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MrRissso [65]
3 years ago
13

A father possesses an X-linked disorder. What is the chance that his son will inherit this genetic abnormality from the father?

Biology
2 answers:
kirza4 [7]3 years ago
8 0

Answer:

Option B

Explanation:

A father genotype for male sexuality is X'Y

Since he possess an X - linked disorder, thus in case if it is crossed with a mother with XX genotype

Thus, X'Y * XX

X'X, X'X, XY, XY

Thus, it is clear that a male child will carry Y chromosomes from his father and X chromosome from her mother. Thus, in order to have X-linked disorder, the male child must receive an affected X allele from his mother.

Hence, the probability of son inheriting an X-linked disorder from his father would be zero.

Option B is correct

Svet_ta [14]3 years ago
7 0

X-linked disorder is a recessive autosomal disorder, which means that

There is a 1 in 4 chance (25%) of having a child who is unaffected.

There is a 1 in 4 chance (25%) of having a child who is affected by the condition. These children have two copies of the gene that do not work correctly.

There is a 1 in 2 chance (50%) of having a child who is a carrier of the condition.  These children have one working copy of the gene and one copy of the gene that does not work correctly.

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Answer:

B

Explanation:

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5 0
3 years ago
Can someone help me with this? My teacher just gives us papers to full out and doesn't teach at all.
Elden [556K]

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8 0
3 years ago
7. In some chickens, the gene for feather color is controlled by codominance. The allele for black is B and the allele for white
kondor19780726 [428]

Answer:

See the answer below

Explanation:

<em>Genotype is the genetic composition of organisms and a gene is made up of two alleles.</em>

a) Since the allele for black is B, <u>the genotype for black chickens would be BB. </u>

b) Since the allele for white is W, <u>the genotype for white chickens would be WW.</u>

c) Since erminette color is heterozygous,<u> the genotype would be BW.</u>

d) Two erminette were crossed:

         BW    x    BW

      BB BW BW WW

i. <em>Probability of having black chick (BB) </em>= 1/4

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iii. <em>Probability of having erminette chicken (BW)</em> = 1/2

5 0
3 years ago
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Murrr4er [49]

Answer:

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6 0
4 years ago
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