A. is the answer, the closer the peaks are to eachother, the higher the frequency and energy.
Answer:
Angiosperm
Explanation:
The mitochondria of flowering Angiosperm plant are the largest. The size of these mitochondria varies from 1.86 million bp to 2.4 million bp. It is also the most complex one as it carries large number of repeating units which allows multiple permutations for recombination. This allows generation of mitochondria of varying sizes with in the same species.
It is least gene dense as the inter regions are highly expanded. One of the most distinguishing feature of these mitochondria's are that they contain polypeptides. These polypeptides are basically non functional but have potential to evolve into functional one.
Answer:
Map distance = Genetic distance, GD = 3.9 MU ≅ 4 MU
Explanation:
The more separated two genes are, the more chances of recombination there will be. The closer they are, the fewer chances of recombination there will be. Two genes that are very close will have a few recombination events and are strongly bounded.
To analyze the recombination frequency, we have to know that:
- 1% of recombination = 1 map unit = 1centi Morgan = 1,000,000 base pairs.
- And that the maximum recombination frequency is always 50%.
The map unit is the distance between the pair of genes for which every 100 meiotic products one of them results in a recombinant one.
Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to identify the genotypes of the parental gametes with the ones of the recombinants. We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the recombinants are the less frequent. So:
Parentals:
- black, curved 337
- yellow, straight 364
Recombinants:
- Black, straight 17
- yellow, curved 12
To calculate the recombination frequency we will make use of the next formula: P = Recombinant number / Total of individuals.
P = Recombinant number / Total of individuals.
P = 17 + 12 / 337 + 364 + 17 + 12
P = 29 / 730
P = 0.039
The genetic distance (GD) will result from multiplying that recombination frequency (P) by 100 and expressing it in map units (MU).
GD = P x 100
GD = 0.039 x 100
GD = 3.9 MU ≅ 4 MU
