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4 years ago

Sort the descriptions based on whether they are related to asexual reproduction or sexual reproduction

Biology

1 answer:

Tpy6a [65]4 years ago

8 0

Asexual: creates genetically identical offspring, organism doesn't have to waste time finding a mate, requires the contribution of a single parent
Sexual: creates genetically unique offspring, organism needs time to reach adulthood to reproduce, requires the contribution of two parents

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In humans there is a dominant allele (A) for the absence of moles; while the recessive allele (a) results in the presence of mol

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Answer:

a. $(p + q)^7$ = $p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7$

b. i. 0.75

ii. 0.000061

iii. 0.012

iv. 0.17

c. 0.67

Explanation:

a. The expansion of the binomial (p + q)7 would be such that:

$(p + q)^7$ = $p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7$

b. Both couples are heterozygous:

Aa x Aa

AA Aa Aa aa

Since A is dominant over a,

probability of having mole (aa) = 1/4

probability of not having moles = 3/4

Therefore, the probability of the first child not having moles = 3/4 or 0.75

ii. Let the probability of not having mole = p and the probability of having mole = q. From the binomial expansion:

$(p + q)^7$ = $p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7$

Probability that all of the children will have moles = $p^0q^7$

since p = 3/4 and q = 1/4

$p^0q^7$ = $(3/4)^0(1/4)^7$ = 0.000061

iii. Probability that the first two children will have no moles and the last five will have moles = $21p^2q5$

= $21(3/4)^2(1/4)^5$

= 0.012

iv. Probability that 4 will have no moles and 3 will have moles out of the 7 children = $35p^4q^3$

= $35(3/4)^4(1/4)^3$

= 0.17

c. Probability that the child born without moles is a carrier of the a-allele = probability of heterozygous.

From the cross in (b), the genotypes of those born without moles are AA and 2Aa. Therefore, the probability of not having moles and be Aa is:

= 2/3 or 0.67

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A man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height. The man's father was six

Degger [83]

Answer:

1/4.

Explanation:

The autosomal dominant trait means that the trait is visible in homozygous dominant and in heterozygous condition. The X linked recessive trait will be pass down from mothers to their sons and daughters will show the trait in homozygous recessive condition only.

The man is achandroplastic dwarf ( Aa) with normal vision (XY ) is married with woman that has normal height (aa) with color blind (XhXh). They have a daughter who is dwarf ( Aa). The probability of being dwarf can be calculated by the cross Aa × aa. The offspring are Aa, Aa, aa, aa. Means 1/2 are the probability of being hetterozygous for achondroplastic . The vision probability is calculated by cross XY and XhXHh. The probability that female is heterozygous is 1/2.

So, the heterozygous probability for both trait is 1/2× 1/2 = 1/4.

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