QUESTION 15 (8 points) Sickle-cell anemia is an interesting genetic disease. Nor

Question

QUESTION 15 (8 points)

Sickle-cell anemia is an interesting genetic disease. Normal homozygous individuals (SS) have normal blood cells that are easily infected with the malarial parasite. Individuals homozygous for the sickle-cell trait (ss) have red blood cells that readily collapse when deoxygenated. Although malaria cannot grow in these red blood cells, individuals often die because of the genetic defect. However, individuals with the heterozygous condition (Ss) have some sickling of red blood cells, but generally not enough to cause mortality. In addition, malaria cannot survive well within these "partially defective" red blood cells. Thus, heterozygotes tend to survive better than either of the homozygous conditions. 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?

Explanation / Answer

Hardy Weinberg equation:

It is used to measured genotype frequencies in a population.

For traits in a simple dominant – recessive system, by using hardy Weinberg equation biologists can predict the genotype frequencies.

p2 + 2pq + q2 =1

P = Dominant alleles (AA) frequency

q= Recessive alleles (aa) frequency.

2pq = Frequency of (Aa)

Hardy Weinberg equation always equals to 1, if it is not equals to 1, evolution is occur.

According to the given problem 9% =.09 = ss = q2.

To find q, simply take the square root of 0.09 to get 0.3.

Since p = 1 - 0.3, then p must equal 0.7.

2pq = 2 (0.7 x 0.3) = 0.42 = 42% of the population are heterozygotes (carriers).

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