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3 years ago

7

Why does sexual reproduction produce offspring with characteristics that are different from their parents, where as offsprings p

roduced through asexual reproduction are identical to their parents?

1 answer:

Gnesinka [82]3 years ago

8 0

They are different because one is stronger than the other

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A fish detects vibrations in the water around it by means of its lateral lines, rows of sensory receptors along each side of the

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3 years ago

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2 years ago

Each of the four pedigrees that follow represents a human family within which a genetic disease is segregating. Affected individ

Ne4ueva [31]

Answer:

<u> The following four traits are -: </u>

<u>Pedigree 1 -</u> A recessive trait (autosomal recessive) is expressed by pedigree 1.
<u>Pedigree 2- Recessive inheritance is defined by Pedigree 2. </u>
<u>Pedigree 3</u> - The inheritance of the dominant trait (autosomal dominant) is illustrated by Pedigree 3.
<u>Pedigree 4-</u> An X-like dominant trait is expressed by Pedigree 4.

Explanation:

<u>Explaination of each pedigree chart</u>-

Pedigree 1 demonstrates the <u>recessive trait </u>since their children have been affected by two unaffected individuals. If the characteristics were X-linked, in order to have an affected daughter, I-1 would have to be affected. $X^A$ In this, both parents are autosomal recessive trait carriers, so the child will be affected by a 1/4 (aa)
<u> Recessive inheritance</u> is defined by <u>Pedigree 2</u>. This is<u> X-related inheritance as autosomal recessive</u> inheritance has already been accounted for in part 1. This inference is confirmed by evidence showing that the father (I-1) is unaffected and that only the sons exhibit the characteristic in generation II, suggesting that the mother must be the carrier. The individual I-2 is a carrier for this X-linked trait. A typical Xa chromosome is attached to the unaffected father (I-1), so the chance of carrier II-5 is 1/2. Probability of an affected son = 1/2 (probability II-5 is a carrier) x 1/2 (probability II -5 contributes ( $X^A$ ) x 1/2 (probability of Y from father II-6) = 1/8. An affected daughter's likelihood is 0 because a typical $X^A$ must be contributed by II-6.
The inheritance of the<u> dominant trait</u> is demonstrated by <u>Pedigree 3 </u>because affected children still have affected parents (remember that all four diseases are rare). The trait must be <u>autosomal dominant</u> because it is passed down to the son by the affected father. There is a 1/2 risk that the heterozygous mother (II-5) would pass on mutant alleles to a child of either sex for an autosomal dominant feature.
<u>Pedigree 4</u> is an <u>X-linked dominant function</u> characterized by the transmission to all of his daughters from the affected father but none of his son. On the mutant X chromosome, the father (I-1) passes on to all his daughters and none of his sons. As seen by his normal phenotype, II-6 therefore does not bear the mutation. An affected child's likelihood is 0.

In the question the pedigree chart was missing ,hence it is given below.

7 0

3 years ago