Consider two genes in a large, randomly mating population of turtles with no mov
ID: 94092 • Letter: C
Question
Consider two genes in a large, randomly mating population of turtles with no movement of individuals in and out of the population and no mutation. The two alleles at one gene, L and M, do not affect fitness. The two alleles at the other gene, T and t, do affect fitness -- they affect the thickness of turtle shells, and the degree to which they are protected from predation. TT individuals have thick shells, which repel predators, and survive best. Tt individuals have medium shell thickness and survive 88% as well as do TT individuals. tt individuals have thin shells and are easy for predators to eat; they survive only 14% as well as do TT individuals.
a) In a population of ADULT turtles, the frequency of individuals with the MM genotype is 0.06. (Calculate the frequency of individuals with the LM genotype. )
b) In a population of gametes at the start of a generation, the frequency of the t allele is 0.23. (Calculate the frequency of zygotes with the Tt genotype.)
c) Continuing from part (b), calculate the frequency of the adults that survive from the zygotes that have the Tt genotype.
d) Continuing from part (c), calculate the frequency of the t allele in the gametes that will produce the next generation.
Explanation / Answer
a)The possible genotypes are MM,ML,LM,LL. Let the frequency of M be p and L be q. Therefore, the given frequency of MM genotype is 0.06. Now by using hardy weinberg equation. Lets say p^2=0.06
Thus, p=0.06=0.245 and q=1-p so, q= 1-0.245=0.755
Now, from thr equation
(pA+qa)^2=p^2AA+2pqAa+q^2aa
So, 2pq= 2*0.245*0.755=0.369
Thus, the frequency of individuals from LM genotype is 0.369.
b)Similarly, let the frequency of t allele be q.
Therefore, q=0.23 and thus p=1-q=1-0.23=0.77.
Now, (pT+qt)^2=p^2TT+2pqTt+q^2tt.......from the equation as stated above.
Therefore, 2pq=2*0.77*0.23=0.354. Thys, the frequency of Tt genotype is 0.354.
c)88% of the adults from Tt genotype survive as mentioned in the question.
So, the frequency of adults that survive amongst them is,
88% of0.354=0.88*0.354=0.311.
d)The t allele is present in Tt and tt only which can produce next generation.
The frequency of tt is,
q^2= 0.77^2=0.59 and only 14% of these survive so,
0.14* 0.59=0.08.
The frequency of t allel in Tt is 0.311/2=0.15.......equal frequency for both T and t.
Thus,the frequency of t allele in the gametes that will produce next generation is 0.15+0.08=0.23.