Question

questions 9 & 10 refer to the following situation: two siamese and three persian cats survive a shipwreck and are carried on driftwood to a previously uninhabited tropical island. all five cats have normal ears, but one carries the recessive allele f for folded ears (his genotype is ff). 9. calculate the frequencies of alleles f and f in the cat population of this island. 10. if you assume hardy - weinberg equilibrium for these alleles (admittedly very improbable), about how many cats would you expect to have folded ears when the island population reaches 20,000? 11. in a certain african population, 4 % of the population is born with sickle cell anemia (aa). individuals that are heterozygous have the advantage of increased resistance to malaria. calculate the percentage of individuals who are heterozygous. 12. in the united states, approximately one child in 10,000 is born with pku (phenylketonuria), a syndrome that affects individuals homozygous for the recessive allele (aa). (a) calculate the frequency of this allele in the population. (b) calculate the frequency of the normal allele. (c) calculate the percentage of carriers of the trait within the population.

Explanation:

9.

Step1: Calculate total alleles

There are 5 cats, and since each cat has 2 alleles for the ear - shape gene, the total number of alleles in the population is $5\times2 = 10$.

Step2: Calculate frequency of $f$

One cat has the genotype $Ff$, so there is 1 copy of the $f$ allele. The frequency of the $f$ allele $q=\frac{1}{10}=0.1$.

Step3: Calculate frequency of $F$

The frequency of the $F$ allele $p = 1 - q$. Since $q = 0.1$, then $p=1 - 0.1=0.9$.

Step1: Recall Hardy - Weinberg equation

The frequency of the homozygous recessive genotype ($ff$) is $q^{2}$ in Hardy - Weinberg equilibrium. We found $q = 0.1$ in the previous question, so $q^{2}=(0.1)^{2}=0.01$.

Step2: Calculate expected number of cats with folded ears

If the population size $N = 20000$, the expected number of cats with folded ears (genotype $ff$) is $N\times q^{2}=20000\times0.01 = 200$.

Step1: Find $q$

If the frequency of the homozygous recessive genotype ($aa$) is $q^{2}=0.04$, then $q=\sqrt{0.04}=0.2$.

Step2: Find $p$

Since $p + q=1$, then $p = 1 - q=1 - 0.2 = 0.8$.

Step3: Calculate frequency of heterozygotes

The frequency of heterozygotes is $2pq$. Substituting $p = 0.8$ and $q = 0.2$, we get $2\times0.8\times0.2=0.32$ or 32%.

Answer:

The frequency of allele $F$ is $0.9$ and the frequency of allele $f$ is $0.1$.

10.