11:46 1 ? dl-mail.ymail.com 1) If ninety plants among the segregating progeny de
ID: 217431 • Letter: 1
Question
11:46 1 ? dl-mail.ymail.com 1) If ninety plants among the segregating progeny derived from a self-cross exhibit a dominan phenotype, in theory how many of those 90 plants are pure if the phenotype is controlled by onc gene? (A) 10 (B) 20, (C) 30. (D) 60 (E) None of the above 2) A cross Aa bb x Aa Bb was performed. Assuming the genes A and B are independenly assorted, what is the probability for the Fl showing the AA BB genotype? (A) 18 (B) 3/8 C)4/8 (D) 9/16 E)None of the above F1 progeny? 3) As in the question above (#2), how many different phenotypes will you expect in (A)2. B) 3 (C)4 D) 6 (E) None of the above. ) In the following cross: Aa bb Cc Dd Ec XAa BB Cc dd EE, what is the probability fo the FI to resemble the phenotype of the sccond parent (shown on the right in the cross)? Assume these five genes are independently assorted (B) 3/16 (C) 3/32 D)9/32 (E) None of the above. 5) As in the question above (84), among 96 Fl plants, in theory how many will have the aa BB CC dd EE genotype? B)2 C) 3. D) 4. E) None of the above. 6) What does 70% penetrance mean for certain phenotype? (A) 70% of any individuals show the expected phenotype (B) Only 30% of any individuals show the expected pbrotype. (C) 70% of all individuals with the genotype of interest did not show the expected phenotype. (D) 30% of all individuals with the genotype of interest did not show the expected phenotype (E) None of the aboveExplanation / Answer
Please find the answers below:
Answer 1: Choice E (Accordingly, the genotypic ratio for self-crossing plant with monohybrid genesis given by 1:2:1, i.e. 1+2/4 or 3/4th of the plants carry the parental dominant gene. Hence, if 90% of the plants are showing dominant phenotype, then 1/4 of them or nearly 1/4*90 i.e. 25 plants would be pure-breeding)
Answer 2: Choice E (None of the offsprings has genotype AABB)
Explanation: The cross is made between Aabb and AaBb. The Punnett square can be formulated as below:
Answer 3: Choice E (The total number of phenotype classes is given by 2n where n is the total value of heterozygous classes. Here n=3 for both parents and hence the total number of possible phenotypes would be 23 or 8)
Answer 4: Choice E (since the second parent contains a homozygous allelic pair for gene B whereas the first parent does not, the progeny would never bear genotype similar to the second parent)
Ab Ab ab ab AB AABb AABb AaBb AaBb Ab AAbb AAbb Aabb Aabb aB AaBb AaBb aaBb aaBb ab Aabb Aabb aabb aabbRelated Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.