NEET UG 2019 · Zoology · Question 151
NEET 2019 Zoology, question 151
Question 151 of the NEET 2019 paper (code P1) is a Zoology question from Evolution. Its idea, Hardy-Weinberg Equation, was asked twice, in 2019 and 2026, across NEET 2017–2026. The NTA final answer key gives option 3.
Updated 2026-10-08
Q151 · Zoology
A gene locus has two alleles A, a. If the frequency of dominant allele A is 0.4, then what will be the frequency of homozygous dominant, heterozygous and homozygous recessive individuals in the population ?
- 10.36 (AA); 0.48 (Aa); 0.16 (aa)
- 20.16 (AA); 0.24 (Aa); 0.36 (aa)
- 30.16 (AA); 0.48 (Aa); 0.36 (aa)Official answer
- 40.16 (AA); 0.36 (Aa); 0.48 (aa)
Official answer: option 3 · NTA final answer key
Same idea
Hardy-Weinberg Equation, in other papers
NEET asked this idea twice, in 2019 and 2026. The other question below.
- 2026 Q147A population of diploid organisms is at Hardy-Weinberg equilibrium. If the frequency of allele A is 0.1, the frequency of AA is __________.Numerical · Medium
Hardy-Weinberg Equation: every question and how it was framed
The chapter
Evolution in ten years of NEET
28 questions from 2017–2026, asked in 10 of 10 papers, about 2.7 a paper. Natural Selection alone had 11.