The answer is <span>Chytridomycota.</span>

<em><u>Nitrogen</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>not</u></em><em><u> </u></em><em><u>a</u></em><em><u> </u></em><em><u>product</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>photosynthesis</u></em>
Answer:
a) How many genotypes will occur in F2
--The genotypes would be- AA Aa aa BB Bb bb CC Cc cc DD Dd dd EE Ee ee (15 genotypes)
Reason-
This is because, after crossing the parents, the F1 generation would all be Heterozygous dominant since one parent is homozygous dominant, while the other is homozygous recessive.
b) 25% of all F2 genotypes would be recessive across the 5 loci
Reason-
This is because each loci would produce only a 25% chance for a recessive trait, which would results to a general probability of 25% also.
c) Yes i would change my answer in A and B if the parent cross was- AABBCCDDEE X aaBBccDDEE
d) Yes i would change my answer in A and B if the parent cross was- AABBCCDDEE X aabbccddEE
The plants that were allowed to self pollinate were the F1 plants.
The plants that are true breeding are P generation plants.
The plants where there were 3times as many tall plants as short plants are in F2 generation.
<h3><u>Explanation:</u></h3>
This question is based on the Mendel’s Experiment. Sir Gregor Johann Mendel was the father of genetics who experimented on garden pea plants <em>Pisum</em> <em>sativum</em> to see whether the characters got mixed or not and to know the real cause behind different traits of same character in plants.
He took the pure homozygous tall and short plants separately which he called as parental generation or P generation. These plants were homozygous, hence pure breeding.
As these plants were crossed between themselves, then the F1 generation showed all tall plants. This is because of the heterozygous plants which showed character of dominant trait. These plants were allowed to self pollinate.
As a result of self pollination of the F1 plants, the F2 plants were 75% tall in number whereas the other 25% short, which gave the phenotypic ratio of 3:1.