Answer:
35%
Explanation:
If two genes are 30 map units apart, 30% of the produced gametes will be recombinant.
A mating between an individual homozygous dominant for both traits (AB/AB) and one homozygous recessive for both traits (ab/ab) is conducted.
The F1 will be heterozygous for both genes: AB/ab.
The F1 progeny is then test-crossed to a homozygous recessive individual:
<h3>AB/ab X ab/ab</h3>
<u>The possible offspring will be:</u>
- Parental (70%): AB/ab and ab/ab
- Recombinant (30%): Ab/ab and aB/ab
Since 30% of all the gametes produced by the F1 individual will be recombinant, 70% will be parental. As there are two types of parental gametes, each of them will have a frequency of 35%.
<u>The offspring that will have a dominant phenotype for both traits has the genotype AB/ab with a proportion of 35%.</u>
<u>Answer</u>: A) Africa and South America only
As shown in the map, the fossil evidence suggests that Cynognathus lived on the modern day continents of South America and Africa. Thus, from this distribution and the fragmentation of the ancient landmass into today's continents, result in the distribution of Cynognathus offspring species also only within the continents of Africa and South America.
Germ cell's I believe. These are cells that includes half your genome in order to reproduce.
I think the answer is B because it happens slowly and through natural selection
Answer:
A fasting blood sugar test. That measures the amount of glucose in the blood. It ranges from 75g/dl to 100g/dl.
Hyperglycemia a high blood glucose value above the normal range will suggest diabetes mellitus.
Explanation: