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77julia77 [94]
3 years ago
8

Red-green color blindness in humans is inherited as an X-linked recessive trait. Usually, heterozygous females are not colorblin

d. However, a doctor identified an unusual case in which a heterozygous female was colorblind in her right eye but had normal color vision in her left eye. If you assume that this phenotype was not due to a new mutation, explain how this might have occurred.
Biology
1 answer:
Maksim231197 [3]3 years ago
6 0

Answer:

The mentioned characteristics of variability are witnessed because of the phenomenon known as X chromosome inactivation in heterozygous females. As females possess two X chromosomes, they actually should not possess any color blindness.  

The phenomenon of X-chromosome inactivation takes place in a random manner at the time of embryo development. During this only one X chromosome is inactivated. If the chromosome comprising the defective gene is inactivated, the individual will possess normal vision.  

If the chromosome comprising the usual gene gets inactivated than the individual will be color blind. In the given case of female, X inactivation took place in one eye. The inactivation of normal alleles has taken place and thus she possesses normal vision with one eye and is color blind in the other eye.  

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3 0
3 years ago
Two humans mate who are heterozygous for earlobe shape where attached earlobes (A) are dominant to free earlobes (a), and earwax
MariettaO [177]

Answer:

3/16

Explanation:

This question involves two different genes coding for earlobe shape and earwax quality. The alleles for attached earlobes (A) and wet earwax (E) are dominant over the alleles for free earlobes (a) and dry earwax (e).

According to this question, two humans mate who are heterozygous for both genes i.e. AaEe × AeEe. Each parent will produce the following gametes: AE, Ae, aE, ae.

Using these gametes in a punnet square (see attached image), the following proportion of offsprings will be produced:

* Attached earlobe, wet earwax (A_E_) = 9/16

* Attached earlobe, dry earwax (A_ee) = 3/16

* Free earlobe, wet earwax (aaE_) = 3/16

* Free earlobe, dry earwax (aaee) = 1/16

Hence, the probability of these parents having their first child with dry earwax (ee) and attached earlobes (A_) i.e. genotype A_ee is 3/16.

7 0
3 years ago
If the normal nucleotide sequence was TACGGCATG, what type of gene mutation is present if the resulting sequence becomes TAGGCAT
koban [17]

Answer:

B. Deletion

Explanation:

the 3rd letter (C) is missing (deleted)

7 0
3 years ago
Plz answer and tell me what number
Sergeu [11.5K]
I’m confused what’s the question?
5 0
3 years ago
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