Answer:
2% of the progeny will be double crossovers for the trihybrid test cross
Explanation:
By knowing the positions of genes, we can estimate the distances in MU between them per region.
- Genes A and B are 10 map units apart (Region I)
- Genes B and C are 20 map units apart (Region II)
- Genes A and C are 30 map units apart
----A-------10MU--------B-------------20MU-------------C---
Region I Region II
We can estimate the recombination frequencies by dividing each distance by 100.
• recombination frequency of A-B region = 10MU / 100 = 0.10
• recombination frequency of B-C region = 20MU / 100 = 0.20
Now that we know the recombination frequencies in each region, we can calculate the expected double recombinant frequency, EDRF, like this:
EDRF = recombination frequency in region I x recombination frequency in region II.
EDRF = 0.10 x 0.20 = 0.02
2% of the progeny will be double crossovers for the trihybrid test cross
Nitrogen fixation would not occur and that would mean that there would be a lack of nitrogen and it wouldn't be able to sustain life. Since nitrogen is an essential element in every amino acid, that also means in every protein. It essential to life and without it life would mostly likely not exist.<span />
According to Mendel’s laws of dominance, when a plant with dominant trait is crossed with plant with recessive trait, it results into dominant phenotype in F1 offspring. In the given question, a plant with pointed leaves (P) are dominant and plant with round leaves (p) is recessive. On crossing a pointed leaves (PP) with a plant with round leaves (pp) a heterozygous (Pp) pant with pointed leaves will be produced.
There are four total electrons to balance out the four protons
