7.1 Inheritance (A Level only) — AQA A Level Biology 7402 Revision Notes & Topical Questions

Section 1 of 22Key terms in genetics

A gene is a DNA sequence that codes for a polypeptide or protein.

Locus

Each gene occupies a specific locus on a chromosome, and a chromosome contains many genes in one long DNA molecule. Alleles are different forms of the same gene.

Homozygous

Individuals with two identical alleles are homozygous; those with different alleles are heterozygous.

The genotype is the allele combination; the phenotype is the observable characteristics, determined by the genotype and the environment.

Horses, for example, have two copies of the coat-colour gene in each cell (diploid).

A dominant allele is expressed in the heterozygote; a recessive allele is expressed only when homozygous.

In a cross between two homozygous parents (AA × aa), all F₁ offspring are heterozygous (Aa) and show the dominant phenotype.

Section 2 of 22Codominance and multiple alleles

In codominance both alleles are expressed in the phenotype of the heterozygote.

1

The gene for human ABO blood type has three alleles: I^A and I^B (codominant) and I^O (recessive).

2

Genotypes and phenotypes: I^AI^A or I^AI^O give group A; I^BI^B or I^BI^O give group B; I^AI^B gives group AB; I^OI^O gives group O.

3

When working with codominant alleles, genetic diagrams are built in the same way,

but genotypes are written with a capital letter for the gene and superscript letters for the alleles (e.g. I^AI^A).

4

There are more possible phenotypes, so the predicted ratios differ.

Section 3 of 22Monohybrid inheritance

Monohybrid inheritance tracks one gene with two alleles. It uses known genotypes, phenotypes and the process of meiosis to predict offspring: during gamete formation, each allele from a homologous pair has an equal chance of being passed on.

Section 4 of 22Sex linkage and haemophilia

Haemophilia is a recessive X-linked disorderthe gene is on the X chromosome and codes for Factor VIII, a protein essential for blood clotting.

Sex linkage and haemophilia

Autosomal genes lie on autosomes (any chromosome that is not a sex chromosome).

Females (XX) have two copies,

so can be unaffected, carriers or affected (e.g. a female heterozygote X^HX^h is a carrier);
- males (XY) have one X,
so a single recessive allele (X^hY) causes the disease.

Males inherit their X from their mother, so the condition is more common in males.

Section 5 of 22Worked examples

Define the terms locus and heterozygous. [2]2 marks
Give all the genotypes that give blood group A, and explain why a person of genotype I^AI^B has group AB. [3]3 marks
Explain why haemophilia is more common in males. [3]3 marks

Section 6 of 22Sex-linked cross for haemophilia

A carrier female (X^HX^h) crossed with a normal male (X^HY) gives a predicted genotype ratio of 1 X^HX^H : 1 X^HX^h : 1 X^HY : 1 X^hY, which is a phenotype ratio of 1 female with normal blood clotting : 1 carrier female : 1 male with haemophilia : 1 male with normal clotting. Method: include the possible gametes produced by each parent; do not skip straight to the Punnett square; state both the genotype and phenotype of the offspring; read carefully to see whether the question refers to all children or a specific sex (e.g. only boys or girls).

Section 7 of 22Dihybrid crosses

Monohybrid crosses follow one gene; dihybrid crosses follow two genes at once.

Dihybrid crosses

The diagram has a similar structure, but each parent produces four gamete types (for AaBb: AB, Ab, aB, ab) and the Punnett square has 16 boxes.

Worked example: a cross can give a predicted ratio of 3 black coat, brown eyes : 3 chestnut coat, brown eyes : 1 black coat, blue eyes : 1 chestnut coat, blue eyes, the 3:1 ratio for one gene (heterozygous × heterozygous) and 1:1 for the other.

Section 8 of 22Linkage

Genes on the same chromosome are linked and inherited together (unless crossing over separates them), so they do not assort independently.

