Genetic crosses and inheritanceEdexcel A-Level Biology B: Revision notes

A gene has different versions called alleles. The genotype is the combination of alleles an organism has and the phenotype is the characteristic that results from the genotype and the environment. An organism with two identical alleles is homozygous and one with two different alleles is heterozygous.

A dominant allele is expressed in the phenotype of a heterozygote. A recessive allele is only expressed when no dominant allele is present, so the organism is homozygous recessive. With codominance, both alleles are expressed in the heterozygote, for example $I^{A}I^{B}$ gives blood group AB. Multiple alleles means a gene has three or more alleles in a population, as in the ABO blood group gene ($I^{A}$, $I^{B}$, $I^{O}$), although any one person has only two.

To construct a genetic cross, write the parental genotypes, the gametes, the possible offspring genotypes (usually in a Punnett square), and the offspring phenotypes and ratio. A cross between two heterozygotes ($Aa \times Aa$) gives a 3:1 phenotype ratio for complete dominance, and a 1:2:1 ratio for codominance.

A pedigree diagram shows the inheritance of a characteristic through a family. If two unaffected parents have an affected child, the allele for the condition must be recessive and both parents must be heterozygous carriers.

When two genes are on different chromosomes, their alleles assort independently in meiosis I. A dihybrid cross between two double heterozygotes (RrYy $\times$ RrYy) gives four phenotypes in a 9:3:3:1 ratio. A test cross between a double heterozygote and a double homozygous recessive (RrYy $\times$ rryy) gives a 1:1:1:1 ratio.

The genes are non-interacting because each one affects a different characteristic. A heterozygote makes four types of gamete (RY, Ry, rY, ry) in equal numbers. To find the probability of a phenotype, multiply the separate probabilities, for example round and green is $\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}$.

Autosomal linkage is when two genes are on the same autosome (a non-sex chromosome), so their alleles tend to be inherited together. They do not assort independently, so a test cross gives mostly parental types rather than a 1:1:1:1 ratio.

Recombinants are offspring with new combinations of alleles. They are produced by crossing over between non-sister chromatids in prophase I, which exchanges alleles between the linked genes.

In Drosophila, grey body (G) is dominant to black body (g), and long wing (L) is dominant to vestigial wing (l). In a test cross of a female that inherited GL on one chromosome and gl on the other with a black vestigial male, most offspring are grey long or black vestigial. The few grey vestigial and black long flies are recombinants. Recombinants are fewer than parental types because crossing over between the genes happens in only some meioses.

A sex-linked gene is on the X chromosome. Females are $XX$ and males are $XY$, and the Y chromosome carries few of the same genes. Males therefore have only one allele for X-linked genes and a single recessive allele gives the condition.

Haemophilia is caused by a recessive allele $X^{h}$. Females need $X^{h}X^{h}$ to have it, but $X^{H}X^{h}$ females are carriers. A carrier mother and an unaffected father have children in the ratio: carrier daughter, unaffected daughter, unaffected son, affected son, each with probability $\frac{1}{4}$. A son always inherits his X chromosome from his mother and never from his father.

The chi-squared test tests whether the difference between observed and expected results is significant or due to chance. Write a null hypothesis (there is no significant difference between observed and expected results), calculate the expected values from the ratio, then use

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

The degrees of freedom equal the number of classes minus one. Compare $\chi^{2}$ with the critical value at $p = 0.05$. If $\chi^{2}$ is greater than the critical value, reject the null hypothesis, because the difference is significant. If it is smaller, the difference is not significant and could be due to chance.

Worked example: a dihybrid cross gives 84, 36, 28 and 12 offspring (total 160) against an expected 9:3:3:1 ratio. Expected: 90, 30, 30 and 10. $\chi^{2} = 0.40 + 1.20 + 0.13 + 0.40 = 2.13$. Degrees of freedom = 3, so the critical value is 7.82. As 2.13 is less than 7.82, the difference is not significant.