4/27/2020 Week 3 - Prac summary: Attempt review
GEN2041 - Foundations of genetics - CLAYTON MALAY
2020
Dashboard / My units / GEN2041-S1 CL MA 2020 / Week 3 / Week 3 - Prac summary
Started on Friday, 3 April 2020, 10:30 AM
State Finished
Completed on Friday, 10 April 2020, 1:42 AM
Time taken 6 days 15 hours
Marks 24.82/26.51
Grade 9.36 out of 10.00 (94%)
Information
Part 3A: Inheritance of the curly wing mutation in Drosophila.
Question 1 Are there any sex differences within the phenotypic classes?
Correct
Mark 0.50 out of Select one:
0.50
No
Yes
Cannot be determined
Your answer is correct.
The correct answer is: No
Question 2
The curly wing mutation is dominant to the wildtype allel
Correct
Mark 0.50 out of
0.50
, 4/27/2020 Week 3 - Prac summary: Attempt review
Phenotypic Observed Expected (E) (O-E) (O-E)2 (O-E)2
class (O)
straight 297 276.67 20.33 413.31 1.49
wings
curly wings 533 553.33 -20.33 413.31 0.75
χ2 value: 2.24
Degrees of freedom = 1 Associated probability = 0.20
<p< 0.10
If you predicted a 1 straight wing: 3 curly wing ratio in (b), use this table -
Computation of χ 2
Phenotypic Observed Expected (E) (O-E) (O-E)2 (O-E)
class (O)
straight wings 297 207.5 89.5 8010.25 38.60
curly wings 533 622.5 -89.5 8010.25 12.87
χ2 value: 51.47
Degrees of freedom = 1 Associated probability = p < 0.001
Question 4 Does the result of the χ2 test support your initial hypothesis? Use the template f
Complete
Explain why it does / does not.
Mark 2.50 out of
2.50 If the χ2 test does NOT support your hypothesis, suggest an alternative genetic
account better for the class data (you do not need to repeat the χ2 test)
Based on the χ2 test for cross B, the computed chi-square value of 2.24 for 1 d
gives a probability of 0.20<p<0.10 that the difference between the observed and
is due to chance alone. As this is greater than 0.05, we support the genetic hyp
GEN2041 - Foundations of genetics - CLAYTON MALAY
2020
Dashboard / My units / GEN2041-S1 CL MA 2020 / Week 3 / Week 3 - Prac summary
Started on Friday, 3 April 2020, 10:30 AM
State Finished
Completed on Friday, 10 April 2020, 1:42 AM
Time taken 6 days 15 hours
Marks 24.82/26.51
Grade 9.36 out of 10.00 (94%)
Information
Part 3A: Inheritance of the curly wing mutation in Drosophila.
Question 1 Are there any sex differences within the phenotypic classes?
Correct
Mark 0.50 out of Select one:
0.50
No
Yes
Cannot be determined
Your answer is correct.
The correct answer is: No
Question 2
The curly wing mutation is dominant to the wildtype allel
Correct
Mark 0.50 out of
0.50
, 4/27/2020 Week 3 - Prac summary: Attempt review
Phenotypic Observed Expected (E) (O-E) (O-E)2 (O-E)2
class (O)
straight 297 276.67 20.33 413.31 1.49
wings
curly wings 533 553.33 -20.33 413.31 0.75
χ2 value: 2.24
Degrees of freedom = 1 Associated probability = 0.20
<p< 0.10
If you predicted a 1 straight wing: 3 curly wing ratio in (b), use this table -
Computation of χ 2
Phenotypic Observed Expected (E) (O-E) (O-E)2 (O-E)
class (O)
straight wings 297 207.5 89.5 8010.25 38.60
curly wings 533 622.5 -89.5 8010.25 12.87
χ2 value: 51.47
Degrees of freedom = 1 Associated probability = p < 0.001
Question 4 Does the result of the χ2 test support your initial hypothesis? Use the template f
Complete
Explain why it does / does not.
Mark 2.50 out of
2.50 If the χ2 test does NOT support your hypothesis, suggest an alternative genetic
account better for the class data (you do not need to repeat the χ2 test)
Based on the χ2 test for cross B, the computed chi-square value of 2.24 for 1 d
gives a probability of 0.20<p<0.10 that the difference between the observed and
is due to chance alone. As this is greater than 0.05, we support the genetic hyp