1

Example: a newt has normal tail (T) or shorter tail (recessive t), and green scales (dominant G) or white scales (recessive g).

2

A newt heterozygous for both (alleles TG on one chromosome and tg on the other) is crossed with a short-tailed white-scaled newt.

3

Parental phenotypes: normal tail, green scales × short tail, white scales.

4

Parental genotypes: (TG)(tg) × (tg)(tg).

5

Parental gametes: (TG) or (tg) from the first, (tg) from the second.

6

With complete linkage the offspring are 1 normal tail, green scales : 1 short tail, white scales, not the 1:1:1:1 expected from independent assortment.

Section 9 of 22Epistasis: a pigeon example

Epistasis is when the allele of one gene masks or modifies the expression of another gene.

1

In pigeons, gene R/r: R codes for a pigment giving grey feathers; r does not make pigment, so feathers are white.

2

A second gene F/f: the dominant allele F stops grey feathers being produced even if R is present.

3

So grey birds are R_ ff; all F_ birds are white; rr birds are white.

4

For two double heterozygotes (RrFf), this gives 3 grey : 13 white.

Section 10 of 22Test crosses

A test cross deduces the genotype of an individual showing the dominant phenotype by crossing it with a homozygous recessive individual.

Test crosses

If the unknown is homozygous dominant (DD × dd) all offspring are Dd with the dominant phenotype; if heterozygous (Dd × dd), offspring are 1 Dd (long ears) : 1 dd (short ears).

Example: a breeder observes that at least one offspring has the recessive short ear phenotype,

so the dominant-eared rabbit Floppy must be heterozygous (Dd).

When analysing any inheritance question, identify: how many alleles exist for each gene; which are dominant and recessive; the type of dominance (complete, codominance, incomplete); whether there is linkage (autosomal or sex-linked); and whether there is epistasis (interaction between genes affecting the same trait).

Section 11 of 22Worked examples

A carrier woman and a man with normal blood clotting have children. Give the probability that a son has haemophilia. [3]3 marks
Explain why a test cross can be used to determine the genotype of an individual with the dominant phenotype. [3]3 marks
Explain why the cross (TG)(tg) × (tg)(tg) does not give a 1:1:1:1 ratio. [2]2 marks
Explain what epistasis is, using the pigeon example. [2]2 marks

Section 12 of 22Predictions are probabilities

Genetic diagrams, such as Punnett squares, predict offspring genotypes and phenotypes from known parental genotypes and meiosis. In sexual reproduction each allele from a homologous pair has an equal chance of entering a gamete and being inherited. Predictions are based on probability, so actual ratios may differ because of random fertilisation of gametes.

Section 13 of 22Worked examples

Homozygous parents

when both parents are homozygous no Punnett square is needed;
- offspring genotypes are directly predictable (e.g. a purple-stemmed homozygous Fast Plant crossed with another homozygous purple gives all purple stems).

Heterozygous cross

two heterozygous Fast Plants interbreeding give a predicted phenotype ratio of 3:1.
Use a Punnett square when both parents are heterozygous and always include genotype and phenotype labels.

Drosophila

parental phenotypes wild type wing × wild type wing;
- parental genotypes Ll × Ll;
- gametes L or l from each;
- offspring LL, Ll, Ll, ll giving 3 wild type : 1 mutant wing.

Dihybrid cross

a 9:3:3:1 ratio is typical when two individuals heterozygous for both genes are crossed.
There are 4 possible phenotypes;
- out of 16 the expected number is 9, 3, 3 and 1, corresponding to percentage probabilities of 56.25 %, 18.75 %, 18.75 % and 6.25 %.

Sex-linked example

a woman is a carrier (X^FX^f).
Step 2: work out her partner Mike's genotype: as Mike does not suffer from the disease, he must have the dominant allele on his single X chromosome, X^FY.
Step 3: cross the two genotypes using a genetic diagram.

