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Snezhnost [94]
3 years ago
11

Why is having an extra chromosome 21 tolerable to the point that someone with this condition can survive to maturity?

Biology
1 answer:
Vilka [71]3 years ago
6 0
I think that having an extra chromosome 21 is tolerable to the point that someone with this condition can survive to maturity because it does not replicate.

There should only be a pair of chromosome 21. In the event that there are 3 chromosome 21, the person that has this extra chromosome has Down syndrome. Changes in the number and structure of this chromosome may result to delayed development, intellectual disability, but not immediate death.
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the couple is also concerned about their child inheriting hemophilia, another rare disease, which is x-linked. natalie’s brother
IgorC [24]

Hemophilia is a recessive X-linked condition and is inherited from parents. . The probability that Natalie and Daniel’s first child has hemophilia is<u> 1/4</u>, not 1/8. The statement is<u> false.</u>

<h3>What is an X-linked gene?</h3>

When talking about sex-linked genes, we refer to all those genes located in one of the sex-chromosomes and not in the somatic ones.

X-linked genes are the genes located specifically in the X sex chromosome.

Traits coded by these genes are inherited. This is, X-linked genes are transmitted from parents to children through generations.

According to this, X-linked genetic disorders will be inherited as well.

  • Men have a high probability of being affected since they only have one copy of the X chromosome.
  • In women, the effect of the mutation might be musk in heterozygous state if the mutation is recessive.

Hemophilia is a recessive X-linked disorder, which means it is coded by the recessive allele.

  • X+ ⇒ X chrosomosome with Dominant allele
  • X- ⇒ X chrosomosome with Recessive allele

<u>Genotype              Phenotype</u>

X+X+                   Healthy woman

X+X-                    Healthy carrier woman

X-X-                     Hemophilic woman

X+Y                      Healthy man

X-Y                       Hemophilic man

Let us make a pedigree of these families,

<u>Natalie's family</u>:

Mother         healthy carrier          X+X-

Father          healthy                      X+Y

Brother        hemophilic                X-Y  

Natalie         Healthy                     X+X- or X+X+  

Natalie has 1/2 chances of being carrier X+X- and 1/2 of being X+X+.

<u>Daniel's family</u>:

Mother         healthy carrier          X+X-

Father          hemophilic                X-Y  

Sister           hemophilic                X-X-

Daniel          Healthy                      X+Y

He has 1/2 chances of being healthy X+Y and 1/2 of being affected X-Y.

According to the product probability rule, if events A and B are independent from each other, the probability of occurrence of both events together is

P(A∩B) = P(A) x P(B)

In this case,

  • event A is Daniel being affected ⇒ 1/2
  • event B is Natalie being carrier ⇒ 1/2

P(A∩B) = P(A) x P(B) = 1/2 x 1/2 = 1/4

The probability that Natalie and Daniel’s first child has hemophilia is 1/4, not 1/8. The statement is<u> false.</u>

You can learn more about X-linked genes at

brainly.com/question/12420228

brainly.com/question/14704182

brainly.com/question/1387724

#SPJ1

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