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A B C D F G H I J K 1 Bb Bb BB bb Bb Bb BB BB Bb bb 2 Bb BB Bb bb BB BB bb BB Bb

ID: 224480 • Letter: A

Question

A

B

C

D

F

G

H

I

J

K

1

Bb

Bb

BB

bb

Bb

Bb

BB

BB

Bb

bb

2

Bb

BB

Bb

bb

BB

BB

bb

BB

Bb

Bb

3

BB

Bb

bb

BB

BB

Bb

bb

Bb

Bb

BB

4

BB

bb

Bb

bb

BB

BB

Bb

Bb

Bb

BB

5

bb

bb

Bb

BB

BB

BB

Bb

Bb

BB

Bb

6

Bb

Bb

BB

BB

BB

Bb

bb

bb

BB

BB

7

Bb

Bb

Bb

Bb

BB

BB

BB

Bb

bb

bb

8

BB

Bb

BB

Bb

Bb

BB

bb

Bb

BB

BB

9

Bb

bb

Bb

Bb

BB

Bb

bb

bb

BB

BB

10

bb

Bb

bb

bb

BB

BB

bb

Bb

bb

Bb

A small subgroup of these animals takes off and moves across a wide river to start a new population. In the table above, choose 10 animals at random to move to this island (underline the alleles of the 10 selected). (2 points)

How many BB genotypes are there in the group?_______________

How many Bb genotypes are in the group? ___________________

How many bb genotypes are in the group? ____________________

Work out the allele frequencies for this new group of animals

Allele Frequency for B allele?_______________

Allele Frequency for b allele? ___________________

A

B

C

D

F

G

H

I

J

K

1

Bb

Bb

BB

bb

Bb

Bb

BB

BB

Bb

bb

2

Bb

BB

Bb

bb

BB

BB

bb

BB

Bb

Bb

3

BB

Bb

bb

BB

BB

Bb

bb

Bb

Bb

BB

4

BB

bb

Bb

bb

BB

BB

Bb

Bb

Bb

BB

5

bb

bb

Bb

BB

BB

BB

Bb

Bb

BB

Bb

6

Bb

Bb

BB

BB

BB

Bb

bb

bb

BB

BB

7

Bb

Bb

Bb

Bb

BB

BB

BB

Bb

bb

bb

8

BB

Bb

BB

Bb

Bb

BB

bb

Bb

BB

BB

9

Bb

bb

Bb

Bb

BB

Bb

bb

bb

BB

BB

10

bb

Bb

bb

bb

BB

BB

bb

Bb

bb

Bb

Explanation / Answer

Selected 10 animals were : 1A:Bb, 2B:BB, 3C:bb, 4D: bb, 5F:BB, 6G:Bb, 7H: BB, 8I: Bb, 9J: BB, 10K: Bb.

because total 100 population, BB:Bb:BB are in 2:2:1 ratios. according to this

No.of BB genotypes in group are 4

No.of Bb genotypes in group are 4

No.of bb genotypes in group are 2

Allel frequency for B allel:According to hardy weinberg law p+q=1,(p+q)2=1

p2+2pq+q2=1

B allel frequency:[ p2+1/2(pq)]/total= 4+1/2(4)/10=6/10, 0.6

b allel frequency: [q2+1/2(pq)]/total=2+1/2(4)/10=4/10, 0.4

p(B)+q(b)=1, 0.6+0.4=1

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