Map Learning Suppose a man is heterozygous for heterochromia, an autosomal domin

Question

Map Learning Suppose a man is heterozygous for heterochromia, an autosomal dominant disorder which causes two different-colored eyes in an individual, produced 25 offspring with his normal-eyed wife. Of their children, different-colored eyes in an individual, produced 25 offspring with his normal-eyed wife. Of their childre, 16 were heterochromatic and 9 were normal. Calculate the Chi-square value for this observation. Incorrect. You may have only calculated the chi-square value for the heterochromatic children. To obtain the c to calculate a chi-square value for both phenotypes and sum them together. Number y have only calculated the chi-square value for the heterochromatic children. To ing heterochromia and i-square value for 16 9 being normal, you need 0.98 Identify the statement that best interprets the results of the Chi-square analysis. Refer to the Chi-Square distribution table to identify the statement that best interprets the Chi-square results 0 O The total number of normal offspring is significantly lower than expected. It is unusual that a man with heterochromia had 16 out of 25 offspring with heterochromia There is a significant difference between the observed number of offspring with heterochromia and the number of offspring that were expected to have heterochromia. It is not unusal that a heterozygous man produced 16 out of 25 offspring with heterochromia. Previous Give Up & View Solution Try Again (i) Next

Explanation / Answer

Consider

Dominant trait=H

Recessive     = h

Heterogygous dominant autosomal disorder=Hh

Normal individual = HH

Therefore a monohybrid cross between Hh X HH

Results in 50% Heterochromia

                   50% normal individual

Therefore out of the 25 individual, 12.5 should be heterochromia and 12.5 should be normal (expected)

There are two phenotypes

Therefore no. of class (n) =2

f=n-1

f= 1

² = [(o-e)²/e] = [d²/e], where d=deviation; o= observed; e=expected

for heterchromia indiviual

d=o-e =16- 12.5

d2= (3.5) 2

d2/e= 12.25/12.5

= 0.98

for normal indiviual

d=o-e =9- 12.5

d2= (-3.5) 2

d2/e= 12.25/12.5

= 0.98

Now, ² = [d²/e] = 0.98+0.98 =1.96

Now we have chi-square value (1.33) and degree of freedom (1). Now use the chart to determine the p-value.
Look in the first horizontal line (for ƒ=1), you are getting chi square as 1.32 at p value = 0.25.

Since the p value (0.25) is greater than 0.05, the deviation is not of statistical significance and implies that chance deviations are occurring.

There is a significant difference between the observed number of offspring with heterochromia and the number of offspring that were expected to have heterochromia

                   50% normal individual

Therefore out of the 25 individual, 12.5 should be heterochromia and 12.5 should be normal (expected)

There are two phenotypes

Therefore no. of class (n) =2

f=n-1

f= 1

² = [(o-e)²/e] = [d²/e], where d=deviation; o= observed; e=expected

for heterchromia indiviual

d=o-e =16- 12.5

d2= (3.5) 2

d2/e= 12.25/12.5

= 0.98

for normal indiviual

d=o-e =9- 12.5

d2= (-3.5) 2

d2/e= 12.25/12.5

= 0.98

Now, ² = [d²/e] = 0.98+0.98 =1.96

Now we have chi-square value (1.33) and degree of freedom (1). Now use the chart to determine the p-value.
Look in the first horizontal line (for ƒ=1), you are getting chi square as 1.32 at p value = 0.25.

Since the p value (0.25) is greater than 0.05, the deviation is not of statistical significance and implies that chance deviations are occurring.

There is a significant difference between the observed number of offspring with heterochromia and the number of offspring that were expected to have heterochromia

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