Define lethal genes, epistasis, linked genes, and polygenic inheritance. Also, calculate allele frequencies for a trait controlled by dominant/recessive alleles where 4% express the recessive phenotyp

Here are the answers to question 7:

7.a. Define the following terms:

• i. Lethal genes: These are genes whose expression results in the death of the organism, often at an early stage of development, before or shortly after birth.
• ii. Epistasis: This is a phenomenon where the expression of one gene (the epistatic gene) masks or modifies the effect of another gene (the hypostatic gene) at a different locus.
• iii. Linked genes: These are genes that are located on the same chromosome and are therefore inherited together more often than genes on different chromosomes.
• iv. Polygenic inheritance: This is a type of inheritance where a single phenotypic trait is determined by the cumulative effect of two or more genes, often resulting in a continuous range of phenotypes.

7.b. A certain autosomal trait is controlled by a dominant allele A and a recessive allele a. In a large randomly mating population, 4% of individuals express the recessive phenotype.

i. Calculate the frequency of the recessive allele (a) and the dominant allele (A).
The frequency of individuals expressing the recessive phenotype is given as 4%, which corresponds to the frequency of the homozygous recessive genotype ($aa$).
Let $q$ be the frequency of the recessive allele (a) and $p$ be the frequency of the dominant allele (A).
According to the Hardy-Weinberg principle:

$$q^2 = \text{frequency of homozygous recessive genotype (aa)}$$ $$q^2 = 0.04$$

To find the frequency of the recessive allele $q$:

$$q = \sqrt{0.04}$$ $$q = 0.2$$

The sum of allele frequencies is $p + q = 1$.
To find the frequency of the dominant allele $p$:

$$p = 1 - q$$ $$p = 1 - 0.2$$ $$p = 0.8$$

The frequency of the recessive allele (a) is $\boxed{0.2}$.
The frequency of the dominant allele (A) is $\boxed{0.8}$.

ii. Calculate the expected frequencies of the three genotypes in the population.
The three genotypes are AA, Aa, and aa. Their frequencies are given by the Hardy-Weinberg equation: $p^2 + 2pq + q^2 = 1$.

• Frequency of homozygous dominant genotype (AA): $$p^2 = (0.8)^2$$ $$p^2 = 0.64$$
• Frequency of heterozygous genotype (Aa): $$2pq = 2 \times 0.8 \times 0.2$$ $$2pq = 0.32$$
• Frequency of homozygous recessive genotype (aa): $$q^2 = (0.2)^2$$ $$q^2 = 0.04$$

The expected frequency of genotype AA is $\boxed{0.64}$. The expected frequency of genotype Aa is $\boxed{0.32}$. The expected frequency of genotype aa is $\boxed{0.04}$.

iii. State four conditions necessary for this population to remain in Hardy-Weinberg equilibrium.

1. No mutation: There should be no new alleles introduced or existing alleles changed.
2. No gene flow: There should be no migration of individuals into or out of the population.
3. Random mating: Individuals must mate randomly, without any preference for specific genotypes.
4. No natural selection: All genotypes must have equal survival and reproductive rates.
5. Large population size: The population must be large enough to avoid random fluctuations in allele frequencies (genetic drift).