HELP with this genetic question. The frequency in the US of of eyecolor is as fo

ID: 253251 • Letter: H

Question

HELP with this genetic question.

The frequency in the US of of eyecolor is as follows:

Your sample 1033 people in a mall in minnesota and get the following results for eye color:

Does your sample have the same distribution of eye colors as the overall population?

1. What are the hypotheses (H0 and HA ) for the statistical test?

2. What is the value of chi statistic? How many degrees of freedom?

3. Which hypothesis can be accepted or rejected?

4. What can you conclude about the distribution of eye colors in the sample?

Phenotype Frequency Brown eyes 70% Blue eyes 15% Hazel/Green Eyes 10% Other 5% Probability n .05 .02 .01 .001 1 3.841 5.412 6.635 10.827 5,991 7.824 9.210 13.815 7,815?9.837 11345-16266 4 9.488 11.668 13.277 18.467 5 11.070 13.388 15.086 20.515 6 12.592 15.033 16.812 22.457 7 14.067 16.622 18.475 24.322 8 15.507 18.168 20.090 26.125 ????- 9 16.919 19.679 21.666 27.877 10 18.307 21.161 23.209 29.588 ilo el 11 19.675 22.618 24.725 31.264 12 21.026 24.054 26.217 32.909 13 22.3622 5.472 27.688 34.528 14 23.685 26.873 29.141 36.123 15 24.996 28.259 30.578 37.697 16 26.296 29.633 32.000 39.252 17 27.587 30.995 33.409 40.790 18 28.869 32.346 34.805 42.312 19 30.144 33.687 36.191 43.820 20 31.410 35.020 37.566 45.315

Explanation / Answer

Observed (O)

Expected (E)

(O-E)^2

?2= (O-E)^2/E

brown eyes

613

723.1

12122

16.7639469

blue eyes

302

154.95

21624

139.5527751

hazel green eyes

103.3

497.29

4.814036786

others

51.65

214.62

4.155324298

1033

?2=

165.2860831

Answer a.

The null hypothesis should be: Ho: the sample is having the same distribution of eye color as the overall population.

The alternate hypothesis should be H1: the sample is not having the same distribution of eye color as the overall population.

Answer b.

Chi-square value = 165.28

Answer c.

Null hypothesis is accepted if calculated value is less than the tabulated value.

Degree of freedom = (rows-1) (columns-1) = (4-1) (4-1) = 9

So, you should look for the tabulated value in the 9th row.

Tabulated value (at 0.05 level of significance) = 16.9

So, null hypothesis is rejected.

Answer d.

The distribution of eye color is not according to the expected values. Distribution of hazel green colors and others is more than expected. This means some new recessive alleles are showing their effects in this population.

May be this population had experienced a genetic drift from the mainland, resulting in slight loss of brown eye genotype.