Question

dom homo = aa
dom hetero = aa
rec homo = aa
p = dom
q = rec
p + q = 1
use the following information to answer the next two questions.
the ancon sheep has unusually short legs. through progressive inbreeding of this strain of sheep, new england farmers were able to produce sheep that are too short to jump the walls that enclose pastures. the characteristic of short legs in sheep is an autosomal recessive trait.

1. establishing a population of short - legged sheep from a single sheep that has short legs illustrates the concept of

a. gene flow
b. succession
c. genetic drift
d. non - random mating

1. 36 sheep in a flock of 100 are ancon sheep

a. how many of the sheep in the flock can be expected to be heterozygous for the ancon trait?
b. how many of the sheep in the flock will not be suitable for breeding future generations of sheep with the ancon trait?

Explanation:

Question 4

Step 1: Recall Concepts

Gene flow is transfer of genes between populations. Succession is ecological change over time. Genetic drift is random change in allele frequencies (e.g., founder effect/bottleneck). Non - random mating is when mating isn't random (e.g., inbreeding, assortative mating).

Step 2: Analyze the Scenario

Establishing a population of short - legged sheep from a single sheep is a founder effect, which is a type of genetic drift. The farmers used inbreeding (non - random mating) to maintain the trait, but the initial establishment of the population from a single sheep is an example of genetic drift. Among the options, genetic drift (option c) fits.

Step 1: Identify Genotype Frequencies

Let the recessive allele be \(a\) (short legs, Ancon trait) and dominant allele be \(A\). We know that the frequency of homozygous recessive (\(aa\)) is \(q^{2}\). Given that 36 out of 100 are Ancon (aa), so \(q^{2}=\frac{36}{100} = 0.36\), then \(q=\sqrt{0.36}=0.6\), and \(p = 1 - q=1 - 0.6 = 0.4\).

Step 2: Calculate Heterozygous Frequency

The frequency of heterozygous (\(Aa\)) is \(2pq\). Substitute \(p = 0.4\) and \(q = 0.6\) into the formula: \(2\times0.4\times0.6=0.48\).

Step 3: Find the Number of Heterozygous Sheep

The total number of sheep is 100. So the number of heterozygous sheep is \(0.48\times100 = 48\).

Step 1: Understand the Trait and Breeding

The Ancon trait is recessive. If we keep breeding Ancon (aa) sheep with each other, we will mostly get aa offspring. But if we want to breed for future generations, we need genetic variation. Since the population is highly inbred (because we started from a single sheep and used inbreeding), there is little genetic variation. Also, the Ancon trait is recessive, and if we only breed Ancon sheep (aa), we are not introducing new alleles. Over time, the lack of genetic variation and the inbreeding will lead to inbreeding depression, and the flock will not be suitable for breeding future generations of sheep with the Ancon trait because of reduced fitness and lack of genetic diversity. But if we just calculate the number of heterozygous: we know that \(p = 0.4\), \(q = 0.6\), \(2pq = 0.48\), number of heterozygous is 48. But the question is about suitability for breeding. However, if we consider the number of sheep that can be heterozygous (carriers) for future breeding: the heterozygous frequency is 0.48, so in 100 sheep, 48 are heterozygous. But if we think about the long - term, due to inbreeding, the number of sheep suitable for breeding (with genetic diversity) will decrease. But based on the Hardy - Weinberg calculation for heterozygous (carriers) which are important for breeding (as they can pass on the recessive allele), the number is 48. But if we consider the question "How many of the sheep in the flock will not be suitable for breeding future generations of sheep with the Ancon trait?" The homozygous dominant (\(AA\)) frequency is \(p^{2}=0.4^{2}=0.16\), number of \(AA\) is \(0.16\times100 = 16\). The homozygous recessive (aa) are Ancon, but if we breed aa with aa, we get aa, but they may have inbreeding problems. However, the \(AA\) sheep do not carry the Ancon allele, so they are not suitable for breeding Ancon sheep. Also, if we consider that only heterozygous (Aa) and homozygous recessive (aa) are relevant for Ancon trait breeding, the homozygous dominant (AA) are not suitable. But if we use the Hardy - Weinberg: \(p^{2}=0.16\), so 16 sheep are AA (not carrying the recessive allele), and if we also consider that inbreeding will reduce fitness of aa over time, but based on the given data, the number of sheep not suitable (AA) is 16, but if we consider the heterozygous are suitable (48) and aa are 36, the non - suitable (AA) is 16. But maybe the question is about heterozygous as the ones that can be used for breeding (since aa x aa gives aa, but may have inbreeding issues, and Aa x Aa gives aa, Aa, AA). But the initial calculation for heterozygous is 48, but if we think about the ones not suitable: AA is 16, and if we consider that aa may have reduced fitness over time, but based on the formula, the number of heterozygous (suitable as they can pass on the allele) is 48.

Answer:

c. genetic drift

Question 5a