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Paha777 [63]
3 years ago
10

In guinea pigs, smooth coat (S) is dominant over rough coat (s), and black coat (B) is dominant over white coat (b). In the cros

s SsBb × SsBb, how many of the offspring will have a smooth black coat, on average?
a. 1⁄4
b. about 9⁄16
c. 1⁄16
d. 6⁄16
e. 2⁄6
Biology
1 answer:
yaroslaw [1]3 years ago
6 0

Answer:

The correct answer is - b. about 9⁄16

Explanation:

I the given question is given:

Smooth coat (S) is dominant over rough coat (s), and black coat (B) is dominant over white coat (b)

Then cross betwen SsBb × SsBb

gametes:

SB, Sb, sB, and sb for each parent:

        SB       Sb      sB      sb

SB   SSBB  SSBb   SsBB  SsBb

Sb   SSBb  SSbb   SsBb  Ssbb

sB    SsBB  SsBb    SsBB  ssBb

sb    SsBb   Ssbb    ssBb   ssbb

every offspring with SB in the combination will express a smooth black coat that is 9 according to the Punnett out of 16 offspring.

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Consider a locus with two alleles - B and b. B is dominant, while b is recessive. There is no mutation. B has a selective advant
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Answer:

0.09

Explanation:

<h3><u>Before selection</u></h3>

Total number in population = 1000

Genotype frequencies

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Genotype frequency of bb = 250/1000 = 0.25

Allele frequencies

Allele frequency of B = BB genotype frequency + half of the Bb genotype frequency = 0.5 + (0.25/2) = 0.625

Allele frequency of b = bb genotype frequency + half of the Bb genotype frequency = 0.25 + (0.25/2) = 0.375

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We are told that after selection, the genotype frequency of bb is changed as they become 50% less fit. This means the frequency of bb individuals changes from 250 to 125 individuals (50% reduction).

Now the total number of individuals is 500 + 125 + 250 = 875.

Genotype frequencies

Genotype frequency of BB = 500/875 = 0.57

Genotype frequency of Bb = 250/875 = 0.29

Genotype frequency of bb = 125/875 = 0.14

Allele frequencies

Allele frequency of B = BB genotype frequency + half of the Bb genotype frequency = 0.57 + (0.29/2) = 0.715

Allele frequency of b = bb genotype frequency + half of the Bb genotype frequency = 0.14 + (0.29/2) = 0.285

<h3><u>Change in frequency of B after 1 generation</u></h3>

0.715 - 0.625 = 0.09

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Answer:

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(not sure what the "O" was at the end but- it's ok)

Hope this helps! :)

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