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mafiozo [28]
3 years ago
9

If your

Biology
1 answer:
tensa zangetsu [6.8K]3 years ago
7 0
Your answer is a. Tyr
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Assume a population of purple and white flower pea plants (2000 total plants) is in Hardy-Weinberg equilibrium. The frequency of
lys-0071 [83]

Answer:

a) 0.48

b)

p= 0.4\\q=0.6

c) 720

Explanation:

Given ,

Frequency of white flower plant (p^2) is 0.16

Thus, frequency of dominant white allele (p) is equal to \sqrt{0.16} \\= 0.4

As we that, as per  Hardy-Weinberg equilibrium equation

p+q=1

Substituting the value of p in above equation, we get

0.4+q=1\\q=1-0.4\\q=0.6

a) As per Hardy-Weinberg equilibrium equation

p^2+q^2+2pq=1\\

Substituting the given and calculated values in above equation, we get -

0.4^2+0.6^2+2pq=1\\0.16+0.36+2pq=1\\2pq=1-0.16-0.36\\2pq=0.48

b) The frequency (decimal form) of the dominant allele

p= 0.4\\q=0.6

c) number of homozygous purple flower plants

= frequency of purple flower plants (q^2) * Number of flowers

= q^2 * 2000\\= (0.6^2)*2000\\= 0.36 * 2000\\= 720

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