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Nana76 [90]
3 years ago
13

Question 5

Biology
2 answers:
timofeeve [1]3 years ago
8 0
They all share at least 50% of the same DNA arrangement. It is correct because i just took the quiz and answered like 1 minute and 26 seconds ago.
GalinKa [24]3 years ago
3 0

Answer:

The correct answer is "They all share at least 50% of the same DNA arrangement".

Explanation:

DNA analysis have allowed to establish phylogenetic trees with reliable information about the evolutionary relationship that species have. In this example Animal A and Animal B share 95% of their DNA sequences, while Animal A and Animal C share 50%. Even though Animal B and Animal C relationship is not established, it is sure to state that all animals share at least 50% of the same DNA arrangement since Animal A and Animal B are almost identical. This support the idea that life changed over time, where it is likely that all animals share a common ancestor that eventually diverged in the three animals at which Animal C is the most divergent.

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In a Hardy-Weinberg population with two alleles, A and a, that are in equilibrium, the frequency of allele a is 0.2. What is the
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In a Hardy-Weinberg population with two alleles, A and a, that are in equilibrium, the frequency of allele a is 0.2. The frequency of individuals that are heterozygous for this allele is 0.32.

<h3>Hardy Weinberg Equilibrium Principle </h3>
  • A stable, idealized population's constant frequency of alleles and genotypes is described by the Hardy-Weinberg equilibrium principle.
  • In a sizable population, random mating, or spontaneous mutations are necessary for Hardy-Weinberg equilibrium.

The Hardy-Weinberg equation dictates that the sum of the allele frequencies for each allele at the locus must equal 1, therefore p + q = 1. The Hardy-Weinberg equation is also written as p2 + 2pq + q2 = 1, where p is the population's frequency of the "A" allele and q is that of the "a" allele.

The homozygous genotype AA frequency is represented by p2, the homozygous genotype aa frequency by q2, and the heterozygous genotype aa frequency by 2pq in the equation.

Here, q = 0.2. Because of this, p = '1- q = 1 - 0.2 = 0.8.

The population of individuals who are heterozygous will now be 2pq, or 2 * 0.8 * 0.2 = 0.32.

Hence, This indicates that the population is heterozygous with a frequency of 0.32.

To learn more about Hardy Weinberg Principle refer to:

brainly.com/question/1365714

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