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Answer:
Red-capped Robin is the group of Robins that share a more recent common ancestor with the Norfolk Island robins.
Explanation:
Red-capped Robin is the group of Robins that share a more recent common ancestor with the Norfolk Island robins, this can be seen by the degree of genetic similarity between these two species, since the Red-capped Robin has 98.2% of genetic similarity .
When two species have the same common ancestor, these species have great genetic similarity and the more recent this ancestry is, the greater the genetic similarity between the species.
Prior knowledge
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Answer:
D. antigen only come from microbes
Explanation:
Antigens are substances which elicits a response by the antibodies. Antigens Antigens include proteins, nucleic acids, lipoproteins, glycoproteins, and certain large polysaccharides.
They parts of antigen molecules that initiate immune responses are called epitopes or antigenic determinants.
The species of <em>Anthoxanthum odoratum</em> is an example of parapatric species.
- Speciation is a process in which a new species form from the existing species due to reproductive isolation.
- In parapatric speciation there is no barrier between the gene flow between similar species.
- In the parapatric speciation populations are adapted to different ecological conditions although the populations of the related species differ in terms of mutation, and due to the effect of drift.
- But they are connected by the random flow of genes between the neighboring organisms irrespective of the changes in their genetic and physiological form.
Learn more about speciation:
brainly.com/question/9288164
The genealogy tree which is missing from the question is in the attachment.
Answer: 1) P1 = 
; 2) P2 = ; 3)P3 = 
Explanation: 1) According to the genealogy tree, the individual III-3 has a P1 = 1/2, because each of the parents are heterozygous for the condition, since one of the ofspring is affected.
2) For individual III-4, one of the parents is heterozygous for the condition, so the probability of being a carrier is P2 = 1/2.
3) Now, for individual IV-1 be affected, the parents has to be carrier. So, this probability, depends on the probability of each parent being heterozygous and that the individual will be carrying the condition, which means:
P3 = 1/2 · 1/2 · 1/4 = 1/16.
The probability of having the condition is 1/16 or 6.25%.
