Homework SourseCh1329 If possible can you please answer as many as possible
Ch13..29>>> If possible can you please answer as many as possible. Your help is greatly appreciated. Have a great Day!
| 29. | (Problem 29a) In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2. What will be the phenotypic ratio in the F2? | ||||||||||
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Solution
29> Answer is D: In dihybrid cross the phenotypic ratio is 9:3:3:1 (Total 16)
9/16 = Bitter & Yellow spots; 3/16 = Bitter & No spots; 3/16 = Sweet & Yellow; 1/16 = Sweet & No spots
30. In back cross, if the parent is homozygous for dominant, all the progeny are with parental characters. That is all the progeny are Bitter & yellow. So the answer is A
31. Answer is C: 1/4 bitter fruit, yellow spots; 1/4 bitter fruit, no spots; 1/4 sweet fruit, yellow spots; and 1/4 sweet fruit, no spots. Because it is similar to test cross
33. AaBbCcddEe × AabbCcDdEe ========>
Aa x Aa in this aa proportion is 1/4
Bb x bb in this bb proportion is 1/2
Cc x Cc in this cc proportion is 1/4
dd x Dd in this dd proportion is 1/2
Ee x Ee in this ee proportion is 1/4
According to Multiplicative law of Probability = 1/4 x 1/2 x 1/4 x 1/2 x 1/4 = 1/ 256
So the proportion of aabbccddee = 1/256. Answer is C
