h indlivisdual tras to behave Accept or reject hypothesis value P value 2 The fo

Question

h indlivisdual tras to behave Accept or reject hypothesis value P value 2 The following are dilntorid testcross data. Do these data approximate the ratio one would expeci for independent at to your cakculations the table, calculate the , and answer the questions 0018 Phenotype 4 11023215 15 ose 442-10.2 aas a. In interpreting this value, you have b. In this case, do you accepu/reject the hypothesis that these data approximate the dibybxid test degrees of freedom. cross ratio with independent assortment? c. What is the probability that the deviations are due to chance alone? d. Complete the following table for the data given in this problem. Note that the data in this prob- dihybrid testcross data. Now, determine whether each gene pair is behaving individu- lem are ally as you would expect in a monohybrid testcross. Each trait considered individually should be expected to approximate a 1:1 ratio as a consequence of Mendel's law of se Accept or reject hypothesis value P value Hypothesis 1Aa:1 aa 1Bb:1 bb e. In view of the x' values obtained for each trait individually, how might you account for the dihybrid testcross ratio obtained? i Cn 35

Explanation / Answer

2) a) The degree of freedom is calculated by the formula,

df = (number of rows -1) (number of columns - 1)

= (2 -1) (4 - 1)

= (1)(3)

= 3

b) The chi-square value is higher than the critical value. So, we reject the null hypothesis.

c) When null hypotheis is rejected, the probability that the deviations are due to chance alone is low.

d) The dihybrid test cross ratio obtained each has a equal chance.

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