Section 14 of 22The chi-squared test

The chi-squared test is a statistical test that determines whether there is a significant difference between observed and expected results. If the difference is statistically significant this suggests the presence of a factor that is not being accounted for (for example linkage or epistasis). The null hypothesis is that there is no significant difference between observed and expected results.

Section 15 of 22Worked examples

Two heterozygous Fast Plants are crossed. State the expected phenotype ratio and explain why the observed ratio may differ. [3]3 marks
Give the expected genotypes of offspring from Ll × Ll in Drosophila and the percentage with the mutant wing. [2]2 marks
Calculate the probability that a child of X^FX^f and X^FY is an affected boy. [2]2 marks

Section 16 of 22Method and formula

The chi-squared value is calculated as χ² = Σ (O − E)² ÷ E, where O is the observed and E the expected number. Steps: state the null hypothesis; work out expected numbers from the predicted ratio; use a table to calculate (O − E)²/E for each class and add them up; find the degrees of freedom; compare the value with the critical value at p = 0.05.

Section 17 of 22Degrees of freedom and probability

The degrees of freedom take into account the number of comparisons made: degrees of freedom = number of classes − 1. Biologists work at a probability level of 0.05 (5 %), meaning there is only a 5 % probability that any difference between O and E has occurred by chance.

Result	Conclusion
χ² ≥ critical value	Significant difference; not due to chance; reject the null hypothesis
χ² < critical value	No significant difference; difference due to chance; accept the null hypothesis

Section 18 of 22Worked example with rabbits

A dihybrid cross between rabbits heterozygous for both genes gave 128 offspring in 4 phenotype classes.

Worked example with rabbits

The expected ratio is 9:3:3:1, so the expected numbers are found by dividing the total by 16: 128 ÷ 16 = 8 per part, giving 72, 24, 24 and 8.

Step 2: use a table to calculate the chi-squared value from the observed and expected numbers.

There are 4 phenotypes, so degrees of freedom = 4 − 1 = 3.

At p = 0.05 and 3 degrees of freedom the critical value is 7.82.
If the calculated value is below this, differences are due to chance;
- if it is above, they are significant and the genes may be linked or an epistatic interaction may be present.

Section 19 of 22Worked layout for the calculation

Use a table with the columnsphenotype, observed (O), expected (E), O − E, (O − E)², (O − E)²/E.

Worked layout for the calculation

Add the final column to get χ².

State the null hypothesis first (there is no significant difference between the observed and expected results),

then the conclusion in context: for example, if the calculated value is smaller than 7.82 the results fit the 9:3:3:1 ratio,
so the two genes are on different chromosomes and assort independently;
- if larger, the genes may be linked or epistasis may be involved.

Chi-squared requires frequencies (counts), a large enough sample (expected values of at least 5) and categories that do not overlap.

Section 20 of 22Worked examples

Calculate the expected numbers for a 9:3:3:1 ratio from 128 offspring. [2]2 marks
State the degrees of freedom for a test with four phenotypes. [1]1 mark
A chi-squared value is 5.2 and the critical value (3 d.f., p = 0.05) is 7.82. State the conclusion. [2]2 marks

Section 21 of 22Key terms

Term	Meaning
Locus	Fixed position of a gene on a chromosome.
Homozygous	Having two identical alleles of a gene.
Heterozygous	Having two different alleles of a gene.
Codominance	Both alleles are expressed in the phenotype of the heterozygote.
Sex-linked	A gene located on a sex chromosome (usually X).
Carrier	Heterozygous individual with a recessive allele that does not show the condition.

Term	Meaning
Dihybrid cross	A cross following the inheritance of two genes.
Linkage	Genes on the same chromosome inherited together.
Epistasis	One gene's alleles affect the expression of another gene.
Test cross	Cross with a homozygous recessive to find an unknown genotype.
F1	First filial generation: offspring of the parental cross.

