2. Problem: In a certain reptile, eyes can be either black or yellow. Two black
ID: 189446 • Letter: 2
Question
2. Problem: In a certain reptile, eyes can be either black or yellow. Two black eyed lizards are crossed, and the result is 72 black eyed lizards, and 28 yelloweyed lizards Your Tentative Hypothesis: The black eyed parents were Bb x Bb. Objective: Test your hypothesis using chi square analysis. In this set, because only two values (traits) are examined, the degrees of freedom (df) is 1. SHOW ALL WORK! 3. Problem: A sample of mice (all from the same parents) shows 58 Black hair, black eyes 1 16 Black hair, red eyes I 19 White hair, black eyes 7 I White hair, red eyes Your tentative hypothesis: (what are the parents?) bjective: Use a chi square analysis to support your hypothesisExplanation / Answer
2) Expected phenotypic ratio is 3:1 (Black eyes: Yellow eyes). To find expected values, multiply the total number of outcomes by the probability of each outcome.
Observed
Expected
(o-e)2
e
Black Eyes
72
(100)(3/4)=75
.12
Yellow eyes
28
(100)(1/4)=25
.12
Total
100
100
2 = (o-e)2 / e
2 = .12 + .12
2 =.24
df = 1
p>.05 (p is between .7 and .5). So, the difference between observed and expected is not significantly different. The differences are observed are small enough to explain by chance alone. We will not reject the null hypothesis that the black eyed parents are heterozygous for eye color.
3) a) Tentative hypothesis (what are the parents?):
Black hair is dominant to white hair and black eyes are dominant to red eyes. Both the parent’s phenotype is black hair and black eyes, and both are the heterozygous for each trait (BbEe x BbEe).
b) Use a chi square analysis to support your hypothesis.
Expected phenotypic ratio for dihybrid cross is 9:3:3:1 (Purple Starchy: Purple Sweet: Yellow Starchy: Yellow Sweet). To find expected values, multiply total number of outcomes by the probability of each outcome.
Observed
Expected
(o-e)2
e
Black Hair, Black Eyes
58
(100)(9/16)=56
.07
Black Hair, Red Eyes
16
(100)(3/16)=19
.47
White Hair, Black Eyes
19
(100)(3/16)=19
0
White Hair, Red Eyes
7
(100)(1/16)=6
.17
Total
100
100
.71
2 = (o-e)2 / e
2 = .07 + .47 + 0 + .17
2 =.71
df = 3
p>.05 (p is between .95 and .9). So, the difference between the observed and expected is not significantly different. The differences are observed are small enough to explain by chance alone. We will not reject the hypothesis that the both parents were heterozygous for each trait (BbEe x BbEe).
Observed
Expected
(o-e)2
e
Black Eyes
72
(100)(3/4)=75
.12
Yellow eyes
28
(100)(1/4)=25
.12
Total
100
100
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