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# Please help me with this punnet square problem?

I need help with the following question: If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene?

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please list steps so I can understand this better?

### 2 Answers

- 1 decade agoFavorite Answer
I have not done genetics in a few years, and I do not have a book with me, but I think I remember the equations you are looking for:

p + q = 1

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

(p^2 means p squared.) In these equations, "p" is the dominant gene, and "q" is the recessive gene. The first equation is basically saying that if you take the percentages of the occurrence of each gene, they must add up to 1. More simply, only the two types of the gene exist in the equation. The second equation is just saying that if you add up the percentages of the organisms that have both dominant genes (p^2), plus the organisms that are heterozygous (2pq), plus the ones that have both recessive genes (q^2), they must add up to 1.

What we know from your question is that 9% of the population is born with sickle-cell anemia. This is just saying that 9% of the population is born homozygous recessive (q^2). What we need to find out is what percentage of the population is born with the heterozygous advantage (2pq). To find this, we first need to find out the occurrence of “p” and “q” in the system:

q^2 = .09

√q^2 = √.09 = .3 = q

p + q = 1

p + .3 = 1 ~ p = 1 - .3 = .7

So “q” is 0.3 and “p” is 0.7. To find out what percentage of the population has the heterozygous advantage (2pq), we just need to plug it into that value.

2pq = 2 x .3 x .7 = .42 = 42%

I am pretty sure that is what you are looking for.

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- 1 decade ago
SS Ss

Ss ss

X% Y%

Y% 9%

***more information is needed

X+2Y=91%

* I would probably say that the breakdown would have X be 9% as well (although not completely sure)

thus 2Y is probably 82

82% of the population.

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