Term	Meaning
Punnett square	Diagram predicting offspring genotypes from parental gametes.
Probability	The chance of an outcome; the basis of predicted ratios.
Null hypothesis	Statement of no significant difference between observed and expected results.
F2	Second generation produced by crossing F1 individuals.
Random fertilisation	Any gamete has an equal chance of fusing with any other.

Term	Meaning
Chi-squared test	Test for significant difference between observed and expected frequencies.
Degrees of freedom	Number of classes minus 1.
Critical value	Value of chi-squared at p = 0.05 above which a difference is significant.
Significant difference	A difference unlikely to be due to chance (probability below 5 %).
Null hypothesis	There is no significant difference between observed and expected.

Section 22 of 22Exam technique

1

Confusing gene and allele.

2

Writing the carrier female as XhXh.

3

Saying the phenotype is determined only by the genotype.

4

Treating codominance as blending.

1

Skipping gametes before the Punnett square.

2

Using a 9:3:3:1 ratio for linked genes.

3

Forgetting whether the question asks about males, females or all children.

4

Confusing epistasis with dominance (which concerns alleles of the same gene).

1

Believing a 3:1 ratio will always be observed exactly.

2

Not labelling genotypes and phenotypes in the Punnett square.

3

Using a Punnett square unnecessarily for two homozygous parents.

4

Giving the dihybrid ratio as 9:3:3:1 for any cross.

1

Using percentages instead of actual numbers in the test.

2

Using the number of classes rather than classes − 1 as degrees of freedom.

3

Saying the result 'proves' the hypothesis.

4

Mixing up which side of the critical value is significant.

Section 1 of 22Key terms in genetics

Section 2 of 22Codominance and multiple alleles

Section 3 of 22Monohybrid inheritance

Section 4 of 22Sex linkage and haemophilia

Section 5 of 22Worked examples

Define the terms locus and heterozygous. [2]2 marks

Give all the genotypes that give blood group A, and explain why a person of genotype IAIB has group AB. [3]3 marks

Explain why haemophilia is more common in males. [3]3 marks

Section 6 of 22Sex-linked cross for haemophilia

Section 7 of 22Dihybrid crosses

Section 8 of 22Linkage

Section 9 of 22Epistasis: a pigeon example

Section 10 of 22Test crosses

Section 11 of 22Worked examples

A carrier woman and a man with normal blood clotting have children. Give the probability that a son has haemophilia. [3]3 marks

Explain why a test cross can be used to determine the genotype of an individual with the dominant phenotype. [3]3 marks

Explain why the cross (TG)(tg) × (tg)(tg) does not give a 1:1:1:1 ratio. [2]2 marks

Explain what epistasis is, using the pigeon example. [2]2 marks

Section 12 of 22Predictions are probabilities

Section 13 of 22Worked examples

Section 14 of 22The chi-squared test

Section 15 of 22Worked examples

Two heterozygous Fast Plants are crossed. State the expected phenotype ratio and explain why the observed ratio may differ. [3]3 marks

Give the expected genotypes of offspring from Ll × Ll in Drosophila and the percentage with the mutant wing. [2]2 marks

Calculate the probability that a child of XFXf and XFY is an affected boy. [2]2 marks

Section 16 of 22Method and formula

Section 17 of 22Degrees of freedom and probability

Section 18 of 22Worked example with rabbits

Section 19 of 22Worked layout for the calculation

Section 20 of 22Worked examples

Calculate the expected numbers for a 9:3:3:1 ratio from 128 offspring. [2]2 marks

State the degrees of freedom for a test with four phenotypes. [1]1 mark

A chi-squared value is 5.2 and the critical value (3 d.f., p = 0.05) is 7.82. State the conclusion. [2]2 marks

Section 21 of 22Key terms

Section 22 of 22Exam technique

Give all the genotypes that give blood group A, and explain why a person of genotype I^AI^B has group AB. [3]3 marks

Calculate the probability that a child of X^FX^f and X^FY is an affected boy. [2]2 marks