A newly discovered species of snakes in the American northwest have been discove

Question

A newly discovered species of snakes in the American northwest have been discovered. A researcher at UTSA counted the different 2 varieties and observed that out of 500 snakes, 295 were black with a thin stripe (BB), 180 were black with 2 thin stripes (Bb), and 25 were all black with no stripes (bb). Report to 2 decimal places. For example, 1.2345 = 1.23 What is the chi-square value if you are testing whether this population is in Hardy-Weinberg Equilibrium? [a] Selected Answer: 0.13 Correct Answer:

Explanation / Answer

Answer:

GENOTYPE FREQUENCIES:

BB (p2) = 295/500 = 0.59

Bb (2pq) = 180/500 = 0.36

bb (q2) = 25/500 = 0.05

ALLELE FREQUENCIES:

Freq of B = p = p2 + 1/2 (2pq) = 0.59 + 1/2 (0.36) = 0.59 + 0.18 = 0.77

Freq of b = q = 1-p = 1 - 0.77 = 0.23.

EXPECTED GENOTYPE FREQUENCIES (assuming Hardy-Weinberg):

BB (p2) = (0.77)2 = 0.5929

Bb (2pq) = 2 (0.77)(0.23) = 0.3542

bb (q2) = (0.23)2 = 0.0529

EXPECTED NUMBER OF INDIVIDUALS of EACH GENOTYPE:

# BB = 0.5929 X 500 = 297

# Bb = 0.3542 X 500 = 177

# bb = 0.0529 X 500 = 26

CHI - SQUARE (X2):

X2 = (O - E)2 / E

X2 = (295-297)2 / 297 + (180-177)2 / 177 + (25-26)2 / 26

= (-2)2 / 297 + (3)2 / 177 + (-1)2 / 26

= 0.0134 + 0.0508 + 0.0384

= 0.1026 = 0.10 (reported to 2 decimal places, as required by the question)___________Answer

X2(calculated) < X2(table) [3.841, 1 df, 0.05 ls].

Therefore, conclude that there is no statistically significant difference between what you observed and what you expected under Hardy-Weinberg. That is, you fail to reject the null hypothesis and conclude that the population is in HWE.

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