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swat32
4 years ago
11

Read each example and identify it with one of the mechanisms that influence gene pools. A zebra migrates to join a different her

d of zebras. Competition for sunlight leads to taller trees. The DNA of a snake changes to make its venom stronger. A grassfire randomly sweeps through a population of buffalo and kills most of the animals.
Biology
2 answers:
tatiyna4 years ago
6 0

Answer: The correct answer for the given mechanisms influencing gene pool are the mechanisms of evolutionary change, which are described as follows-

1) Gene Flow

- It is described as the transfer of genes from one population to another population so that frequency of gene is increased.

-A zebra migrates to join a different herd of zebras.

2) Natural Selection

- A mechanism of evolutionary change in which individuals with advantageous traits survive and reproduce whereas the others ( those that can not adapt) eliminate from the population.

-Competition for sunlight leads to taller trees.

3) Mutation

- Permanent change in the nucleotide sequence of DNA ( genetic material) that could have harmful or useful consequences.

-The DNA of a snake changes to make its venom stronger.

4) Genetic Drift - It is a random phenomenon ( that occurs by chance such as fire) that results in change in the gene frequency in a small population.    

-A grassfire randomly sweeps through a population of buffalo and kills most of the animals.

sdas [7]4 years ago
4 0
1st: Gene Flow
2nd: Natural Selection
3rd: Mutation
4th: Genetic Drift
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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

<em>Therefore, the probability of the first child not having moles </em>= 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

<em>Probability that all of the children will have moles </em>= p^0q^7

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

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

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

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

                         = 0.012

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

               = 35(3/4)^4(1/4)^3

                      = 0.17

c. <em>Probability that the child born without moles is a carrier of the a-allele  = probability of heterozygou</em>s.

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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