The blood types of the children formed from a woman with type A blood (genotype: AO) is married to a type B person (genotype: BO) is:
<h3>How to cross parents with specific genotype?</h3>
According to this question, a woman with genotype AO is crossed with a man of genotype BO.
This means the following cross will be performed:
AO × BO
The offsprings of this cross will have the following genotype:
Learn more about genotype at: brainly.com/question/12116830
<span>The answer is sympatric
speciation. This is differentiated from allopatric speciation that involves the
divergence of a species due to the separation of parts of the population by a
geographical feature (such as a valley of the mountain). In sympatric
speciation, the divergence may be due to behavioral and reproductive isolation
within the population</span>
The correct answer to this question is B. Some members of the Fugate family had a genetic condition called methemoglobinemia, that gave a blue color to their skin. The gene causing this disease is an autosomal recessive gene, so to have this condition you need to be homozygotes for this recessive gene. Gene flow was prevalent in the Fugate family since not all of the children had blue complexion, but only four of them. The three children that did not have blue complexion, had the dominant allele, probably through gene migration from another population nearby. Also, since it is easier nowadays to travel, two separate populations could meet more easily. This increases the incidence of gene flow and explains the reduction of the occurrence of the condition.
Answer:
480 nucleotides comprise this molecule of dna
Explanation:
According to Chargaff rule,
- Adenine is paired with thymine
- Cytosine is paired with guanine
Given in the question
Adenine residue = 120
Cytosine residue = 120
So, [A] = [T]
i.e. [A =120] = [T=120]
again, [C] = [G]
i.e. [C=120] = [G=120]
hence, total number of nucleotides present
=> [A] + [T] + [C] + [G]
=> 120 + 120 +120 +120
=> 480.
