Cystic fibrosis is a rare recessive genetic condition. Suppose that the allele that causes Cystic fibrosis has a frequency of 0.001 in a certain population. If the population can be assumed to be at Hardy Weinberg equilibrium approximately how many heterozygous carriers of the disease will be born for every individual born with the disease?

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Shariqahmed Shariqahmed · Answer 1 · 2020-03-24T18:40:23+01:00

Answer:

As per the Hardy - weinberg condition

[tex]P^{2} + 2PQ+ Q^{2} = 1[/tex]

Where, P , is the recurrence of prevailing allele in the populace.

Q is the recurrence of passive allele in the populace.

2PQ indicate the recurrence of heterozygous bearers people.

Presently, right now, of allele causing cystic fibrosis is 0.001.

It would be ideal if you note that cystic fibrosis is brought about by passive allele.

Thus, right now,

Recurrence of latent allele(q) is given = 0.001(given)

In this way, we can ascertain the recurrence of prevailing allele is determined utilizing the formulate

P + Q = 1

Hence, P = 1 - Q

P = 1 - 0.001 = 0.999

In this way, to figure, heterozygous people recurrence can be determined by formula

=2pq

= 2 * 0.001 * 0.999 = 0.01998

Hence, number of people will be 0.01998 * 100 = 1.998

Hence, number of individual per thousand individuals would be:

1.998 * 1000 = 1998