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# Hardy-Weinberg equation more accurate for predicting the genotype ?

Which of the following conditions would tend to make the Hardy-Weinberg equation more accurate for predicting the genotype frequencies of future generations in a population of a sexually reproducing species?

Possible Answers

A. a tendency on the part of females to mate with the healthiest males

B. mutations that alter the gene pool

C. a small population size

D. frequent interbreeding with individuals from a second population with different values of p and q

E. little gene flow with surrounding populations

### 2 Answers

- N ELv 71 decade agoFavorite Answer
The Hardy-Weinberg equilibrium depends upon a set of conditions that establish that the allele frequencies don't change from generation to generation. These conditions are: no mutations, large, randomly breeding population (no preferences between mates), no gene flow into or out of the population (no individuals migrating in or out), no genetic drift. So, of the above answers, A is wrong because this would meant this is not a randomly mating population, B is wrong because there are mutations, C is wrong because a small population size increases the likelihood of genetic drift (a couple of individuals die unexpectedly and all of a sudden, the allele frequencies can change drastically), D, is wrong because this would introduce new allele frequencies into the population, thus leaving E as the correct answer-the Hardy-Weinberg equation would be more accurate for predicting the frequency of genotypes if there is little gene flow with surrounding populations.

- 5 years ago
If you have the frequency of the recessive phenotype (for example, the number of people who cannot role their tongue divided by the total number of people you're looking at), you have the frequency of homozygous recessive genotype: q^2. Then, you can take the square root of that to get q, which is the allele frequency of the recessive gene. The equation for allele frequencies is p + q = 1, so you can then rearrange the equation and use what you calculated for q to get p: p = 1 - q. Now you have p and q. p^2 is the frequency of homozygous dominant genotypes. 2pq is is the frequency of the heterozygotes, and as I said earlier, q^2 is the frequency of the homozygous recessive genotype.