Nucleic acid is the macromolecule that carries code for building our body.
The correct answer is insertion!
a G was added in the third sequence!
hope this helps!!
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
The answer would be the ligament.
