This question has multiple parts. Please answer all the parts to receive full cr
ID: 206176 • Letter: T
Question
This question has multiple parts. Please answer all the parts to receive full credit. Consider the following: You are looking at two genes in pea plants. One gene determines whether peas will be yellow or green, and one gene determines whether peas will be round or wrinkled. Yellow is dominant to green and round is dominant to wrinkled. There is no epistasis and genes sort independently. You record the following results in an F2 generation (offspring from a dihybrid cross): 442 yellow, round 167 yellow, wrinkled 139 green, round 52 green, wrinkled Answer the following questions: 1. What is the expected phenotypic ratio of the offspring resulting from a dihybrid cross? 2. How many total offspring resulted from your cross? 3. How many of the offspring do you expect to see at each phenotype? 4. What is your null hypothesis? 5. (worth 5 points) Calculate Chi-Square and report the value. (Show your equations as best you can or attach a document to receive partial credit) 6. What are the degrees of freedom? 7. What p-value do you use? 8. Do you accept or reject your null hypothesis based on your calculated Chi-Square value?
Explanation / Answer
1. What is the expected phenotypic ratio of the offspring resulting from a dihybrid cross?
As the genes are not linked and as there is no epistatis, it will follow the mendilian dihybird cross ratio i.e. 9:3:3:1
2. How many total offspring resulted from your cross?
442 + 167 + 139 + 52 = 800
3, 4, 5, 6 - Find the below analysis.
Null hypothesis: The observed values are not deviating from the expected values.
Test statics –
Category
yellow, round
yellow, wrinkled
green, round
green, wrinkled
Total
Observed values (O)
442
167
139
52
800
Exptected Ratio (ER)
9
3
3
1
16
Exprected Values (E)
450
150
150
50
Deviation (O-E)
-8
17
-11
2
D^2
64
289
121
4
D^2/E
0.142222
1.926667
0.806667
0.08
2.955556
X^2
2.955556
Degrees of freedom
4
-
1
3
Inference: The calculated chisquare value i.e. 2.95 is less than the table value i.e. 7.82 at 3 DF and 0.05 probability. Hence the null hypothesis is accepted.
7. the pavalue is 0.05
8. Null hypothesis accpted. Please find the explanation in inference.
Category
yellow, round
yellow, wrinkled
green, round
green, wrinkled
Total
Observed values (O)
442
167
139
52
800
Exptected Ratio (ER)
9
3
3
1
16
Exprected Values (E)
450
150
150
50
Deviation (O-E)
-8
17
-11
2
D^2
64
289
121
4
D^2/E
0.142222
1.926667
0.806667
0.08
2.955556
X^2
2.955556
Degrees of freedom
4
-
1
3
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