Question

nitance of ive l eopard frogs their name. In heno erns in rast, the burnsi phenotype lacked spots on its carried out the following cross, producing the the Parent burnsi x burnsi progeny indicated. 39 burnsi, 6 pipiens are the likely genotypes of the two parents? Be sure to specify the dominant a a. What and the recessive trait. are test to evaluate the fit of the observed numbers Use a chi-squ basis of your proposed genotypes. b. of progeny to the number expected on the 5. Using the same information from question 4, evaluate the following cross: Parent phenotypes burnsi x pipiens Progeny phenotypes 23 bunsi, 33 pipiens What are the likely genotypes of the two parents? Be sure to specify the dominant and the recessive trait. c. Use a chi-square test to evaluate the fit of the observed numbers of progeny to the number expected on the basis of your proposed genotypes. d. 6. An ordinary 6-sided die is rolled repeatedly with the following outcome: 63 ones, 31 twos, 54 threes, 406 fours, 51 fives, and 55 sixes. Would you conclude from this data that it is a fair die? Give statistical support (chi- square test) for your conclusion. Moore investigated the inheritance of spott ing patterns in leopard frogs. The pipiens phenotype had the rd frogs their name. In contrast, the burnsi phenotype lacked spots on its back Moore carried out the following cross, producing the progeny indicated Parent phenotypes burnsi x burnsi Progeny phenotypes 39 burnsi, 6 pipiens What are the likely genotypes of the two parents? Be sure to specify the dominant and the recessive trait a. Use a chi-square test to evaluate the fit of the observed numbers of progeny to the number expected on the basis of your proposed genotypes. b. 5. Using the same information from question 4, evaluate the following cross: Parent phenotypes burnsi x pipiens Progeny phenotypes 23 burnsi, 33 pipiens c. What are the likely genotypes of the two parents? Be sure to specify the dominant and the recessive trait. Use a chi-square test to evaluate the fit of the observed numbers of progeny to the number expected on the basis of your proposed genotypes d. 6. An ordinary 6-sided die is rolled repeatedly with the following outcome: 63 ones, 31 twos, 54 threes, 46 fours, 51 fives, and 55 sixes. Would you conclude from this data that it is a fair die? Give statistical support (chi- square test) for your conclusion.

Answer 1

4. Let us assume that the burnsi phenotype is determined by the dominant B allele while pipiens is determined by recessive b allele.

a. The genotype of the parent is Bb and Bb

Burnsi (Bb)                            X                     Burnsi (Bb)                            P1

F1 progeny is described below:

Gametes	B	b
B	BB	Bb
b	Bb	bb

The expected phenotype ratio is Burnsi (BB/Bb) : pipiens (bb) = 3:1

The observed phenotype number is Burnsi (BB/Bb) : pipiens (bb) = 39 : 6

The expected phenotype number is Burnsi (BB/Bb) : pipiens (bb) = 45*3/4 : 45*1/4 ~ 34: 11

b. For the goodness of fit chi-square test, we need to state a null hypothesis.

The null hypothesis: there is no significant difference between the observed and the expected value. Thus the determined genotype of the parent is correct.

If the calculated value < the predicted value in table: we would accept the null hypothesis

If the calculated value > the predicted value in table: we would reject the null hypothesis

Chi square = χ² = ∑ ([(O - E)² / E ]       [∑= summation, O= observed no, E= expected no]

χ²         = (39 - 34)² / 34 + (6 - 11)² / 11

= 25/34 + 25/11

= 0.735 + 2.273

= 3.008

Degree of freedom = n – 1    [N= No of constraints] = 2-1 = 1

Critical value of χ² from chi square table = At p value 0.05 and degree of freedom 1 the Chi square value is 3.841

Accept or reject null hypothesis: Accept

The calculated value (3.008) < the predicted value (3.841) in table at p value 0.05

So, the genotypes of the parents are correct with only 5% (p value 0.05 means 5%) probability of the error.

5. Let us assume that the burnsi phenotype is determined by the dominant B allele while pipiens is determined by recessive b allele.

c. The genotype of the parent is Bb and Bb

Burnsi (Bb)                            X                     Pipiens (bb)                            P1

F1 progeny is described below:

Gametes	B	b
b	Bb	bb
b	Bb	bb

The expected phenotype ratio is Burnsi (Bb) : pipiens (bb) = 1:1

The observed phenotype number is Burnsi (BB/Bb) : pipiens (bb) = 23 : 33

The expected phenotype number is Burnsi (BB/Bb) : pipiens (bb) = 56* ½ : 56 * ½ = 28: 28

d. For the goodness of fit chi-square test, we need to state a null hypothesis.

The null hypothesis: there is no significant difference between the observed and the expected value. Thus the determined genotype of the parent is correct.

If the calculated value < the predicted value in table: we would accept the null hypothesis

If the calculated value > the predicted value in table: we would reject the null hypothesis

Chi square = χ² = ∑ ([(O - E)² / E ]       [∑= summation, O= observed no, E= expected no]

χ²         = (23 - 28)² / 28 + (33 - 28)² / 28

= 25/28 + 25/28

= 0.893 + 0.893

= 1.786

Degree of freedom = n – 1    [N= No of constraints] = 2-1 = 1

Critical value of χ² from chi square table = At p value 0.05 and degree of freedom 1 the Chi square value is 3.841

Accept or reject null hypothesis: Accept

The calculated value (1.786) < the predicted value (3.841) in table at p value 0.05

So, the genotypes of the parents are correct with only 5% (p value 0.05 means 5%) probability of the error.

6. The expected ratio of a fair dice roll is 1:1:1:1:1:1

The observed number is 63: 31: 54: 46: 51: 55

The expected number is = 300*1/6: 300*1/6: 300*1/6: 300*1/6: 300*1/6: 300*1/6 = 50: 50: 50: 50: 50: 50

For the goodness of fit chi-square test, we need to state a null hypothesis.

The null hypothesis: there is no significant difference between the observed and the expected value. Thus the dice is a fair dice.

If the calculated value < the predicted value in table: we would accept the null hypothesis

If the calculated value > the predicted value in table: we would reject the null hypothesis

Chi square = χ² = ∑ ([(O - E)² / E ]       [∑= summation, O= observed no, E= expected no]

χ²         = (63 - 50)² / 50 + (31 - 50)² / 50 + (54 - 50)² / 50 + (46 - 50)² / 50 + (51 - 50)² / 50 + (55 - 50)² / 50

= 169/50 + 361/50 + 16/50 + 16/50 + 1/50 + 25/50

= 3.38 + 7.22 + 0.32 + 0.32 + 0.02 + 0.5

= 11.76

Degree of freedom = n – 1    [N= No of constraints] = 6-1 = 5

Critical value of χ² from chi square table = At p value 0.05 and degree of freedom 1 the Chi square value is 11.070

Accept or reject null hypothesis: Reject

The calculated value (11.76) > the predicted value (11.070) in table at p value 0.05

We can reject the null hypothesis that the observed values of rolled dice are the same as the theoretical distribution of a 1:1:1:1:1:1 ratio.

So, this dice is not a fair dice.

Kindly revert for any queries and concerns.