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Helga [31]
3 years ago
15

Sickle-cell anemia is a genetically inherited disease. Homozygous individuals (SS) have normal blood cells that are easily infec

ted with the malarial parasite. Thus, many of these individuals become ill and may die. Individuals homozygous for the sickle-cell trait (ss) have sickled red blood cells that readily collapse when deoxygenated. Although malaria cannot grow in these red blood cells, individuals often die because of the hemoglobin disorder. However, individuals with the heterozygous condition (Ss) have some sickling of red blood cells, but generally not enough to cause death. In addition, the heterozygotes tend to survive better than either of the homozygous conditions as they are resistant to malaria. If 16% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be heterozygous (Ss) for sickle-cell and therefore resistant to malaria?
Biology
1 answer:
aleksklad [387]3 years ago
6 0

Answer:

Explanation:

it depends on the total percentage of the population. that way we can pick out the percentage of the heterozygous

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Answer:    

The correct order is F, E, C, G, H, A, D, and B (look at the image in the attached files)

Explanation:

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  • <u>Metaphase:</u> The polar and the kinetochore fibers drive each individual chromosome to the equatorial plane. This stage ends when all the chromosomes are completely arranged in the medial area.  (FIGURE C)
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Frogs with deformities that typically include missing limbs, extra limbs, partial limbs, limbs that are bent and contorted, or limbs that have little muscle, are malformed frogs. Most frogs are usually malformed because of the sensitivity to the environmental factors around them while growing out of the tadpole stage

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1. If the frequency of two alleles in a gene pool is 90% A and 10% a, what is the frequency of individuals in the population wit
Illusion [34]

Answer: 0.18

Explanation:

For the alleles, the percentage distribution of each is 'A' (90% = 0.9)

While 'a' (10% = 0.1)

Hence, 0.9 and 0.1 are the respective frequencies of each allele

Now, apply Hardy-Weinberg Equilibrium equation, where heterozygotes are represented by the 2pq term.

Therefore, the number of heterozygous individuals (Aa) is equal to 2pq which equals

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