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
IMMEDIATE ANWERS NEED PLZ HLP Use this information answer the next two questions. A population of...
QUESTION 4 Use this information to answer the next two questions. For a particular population of Drosophila, the following genotypes at the T ocus are observed: TT 480 Tt 460 tt 60 Total 1000 What is the EXPECTED number of Tt individuals if this population is in Hardy-Weinberg equilibrium? a. 206 b. 412 ос. 460 d.546 QUESTION 5 If you were to conduct a chi-square test to determine if the population in the question above were in Hardy-Weinberg equilibrium, how...
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