examined

XY-XX type

12.7Sex linkage

12.7.1Sex linkage in DrosophilaOften examined

Morgan's experiments

Morgan's conclusion

12.7.2Sex linkage in humansOften examined

X-linked recessive inheritance

X-linked dominant inheritance

Y-linked inheritance

Genetics of haemophiliaOften examined

Blood clotting

Genetics of colour-blindnessOften examined

Linkage can be observed if the number of progeny is too largeOften examined

Why a large number of progeny is needed

Techniques employed for embryonic screeningOften examined

Amniocentesis

Chorionic villus sampling (CVS)

Formula sheet

Key terms

Multiple choice questionsp. 341–342

Punnett square of the F $_{2}$