**Hardy Weinberg Explainations**

Index Notes Labs Web Quests Assignments Quizzes Links Student Work

Answers

Note: When the
question tells you about the frequency, they are giving you p or q.
If they give you a percentage or a number like 270 out of 3000, they are
giving you p^{2} or q^{2 }depending on whether it is a dominant
or recessive trait.

1. p+q = 1

p + .1 = 1

p = .9 .9 is the frequency of the A allele in the population.

To determine how many out of 3600 people will be heterozygous for the trait, you must use the other formula to determine the percentage and then multiple.

p^{2} + 2pq
+q^{2} = 1

(.9)^{2} +
2(.9 X .1) + (.1)^{2} = 1

.81 + .18 + .01 = 1 – 18% of the population will be heterozygous and that number is 648 people

2. Telling you the frequency of the each allele, A or a, in the population.

3. This formula gives you the percentage of the population with those traits.

4. 500/2000 = .25 = q2 q = .5 (find the square root of .25)

p + q = 1

p + .5 = 1

p =.25

p2 + 2pq + q2 = 1

.25 + .50 + .25 = 1

50% of the population will be heterozygous or 1000 individuals

75% of the population will have setae (AA and Aa) or 1500 individuals

5. Using the same numbers, the 500 worms that were eaten is referred to as hybrid vigor because the heterozygotes advantage. The frequency would not change, it would still be .5.

6. 81% = .81 or p2 and p = .9

p + q = 1

.9 + q = 1 q = .1 The frequency of A is .9 and the frequency of a = .1

7. 270/3000 = .09% or p2 therefore p = .3

p + q = 1

p + .3 = 1 therefore, p = .7 The frequency of A is .7 and the frequency of a is .3

p2 + 2pq + q2 = 1

.49 + .42 + .09 = 1 42% of the population is heterozygous or 1260 individuals.

91% of the population will have red eyes or 2730 individuals