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Use this information answer the next two questionsIMMEDIATE ANWERS NEED PLZ HLP

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Answer #1

BB..........306.....(p2).........306/447 = 0.6845

Bb.............127......(2pq)...........127/447 = 0.2841

bb..............14........(q2).............14/447 = 0.03131

Total...........447

The frequency of B allele is ......... p = BB + Bb/2 = 0.6845 + 0.2841 /2 = 0.8265 or 0.83

The frequency of b allele is .........q = bb + Bb/2 = 0.03131 + 0.2841/2 = 0.17336 or 0.174

Question 1: the frequency of B allele is ...............0.83

The correct option is ............... b

Question 2: the frequency of bb genotype iss .............0.031

The correct option is ............... b

Question 3........Hardy Weinberg equilibrium (p2 + 2pq +q2 =1)

0.36, 0.48, 0.16

p = 0.6, q = 0.4, 2pq = 2 x 0.6 x 0.4 = 0.48

The correct option is ...............c.C

Question : 4

TT = 480/1000 = 0.48, T = p = p2 + 1/2 (2pq)= 0.48 +1/2 (0.46) = 0.48+0.23 =0.71

Tt = 460/1000 = 0.46, 2pq = 2 x 0.71 x 0.29= 0.4118 or 0.412

tt = 60/1000 = 0.06, t =q= 1-p = 1-0.71 = 0.29

The correct option is ...............b.         0.412

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