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riadik2000 [5.3K]
3 years ago
12

In a hardy-weinberg population with two alleles, a and a, that are in equilibrium, the frequency of allele a is 0.2. what is the

percentage of the population that is heterozygous for this allele
Biology
1 answer:
Allushta [10]3 years ago
8 0
The Hardy-Weinberg equation is as follows:
( {p + q})^{2}  = 1
{p}^{2}  +2pq +   {q}^{2}  = 1
Where:
(convert all % to decimals)

p= homozygous dominant
q= homozygous recessive
pq= heterozygous

While you did not specify whether the 0.2 frequency was for dominant or recessive, we can still figure out the answer.

Using the 1st equation, we can solve for the other dominant/recessive frequency:

1-0.2=0.8

Meaning that:

p= 0.8 & q=0.2

If the heterozygouz frequency is 2pq, then it becomes a simple "plug & chug" sort of approach.

2(0.8)(0.2)= 2(0.16)= 0.32

So, the heterozygous frequency would be:
0.32

Hope this helps!
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The question is incomplete as it does not have the options which are:

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Answer:

a video that shows penguin species that have different mating dances

<em><u>The reproductive isolation is the factor which allows the species to evolve and form a new species, therefore, this factor is used to describe a species.</u></em>

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<em><u>The reproductive isolation refers to a mechanism in which the species are isolated to not interbreed with the individual of other species. The reproductive isolation could be due to behavioural, psychological, physical or genetic differences. evolved as an adaptation.</u></em>

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<em><u>In the given cases, the penguin species which shows different mating dances is considered to be reproductively isolated as the different dance rituals evolved as behaviour to prevent them from mating.</u></em>

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<em><u>Thus, the selected option is correct.</u></em>

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