In an electrophoretic study a species of pine, you can distinguish heterozygotes

Question

In an electrophoretic study a species of pine, you can distinguish heterozygotes and both homozygotes for each of two genetically variable enzymes, each with two alleles (A1,A2 and B1,B2). A sample from a natural population yields the following numbers of each genotype: 8A1A1B1B1, 19 A1A2B1B1, and 10 A2A2B1B1, 42A1A1B1B2, 83A1A2B1B2, 44A2A2B1B2, 48A1A1B2B2, 97A1A2B2B2, 49A2A2B2B2.

C) Assuming linkage equilibrium calculate expected frequency of each of the nine genotypes. (Hint:the expected frequency of A1A1B1B2 is 0.106

D) From the results of part c, calculate the expected number of each genotype in the sample

E) are the loci actually in linkage equilibrium.

Explanation / Answer

There are 400 total individuals,allelic frequencies

PartA- Expected No.Of each genotype in the sample

1.Frequency A1 ,[Pa] equal to [8+ 42+ 48+ 0.50x19 + 0.50x83+ 0.50 x 97] /400 equal to-0.49375.

2.FrequencyA2 [qa] equal to {10+44+ 49+ 0.50 19+0.50x0.83+0.50 x 97] / 400 equal to-0.50625.

3.Frequency B1 [Pb] equal to {8+19 +10+ 0.50x42+ 0.50x83 + 0.50 X 44] /400 equal to-0.30375.

4.FrequencyB2 .[q b] equal to {48 + 97+ 49+ 0.50x42+ 0.50x83+ 0.50x44] / 400 equal to-0.69625.

C.Loci are in equilibrium

According to Hardy-Weinburg equilibrium ,P2; 2pq ; q2 .

Observed frequencies for Locus are , 0.245;0.4975;0.2575

>Expected ratio are 0.2438;0.499;0.2563 , this suggest that genotype frequencies of Locus.

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