According to Hardy-Weinberg’s principle the allele …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Aniket Garg 8 years, 5 months ago
- 2 answers
Naveen Sharma 8 years, 5 months ago
Let the frequency of the dominant allele be p
Let the frequency of recessive allele be q
We know that,
p + q = 1
The frequency of the dominant allele be 0.6 ( Given)
Thus,
0.6 + q = 1
=> q = 0.4
For a population in Hardy Weinberg equilibrium,
p2 + 2pq + q2 = 1
Where,
p2 = frequency of homozygous dominant
2pq = Frequency of heterozygous
q2 = frequency of homozygous recessive
So,
Frequency of heterozygous = 2pq = {tex}2\times 0.6\times 0.4 = 0.48{/tex}
Test
Related Questions
Posted by Sukhmanpreet Kaur 1 year, 8 months ago
- 2 answers
Posted by Bhawna Rohilla 1 year, 8 months ago
- 0 answers
Posted by Joyab Khan 1 year, 8 months ago
- 0 answers
Posted by Arshi Naaz 1 year, 7 months ago
- 0 answers
Posted by Srishti Semwal 1 year, 8 months ago
- 0 answers
Posted by Sabina Naaz 1 year, 8 months ago
- 1 answers
Posted by Gauri Singh 1 year, 8 months ago
- 0 answers
Posted by Arshi Naaz 5 months, 3 weeks ago
- 0 answers
Posted by Nida Shams 1 year, 8 months ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Saksham Sawarna 8 years, 5 months ago
1Thank You