7-9 Recitation Problem Set -Probability and Chi square Chi Square Questions. For

Question

7-9

Recitation Problem Set -Probability and Chi square Chi Square Questions. For each question: a. Define the conditions you are observing b. Write out your null hypothesis c. Write the number of expected and observed individuals for each condition d, write out your equation to calculate chi square: x. e. Determine the degrees of freedom f. Use the chi square chart to identify the p value for your experiment. Interpret your results. 1 5 (observed expected expected Your interpretation should include both whether you reject or fail to reject your null hypothesis as well as whether your predictions about the type of cross involved were supported by the data or not. In flower color inheritance, purple (P) is dominant to white (p). Consider a cross in which you cross two heterozygous flowers. In the resulting progeny, 705 are purple, and 224 are white. Determine whether your data fit your expected results 7. 8. A sample of mice (all from the same parents) shows 58 Black hair, black eyes | 16 Black hair, red eyes19 white hair, black eyes 7awhite hair|red eyes. Propose a genotype and phenotype for the parents, and use chi square to test whethel your hypothesis is correct. Examine the photo of the ear of corn and determine the type of cross and genes responsible for the coloration and texture of the corn kernels like the one show below. There are four grain phenotypes in the ear. Purple and smooth (A), Purple and Shrunken (B), Yellow and Smooth (C), Yellow and Shrunken (D). Propose a genotype and phenotype for the parents, and use chi square to test whether your hypothesis is correct 9. biologycorner com

Explanation / Answer

7a. This cross is monohybride cross. Here, only one character is dominated i.e. the colour of the flower. The result have come properly in 2:1 ratio. Purple color is dominant and write is recessive.

7b. Null hypothesis of the result:

There are no significant difference in between observed and expected results.

7c. RESULTS

7d. From the equation of Chi Square test the results have given bellow:

Purple: X2 = 705-750/750 = -0.06

White X2 = 224-225/225 = -0.004

8a. This cross is dihybride cross. There are two characters i.e. one is hair color and another is eye color. The results have come in 9:3:3:1 ratio.

8b. Null hypothesis of the result:

There are no significant difference in between observed and expected results.

8c. RESULTS

Color

Observed

Expected

Black hair + black eyes

58

55

Black hair + red eyes

16

19

White hair + black eyes

19

19

White hair + red eyes

7

7

8d.

Black hair + black eyes X2 = 58-55/55 =0.05

Black hair + red eyes X2 = 16-19/19 = 0.15

White hair + black eyes X2 = 0

White hair + red eyes X2 = 0

9a. This cross is dihybride cross. There are two characters i.e. one is color and another is structure. The results have come in 9:3:3:1 ratio.

9b. Null hypothesis of the result:

There are no significant difference in between observed and expected results.

9c. RESULT

Color+ texture

Expected

Purple + Smooth

55

Purple + Shrunken

20

Yellow + Smooth

20

Yellow + Shrunken

5

Color Observed Expected Purple 705 750 White 224 225

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