The P_1 cross of starchy purple x sweet colorless resulted in all starchy and pu

Question

The P_1 cross of starchy purple x sweet colorless resulted in all starchy and purple F_1. The F_2 is as follows: 88 starchy and purple, 32 starchy and colorless, 29 sweet and purple, 17 sweet and colorless. How many genes and how many alleles per gene (respectively) are involved? a) 1, 4 b) 1, 2 c) 2, 2 d) 2, 4 e) 4, 2 The number of degrees of freedom is ___ a) 0 b) 1 c) 2 d) 3 e) 4 The Chi-squarer for this data is equal to 4.709. based on the probability table (see last page), the corresponding p value is ___ a) p>0.95 b) 0.7>p>0.5 c) 0.3>p>0.2 d) 0.2>p>0.1 e) 0.05>p>0.01 A different repeat of the experiment from previous question produced a very different set of F_2 counts. For this data set, the corresponding p value was at 0.01>p>0.001. Based on this p value, one must ___. a) reject the null hypotheses b) fail to reject the null hypothesis c) cannot draw any conclusion regarding the null hypothesis Null hypothesis states that ___ a) there is no significant difference between the expected and the observed result

Explanation / Answer

17. each trait is carried by separate gene and each gene is expressed by 2different alleles...so there are 4genes and 2alleles per each gene

18. degree of freedom=3 i.e.(4-1)

19.probability table not provided

20. reject the null hypothesis as p value <0.01 means the difference is statistically strongly significant and hence the null hypothesis needs to be rejected.

21.yes null hypothesis states that there is no significant difference between the expected and observed result

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