series of A - dark ba the A locus governs the distribution of pigment in the coa
ID: 255529 • Letter: S
Question
series of A - dark ba the A locus governs the distribution of pigment in the coat. multiple alleles. These alleles control the following phenotype y (black mane and tail), a seal brown, and a - recessive e order of dominance is A+ >A> a> a -wild type A) If, in a certain horse population, A+ -0.4, A-0.2, a-0. frequencies, then calculate the expected frequencies of the four The following data were collected for two wild horse populations. - 0. 3 are the allelic B) Populations Corolla, NC Coat colors Dark black Seal brown Recessive black 75 53 Assateague Island, MD Note: Herds in Corolla the Assateague Island allows a herd size of up to 150; a monthly census is Assateague and the total population in the above data for Assateague equa report updated on April 4, 2018 are reportedly maintained between 120 and 130 individuals. In contrast, in ls that in the census e chi-square test to determine whether either of these populations is in Hardy-Weinberg both populations equilibrium with respect to the coat color phenotype. For simplicity.assur have the same allelic frequencies given in part A d.f. 1.32 2.77 4.11 5.39 2.71 4.61 6.25 7.78 3.84 5.99 7.81 9.49 5.02 7.38 9.35 11.1 7.88 10.6 12.8 14.9 10.8 13.8 16.3 18.5 6.63 2 9.21 13.3 C) Given the results of your Chi-Square testing, is there likely to be any selection acting on D) What type of selection COULD be acting on these wild horse populations found on barrier E) In what way does (or doesn't) your hypothesis align with the differences between observed either of these populations? Explain your response. islands in North Carolina and Maryland? Explain and expected phenotypes for coat coloration? Explain. : If wild-type females from the population you considered in part A (i.e. you are using BONUS: those frequencies) were captured and repeatedly mated with black males to that 800 offs are produced, how many of the 800 are expected to be black in color? How man brown? "You will not be penalized for not doing the bonus question. (6 p y would be sealExplanation / Answer
Allele freq
A+ = 0.4
A =0.2
a’ = 0.1
a =0.3
Genotypes:
(A+ + A + a' + a)2 = A+2 + A2 +a'2 +a2 + 2A+A +2A+a’ +2A+a + 2Aa’ + 2Aa + 2a’a
Genotype freq
The Expected Phenotype frequencies of
Wild-type (Bay) = 0.16+0.16+0.08+0.24 = 0.64;
Dark Bay = 0.04+0.04+0.12 = 0.2;
Seal Brown = 0.01+0.06 = 0.07; and
Reccessive Black = 0.09
A+2 0.16 A2 0.04 a'2 0.01 a2 0.09 2A+.A 0.16 2A+.a' 0.08 2A+.a 0.24 2A.a' 0.04 2A.a 0.12 2a'.a 0.06Related Questions
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