Answer:
22 mu
Explanation:
Since maximum number of flies are observed with +pb and s++ phenotype, they are the parental combinations.
Minimum number of flies are observed with +p+ and s+b phenotype hence they are the result of double crossover.
Gene order would be +bp and s++ since it is the only case which would lead to production of above mentioned double crossover. Hence gene b is in the middle of genes s and p.
Single cross over between genes s and b will give progeny +++ and sbp.
Map distance between s and b loci = recombination frequency =
(number of recombinants/ total progeny)*100
= [(single cross over between s and b + double crossover)/total progeny]*100
= [(102+106+7+5/1000]*100
=(220/1000)*100
=0.22*100
=22 mu
Answer:
I think it would be they will eventually reach their carrying capacity.
If a man and his wife are both heterozygous Type A, what is the probability of blood types that their offspring would have hope this helps
The answer is glucose.
Hope this helps.
