Dihybrid inheritance, multiple alleles and sex linkageAQA A-Level Biology: Revision notes

Section 1

Dihybrid crosses

A dihybrid cross follows the inheritance of two genes at once. If the genes are on different chromosomes, they assort independently at meiosis, so each gamete receives one allele of each gene in any combination.

An $RrYy$ plant makes four types of gamete, $RY$ , $Ry$ , $rY$ and $ry$ , in equal numbers. Crossing two heterozygotes gives 16 combinations and the phenotypic ratio 9 : 3 : 3 : 1 (double dominant : dominant for gene 1 only : dominant for gene 2 only : double recessive).

Crossing the heterozygote with a homozygous recessive ( $RrYy$ × $rryy$ ) gives 1 : 1 : 1 : 1.

Key termsdihybrid crossindependent assortmentgamete

Exam tip

List the four gametes from each parent first, then fill a 4 × 4 square. Count the phenotypes after the square is complete.

Section 2

Working out numbers and ratios

Use the ratio to find expected numbers. With 640 offspring from a 9:3:3:1 cross, divide by 16 to get 40 per part, so the classes are 360, 120, 120 and 40.

If one gene shows codominance, its ratio from two heterozygotes is 1:2:1 instead of 3:1. Multiply the ratio for each gene to build the combined ratio: (3 : 1) × (1 : 2 : 1) gives 3 : 6 : 3 : 1 : 2 : 1.

Key termsphenotypic ratio

Section 3

Multiple alleles

Some genes have more than two alleles in a population, but an individual diploid still has only two. The ABO blood group gene has alleles $I^A$ , $I^B$ and $I^O$ . $I^A$ and $I^B$ are codominant, and $I^O$ is recessive to both.

Group A: $I^AI^A$ or $I^AI^O$
Group B: $I^BI^B$ or $I^BI^O$
Group AB: $I^AI^B$
Group O: $I^OI^O$

A group A parent ( $I^AI^O$ ) and a group B parent ( $I^BI^O$ ) can produce children of all four groups, each with probability 1/4.

Key termsmultiple alleles

Common mistake

A person can only carry two alleles of a gene, even if the gene has many alleles in the population.

Section 4

Sex linkage

Humans have 22 pairs of autosomes and one pair of sex chromosomes: females are XX and males XY. A sex-linked gene is on the X chromosome. The Y chromosome is shorter and has no corresponding allele for most X-linked genes.

Males have only one X chromosome, so one recessive allele is expressed. Females need two. This is why X-linked recessive conditions such as haemophilia and red-green colour blindness are more common in males.

A son inherits his X chromosome from his mother and cannot get an X-linked allele from his father. A carrier mother ( $X^HX^h$ ) and normal father ( $X^HY$ ) have a 1/2 chance of each son being affected and none of their daughters affected (half are carriers).

Key termssex linkagecarrierhemizygous

Exam tip

Write alleles as superscripts on X, such as $X^h$ , and always show the Y with no allele.

Section 5

Fully labelled diagrams

For every cross, show: parental phenotypes, parental genotypes, gametes, offspring genotypes and offspring phenotypes with the ratio. For dihybrid, multiple allele or sex-linked crosses, state the allele key first. State the ratio in whole numbers and, if asked, convert to a probability by dividing by the total. For sex-linked crosses, state the sex of each phenotype, for example 'red-eyed males'.

Key termsgenetic diagram

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Dihybrid inheritance, multiple alleles and sex linkage

In a species of garden pea, round seeds are produced by the dominant allele $R$ and wrinkled seeds by the recessive allele $r$ . Yellow seeds are produced by the dominant allele $Y$ and green seeds by the recessive allele $y$ . The two genes are on different chromosomes. A plant breeder crosses plants that are heterozygous for both genes ( $RrYy$ ).
An $RrYy$ plant is crossed with a plant that has wrinkled green seeds. State the expected phenotypic ratio in the offspring and explain how it arises.2 marks
The ABO blood group in humans is controlled by one gene, $I$ , with three alleles. The alleles $I^A$ and $I^B$ are codominant, and the allele $I^O$ is recessive to both. A woman with blood group A (genotype $I^AI^O$ ) and a man with blood group B (genotype $I^BI^O$ ) have children. In a separate paternity enquiry, a woman of blood group A has a child of blood group O.
A man with blood group AB is named as the father of the child of blood group O. Explain why he cannot be the father.2 marks
Haemophilia A is a blood-clotting disorder caused by a recessive allele, $h$ , on the X chromosome. The dominant allele, $H$ , codes for a working clotting factor. There is no corresponding gene on the Y chromosome. A woman who is a carrier and a man with normal clotting plan to have children.
Explain why haemophilia is much more common in males than in females.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).