CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

9 percent of all cicadas exhibit the homozygous recessive condition known as "flippant wings," the gene frequency for that gene in the general population is

A
Cannot be determined
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
91 percent
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.9
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
E
0.03
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 0.3
Hardy-Weinberg equation is p2 + 2pq + q2 = 1. Here p is the frequency of the "A" allele and q is the frequency of the "a" allele in the population. Thus, p2 represents the frequency of the homozygous dominant genotype AA, q2 is the frequency of the homozygous recessive genotype aa, and 2pq represents the frequency of the heterozygous genotype Aa. In given question, frequency of homozygous recessive genotype (q2)= 9% or 0.09. Frequency of recessive allele (q)= 0.3. Correct answer is D.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
BIOLOGY
Watch in App
Join BYJU'S Learning Program
CrossIcon