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masha68 [24]
3 years ago
8

One of the first Mendelian traits identified in humans was a dominant condition known as brachydactyly. This gene causes an abno

rmal shortening of the fingers or toes (or both). At the time, some researchers thought that the dominant trait would spread until 75 percent of the population would be affected (because the phenotypic ratio of dominant to recessive is 3:1).
Why was this reasoning incorrect?

A. Rare alleles tend to remain rare even when they are dominant so the dominant trait would spread until 25 percent of the population would be affected.
B. The distribution of a gene among individuals is determined by mating and environmental factors, so only a recessive trait can spread until 75 percent of the population would be affected.
C. Rare alleles tend to remain rare even when they are dominant.The distribution of a gene among individuals is determined by mating and environmental factors.
D. The frequency of an allele is determined only by migration, so it is impossible to determine the distribution of desease in human population using Mendelian crosses.
Biology
1 answer:
kvv77 [185]3 years ago
5 0

Answer:

The correct option is C. Rare alleles tend to remain rare even when they are dominant.The distribution of a gene among individuals is determined by mating and environmental factors.

Explanation:

Most people believe that a rare allele would only be recessive. But this is not correct. A rare allele can be dominant. The frequency of an allele to occur in a population will depend on the environmental factors. The alleles which code for traits that are best suitable for living in an environment will be seen in more abundance. The frequency of an allele to occur in a population also depends on the breeding trends of the population.

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Moss is an example of which type of plant?
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22. A plant geneticist is investigating the inheritance of genes for bitter taste (Su) and explosive rind (e) in watermelon (Cit
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Answer:

See the answer below

Explanation:

Using the formula for calculating Chi square (X^2):

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For the expected frequency, the phenotypes are supposed to assort in 9:3:3:1 according to Mendel ratio.

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phenotype (3) = 3/16 x 293 = 54.94

phenotype (4) = 1/16 x 293 = 18.31

The X^2 is calculated thus:

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  1                          88            164.81            \frac{(88 - 164.81)^2}{164.81} = 35.80

  2                          62             54.94            \frac{(62 - 54.94)^2}{54.94} = 0.91

  3                           62             54.94              \frac{(62 - 54.94)^2}{54.94} = 0.91

  4                            81              18.31               \frac{(81 - 18.31)^2}{18.31} = 214.64

Total X^2 = 35.80 + 0.91 + 0.91 + 214.64 = 252.26

To the nearest tenth = 252

Degree of freedom = n - 1 = 4- 1 = 3

X^2 tabulated  = 7.815

<em>The calculated </em>X^2<em> exceeds the critical value, hence, the null hypothesis is rejected. The two genes did not assort independently.        </em>

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