Answer:
2)
Given that,
lemurs with DD = 40
lemurs with DL = 160
lemurs with LL = 400
total number of lemurs = 600
frequency of allele D = proportion of lemurs with DD + 1/2 proportion of lemurs with DL
= 40/600 + 1/2 (160/600)
= 0.066 + 0.133
= 0.199 = 0.2 approximately
frequency of L allele = proportion of lemurs with LL + half the proportion of lemurs with DL
= 400/600 + 1/2 (160/600)
= 0.666 + 0.133
= 0.799 = 0.8 approximately
Therefore the frequency of allele D in above population is 0.2 and that of allele L is 0.8.
2) given data,
C is dominant over c
frequency of c = 0.02
and the population is in Hardy Weinberg equilibrium
from Hardy Weinberg principle,
p (frequency of allele C) + q (frequency of allele c) = 1
p + 0.02 = 1
p = 1 - 0.02
p = 0.98
again from Hardy Weinberg principle
p2(frequency of genotype CC) + 2pq (frequency of genotype Cc) + q2 (frequency of genotype cc) = 1
from this,
frequency of genotype CC, p2 = 0.98 x 0.98 = 0.96
frequency of genotype Cc , 2pq = 2 x 0.98 x 0.02 = 0.0392
frequency of genotype cc, q2 = 0.02 x 0.02 = 0.0004
Therefore frequency of allele C is 0.98, frequencies of genotypes CC, Cc and cc are 0.96, 0.0392 and 0.0004 respectively if the frequency of allele c is 0.02.
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