Question 3 (8 points) Brown fur stripes (autosomal recessive) is found in 13% of

Question

Question 3 (8 points)

Brown fur stripes (autosomal recessive) is found in 13% of the squirrels in Yellowstone National Park. Black is the dominant stripe color. Answer the following questions about this population, assuming this population of squirrels is in Hardy-Weinberg equilibrium with respect to fur stripes.

What are the frequencies of the brown stripe (b) and black stripe (B) alleles?

b) What is the frequency of heterozygotes?

c) In the next generation, what is the proportion of heterozygote matings? (Hint: You will need to use the product rule.)

d) In the next generation, what proportion of squirrels will be brown stripe progeny resulting from heterozygote matings? (Hint: You will need to use the product rule.)

Explanation / Answer

a) Since brown is an autosomal recessive trait therefore, 13% squirrels with brown stripe means that this percentage of population is brown striped while the rest is all black striped.

Thus, q2 = .13

Now calculating frequency of allele b we have

f(b) = q = square root of .13 = .360

f(B) = p = 1- q = 1- .360 = .64

Thus, the frequencies of brown stripe (b) and black stripe (B) alleles are .360 and .64 respectively.

b) Calculating the expected genotypes under Hardy- Weinberg equilibrium we get,

BB = p2 = (.64)2 = .4096

Bb = 2pq = 2* .360 * .64 = .4608

bb= q2 = (.360)2 = .1296

Thus, from the above genotype calculations the frequency of heterozygotes (Bb) in population is .4608 or 46.08%.

c) Here we need to use the product rule which dictates that independent events tend to be as likely as events which are grouped. Therefore, we have:

We calculated the chance of animal to be heterozygous as 46.08%. Now it is likely that there is 46.08% chance that they have a heterozygous mate. We multiply the given percentages and get 21.23% as the answer. This means that there is a 21.23% chance that there will be heterozygous mating in the next generation.

d) From the above calculations if heterozygotes gave 25% brown stripe progeny then there will be 21.23% heterozygous crosses of all the squirrels.

Now calculating 25% of the same we get 5.3%.

Hence, 5.3% of brown stripe offspring will be heterozygous from such crosses.

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