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

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7-9

Recitation Problem Set- Probability and Chi square Ci 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 utserved perted expected Your interpretation should include both whether you reject or fail to reject your nul hypothesis as well as whether your predictions about the type of cross involved were supported by the data or not. 7. 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, 70% are purple, and 22A are white Determine whether your data fit your expected results 8. A sample of mice (all from the same parents) shows 58 Black hair, black eyes I 16 Black hair, red eyes19 White hair, black eyes the parents, and use chi square to test whethel your hypothesis is correct. white hair red eyes Propose a genotype and phenotype for 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 9. and Shrunken (D). Propose a genotype and phenotype for the parents, and use chi square to test whether your hypothesis is correct bioogycorner com

Explanation / Answer

I am answering the first full question (7).

7) In flower inheritence, purple (P) is dominant over white (p). Two heyerozygous flowers with the genotype Pp are crossed. This is represented in the following Punnett square.

Purple and white flowers are produced in the ratio of 3:1.

The chi square value is 0.367

The critical value for one degree of freedom is 3.84.

The chi square value is less than the critical value. Therefore, the null hypothesis is true.

So, the data obtained is within the expected numbers.

P p P PP Pp p Pp pp

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