Answer:
Aa X aa
Explanation:
Genotype of the offspring is Aa Aa aa aa
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:
DNA
a. deoxyribonucleic acid
b. adenine, guanine, cytosine, thymine
c. deoxyribose
d. polynucleotide chain
e. mostly double stranded
f. carries genetic information
RNA
a. ribonucleic acid
b. adenine, guanine, cytosine, uracil
c. ribose
d. polynucleotide chain
e. mostly single stranded
f. responsible for coding, decoding, regulation, and expression of genes
Answer:
The answer is B
Explanation:
I believe I already answered this question.
