Question

Consider a population of individuals each of whom possesses two genes that can be either type \(A\) or type \(a\). Suppose that in outward appearance type \(A\) is dominant and type \(a\) is recessive. (That is, an individual will have only the outward characteristics of the recessive gene if its pair is aa.) Suppose that the population has stabilized, and the percentages of individuals having respective gene pairs \(A A, a a\), and \(A a\) are \(p, q\), and \(r .\) Call an individual dominant or recessive depending on the outward characteristics it exhibits. Let \(S_{11}\) denote the probability that an offspring of two dominant parents will be recessive; and let \(S_{10}\) denote the probability that the offspring of one dominant and one recessive parent will be recessive. Compute \(S_{11}\) and \(S_{10}\) to show that \(S_{11}=S_{10}^{2} .\) (The quantities \(S_{10}\) and \(S_{11}\) are known in the genetics literature as Snyder's ratios.)

Short answer

The probabilities of having a recessive offspring from two dominant parents (probability \(S_{11}\)) and one dominant and one recessive parent (probability \(S_{10}\)) are calculated as follows: \(S_{11} = \frac{r^2}{4(p+r)^2}\) and \(S_{10} = \frac{r}{2(p+r)}\) By squaring the value of \(S_{10}\), we find that \(S_{10}^2 = \frac{r^2}{4(p+r)^2}\), which shows that \(S_{11} = S_{10}^2\).

Step-by-step solution

Write down the gene pair percentages and the dominant and recessive genetics rule

In a stabilized population, the proportions of individuals with gene pairs \(A A\), \(a a\), and \(A a\) are given as \(p, q\) and \(r\) respectively. We have the following information: dominant gene pair \(A A\), recessive gene pair \(a a\), mixed gene pair \(A a\). The rule for dominant and recessive genes is that an individual will have the outward appearance of the recessive gene if and only if its gene pair is \(aa\).

Find the probability of having a recessive offspring for specific parent combinations

To find the probabilities \(S_{11}\) and \(S_{10}\), we have to analyze the possible gene combinations of the parents in each situation. For a recessive offspring from two dominant parents, we need to consider the possible gene pair combinations for dominant parents. Dominant parents can be either \(AA\) or \(Aa\), so we can have the following: 1. Dominant parent 1 (\(A A\)) and dominant parent 2 (\(A A\)) 2. Dominant parent 1 (\(A A\)) and dominant parent 2 (\(A a\)) 3. Dominant parent 1 (\(A a\)) and dominant parent 2 (\(A A\)) 4. Dominant parent 1 (\(A a\)) and dominant parent 2 (\(A a\)) For a recessive offspring from one dominant and one recessive parent, we can have the following possible combinations: 1. Dominant parent (\(A A\)) and recessive parent (\(a a\)) 2. Dominant parent (\(A a\)) and recessive parent (\(a a\))

Calculate the probability of having a recessive offspring for each situation

Now that we have outlined the possible parent combinations for each situation, we will calculate the probability of having a recessive offspring in each case. For \(S_{11}\), the probability of having a recessive offspring from two dominant parents: 1. Dominant parent 1 (\(A A\)) and dominant parent 2 (\(A A\)): Probability is 0, since there is no recessive gene in their pairs. 2. Dominant parent 1 (\(A A\)) and dominant parent 2 (\(A a\)): Probability is 0, as all offspring will have at least one \(A\) gene. 3. Dominant parent 1 (\(A a\)) and dominant parent 2 (\(A A\)): Probability is 0, as all offspring will have at least one \(A\) gene. 4. Dominant parent 1 (\(A a\)) and dominant parent 2 (\(A a\)): Probability is 0.25, as there is a 25% chance that their offspring will be \(a a\). For calculating \(S_{11}\), we will take the weighted average considering the proportion of each combination: \(S_{11} = 0 \cdot \frac{p^2}{(p+r)^2} + 0 \cdot \frac{p \cdot r}{(p+r)^2} + 0 \cdot \frac{r \cdot p}{(p+r)^2} + 0.25 \cdot \frac{r^2}{(p+r)^2}\) \(S_{11} = \frac{r^2}{4(p+r)^2}\) For \(S_{10}\), the probability of having a recessive offspring from one dominant and one recessive parent: 1. Dominant parent (\(A A\)) and recessive parent (\(a a\)): Probability is 0% since all offspring will have at least one \(A\) gene. 2. Dominant parent (\(A a\)) and recessive parent (\(a a\)): Probability is 50% as their offspring can be \(A a\) or \(a a\). For calculating \(S_{10}\), we will take the weighted average considering the proportion of each combination: \(S_{10} = 0 \cdot \frac{p \cdot q}{(p+r) \cdot q} + 0.5 \cdot \frac{r \cdot q}{(p+r) \cdot q}\) \(S_{10} = \frac{r}{2(p+r)}\)

Show that \(S_{11} = S_{10}^2\)

Finally, to prove that \(S_{11}=S_{10}^2\), we will square the value of \(S_{10}\) and compare it with \(S_{11}\): \(S_{10}^2 = \left(\frac{r}{2(p+r)}\right)^2\) \(S_{10}^2 = \frac{r^2}{4(p+r)^2}\) \(\implies S_{11} = S_{10}^2\) Hence, we have shown that \(S_{11} = S_{10}^2\), as required.