A-Level — biologyHard5 min read

Inheritance and Genetic Crosses

Monohybrid and dihybrid crosses, co-dominance, multiple alleles, sex linkage, and chi-squared test

Tutor AI Team2026-02-25T13:46:21.531Z

# Inheritance and Genetic Crosses

At A-Level, you need to perform and interpret genetic crosses involving multiple patterns of inheritance, including dihybrid crosses, co-dominance, epistasis, and sex linkage. You also need to apply the chi-squared test to evaluate whether observed results match expected ratios.

1. Monohybrid Inheritance — Review

One gene, two alleles
Standard ratios:
- Heterozygous × Heterozygous: 3:1 phenotype ratio (1:2:1 genotype)
- Heterozygous × Homozygous recessive: 1:1 phenotype ratio
- Test cross: Cross unknown genotype with homozygous recessive to determine if it's homozygous or heterozygous

2. Dihybrid Inheritance

Involves two genes on different chromosomes (unlinked genes, assort independently).

Cross: AaBb × AaBb

Gametes from each parent: AB, Ab, aB, ab

Punnett Square (4×4):

	AB	Ab	aB	ab
AB	AABB	AABb	AaBB	AaBb
Ab	AABb	AAbb	AaBb	Aabb
aB	AaBB	AaBb	aaBB	aaBb
ab	AaBb	Aabb	aaBb	aabb

Phenotype ratio: 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb

The 9:3:3:1 ratio is the hallmark of a dihybrid cross between two heterozygotes.

3. Co-dominance

Co-dominance occurs when both alleles are equally expressed in the heterozygote.

Example: Blood Group (ABO System) — Also Multiple Alleles

The ABO blood group is controlled by a single gene with three alleles: $I^A$ , $I^B$ , $I^O$

$I^A$ and $I^B$ are co-dominant with each other
$I^O$ is recessive to both $I^A$ and $I^B$

Genotype	Blood Group (Phenotype)
$I^A I^A$ or $I^A I^O$	A
$I^B I^B$ or $I^B I^O$	B
$I^A I^B$	AB (co-dominance)
$I^O I^O$	O

Cross example: Father $I^A I^O$ × Mother $I^B I^O$

	$I^A$	$I^O$
$I^B$	$I^A I^B$ (AB)	$I^B I^O$ (B)
$I^O$	$I^A I^O$ (A)	$I^O I^O$ (O)

Offspring: 1 AB : 1 A : 1 B : 1 O

4. Sex-Linked Inheritance

Genes located on the X chromosome show sex-linked inheritance patterns.

Example: Haemophilia (X-linked recessive)

Alleles: $X^H$ (normal clotting) and $X^h$ (haemophilia)

Genotype	Phenotype
$X^H X^H$	Normal female
$X^H X^h$	Carrier female (normal clotting)
$X^h X^h$	Haemophiliac female (rare)
$X^H Y$	Normal male
$X^h Y$	Haemophiliac male

Cross: Carrier female × Normal male

	$X^H$	$X^h$
$X^H$	$X^H X^H$	$X^H X^h$
$Y$	$X^H Y$	$X^h Y$

50% of sons will have haemophilia ( $X^h Y$ )
50% of daughters will be carriers ( $X^H X^h$ )
No daughters will have haemophilia

5. Autosomal Linkage

When two genes are on the same chromosome, they are linked and tend to be inherited together.

Linked genes do NOT assort independently → ratios deviate from expected 9:3:3:1
Crossing over during meiosis can separate linked genes → producing recombinant phenotypes
The further apart two genes are on a chromosome, the more likely crossing over will occur between them

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

Produces modified dihybrid ratios (e.g., 9:3:4, 9:7, 12:3:1)
Example: Coat colour in Labrador dogs — the E gene controls whether pigment is deposited; the B gene controls pigment colour (black vs brown). Dogs with genotype ee are yellow regardless of their B genotype.

7. The Chi-Squared ($\chi^2$) Test

The chi-squared test determines whether the difference between observed and expected results is due to chance or is statistically significant.

The Formula

$\chi^2 = \sum \frac{(O - E)^2}{E}$

Where:

$O$ = observed value
$E$ = expected value
$\sum$ = sum for all categories

Steps

State the null hypothesis ( $H_0$ ): There is no significant difference between observed and expected results (any difference is due to chance)
Calculate the expected values from the predicted ratio
Calculate $\chi^2$ using the formula
Calculate degrees of freedom = (number of categories $- 1$ )
Compare $\chi^2$ with the critical value at the chosen significance level (usually $p = 0.05$ )
If $\chi^2$ < critical value: Accept $H_0$ — the difference is not significant (results fit the expected ratio)
If $\chi^2$ > critical value: Reject $H_0$ — the difference IS significant (results do NOT fit the expected ratio)

Worked Example

A cross between two heterozygous pea plants (Tt × Tt) produces:

Tall: 72 (expected 75)
Short: 28 (expected 25)
Total: 100

$\chi^2 = \frac{(72-75)^2}{75} + \frac{(28-25)^2}{25} = \frac{9}{75} + \frac{9}{25} = 0.12 + 0.36 = 0.48$

Degrees of freedom = 2 − 1 = 1

Critical value at $p = 0.05$ , df = 1: 3.84

Since $0.48 < 3.84$ : Accept $H_0$ — the results are consistent with a 3:1 ratio.

Practice Questions

1. In a dihybrid cross (AaBb × AaBb), what phenotype ratio is expected? Explain why. (3 marks)
1. Parents with blood groups A (genotype $I^A I^O$ ) and B (genotype $I^B I^O$ ) have children. What blood groups are possible? (3 marks)
1. Explain why haemophilia is more common in males than females. (3 marks)
1. In a cross, 315 purple and 108 white flowers were observed. The expected ratio is 3:1. Perform a chi-squared test (critical value at p=0.05, df=1 is 3.84). (4 marks)
1. Explain what is meant by autosomal linkage and its effect on phenotype ratios. (3 marks)
Answers

Want to check your answers and get step-by-step solutions?

Summary

Dihybrid crosses with unlinked genes give a 9:3:3:1 ratio.
Co-dominance: both alleles expressed equally (e.g., AB blood group).
Sex-linked: genes on X chromosome; males more likely to show recessive conditions.
Autosomal linkage: genes on the same chromosome; deviates from expected ratios.
Chi-squared test: $\chi^2 = \sum(O-E)^2/E$ ; compare to critical value to accept or reject $H_0$ .

Tags:

#a-level#biology#genetics#inheritance#chi-squared

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