biology : class12 Recombinant Phenotype Classes : Dihybrid Cross, Independent Assortment : Law of Independent Assortment MCQ Questions and Answers
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Recombinant Phenotype Classes
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MCQ Questions
1. A dihybrid F1 individual AaBb is produced by crossing AABB with aabb and is then self-crossed. What fraction of the offspring will display recombinant phenotypes relative to the original parental phenotypes?
A.
1/16
B.
3/16
C.
6/16
D.
9/16
Correct Answer:
6/16
Explanation:
The parental phenotypes are A_B_ and aabb, while the recombinant phenotypes are A_bb and aaB_. Their frequencies are 3/16 and 3/16, respectively, giving a total recombinant phenotype frequency of 6/16, or 3/8.
2. In the testcross AaBb × aabb, which offspring phenotype classes represent recombinants when the AaBb parent originated from AABB × aabb?
A.
A_B_ and aabb
B.
A_bb and aaB_
C.
A_B_ and A_bb
D.
aaB_ and aabb
Correct Answer:
A_bb and aaB_
Explanation:
The original parental combinations were AB and ab, producing the phenotypes A_B_ and aabb in the testcross. The nonparental or recombinant combinations are Ab and aB, producing A_bb and aaB_.
3. In the self-cross AaBb × AaBb, where the F1 arose from AABB × aabb, which genotypes contribute to the recombinant phenotype A_bb?
A.
AABB and AaBB
B.
AABb and AaBb
C.
AAbb and Aabb
D.
aaBB and aaBb
Correct Answer:
AAbb and Aabb
Explanation:
The recombinant phenotype A_bb requires at least one dominant A allele and a homozygous recessive bb genotype. Therefore, its contributing genotypes are AAbb and Aabb.
4. A self-cross AaBb × AaBb is performed after producing AaBb from AABB × aabb. Among 1,600 offspring, how many are expected to show the recombinant phenotype aaB_?
A.
100
B.
300
C.
400
D.
600
Correct Answer:
300
Explanation:
The recombinant phenotype aaB_ has frequency 1/4 × 3/4 = 3/16 in the dihybrid self-cross. Thus, 1,600 × 3/16 = 300 offspring are expected to display this recombinant phenotype.
5. An F1 dihybrid AaBb is obtained from the cross AAbb × aaBB and is self-crossed. Which pair constitutes the recombinant phenotype classes in the F2 generation?
A.
A_B_ and aabb
B.
A_bb and aaB_
C.
A_B_ and A_bb
D.
aaB_ and aabb
Correct Answer:
A_B_ and aabb
Explanation:
The original parental phenotypes were A_bb and aaB_, corresponding to the allele combinations Ab and aB. Therefore, the nonparental recombinant phenotype classes are A_B_ and aabb.
6. For the cross described as AAbb × aaBB followed by selfing of the AaBb F1, what is the combined frequency of the two recombinant phenotype classes?
A.
3/16
B.
4/16
C.
6/16
D.
10/16
Correct Answer:
10/16
Explanation:
With the parental phase AAbb × aaBB, the recombinant phenotypes are A_B_ and aabb. Their frequencies are 9/16 and 1/16, respectively, so the combined recombinant frequency is 10/16, or 5/8.
7. In a testcross AaBb × aabb, the progeny are classified as A_B_, A_bb, aaB_, and aabb. If AaBb was formed from AABB × aabb, which observed result most strongly indicates independent assortment?
A.
Only A_B_ and aabb are produced
B.
A_B_ and aabb greatly outnumber A_bb and aaB_
C.
All four phenotype classes occur in approximately equal numbers
D.
A_bb is absent but aaB_ is present
Correct Answer:
All four phenotype classes occur in approximately equal numbers
Explanation:
Independent assortment in a dihybrid testcross produces the four gametic combinations AB, Ab, aB, and ab in equal proportions. Consequently, the four phenotype classes, including the two recombinant classes A_bb and aaB_, are expected in an approximately 1:1:1:1 ratio.
8. In the self-cross AaBb × AaBb, the two recombinant phenotype classes are A_bb and aaB_. What is the ratio of these two recombinant classes to each other?
A.
1:1
B.
1:3
C.
3:1
D.
9:1
Correct Answer:
1:1
Explanation:
The frequency of A_bb is (3/4)(1/4) = 3/16, and the frequency of aaB_ is (1/4)(3/4) = 3/16. Thus, the two recombinant phenotype classes occur in a 1:1 ratio.
9. A dihybrid self-cross yields 720 offspring. The F1 parent was produced from AABB × aabb. How many offspring are expected to display either of the recombinant phenotypes?
A.
135
B.
180
C.
270
D.
405
Correct Answer:
270
Explanation:
For an F1 derived from AABB × aabb, the recombinant phenotypes are A_bb and aaB_, each occurring at 3/16. Their combined frequency is 6/16 = 3/8. Therefore, 720 × 3/8 = 270 offspring.
10. A self-cross of AaBb produces 9/16 A_B_, 3/16 A_bb, 3/16 aaB_, and 1/16 aabb. If the original cross was AAbb × aaBB, which conclusion is correct?
A.
A_B_ and aabb are parental phenotypes, each totaling 10/16
B.
A_bb and aaB_ are recombinant phenotypes, each totaling 6/16
C.
A_B_ and aabb are recombinant phenotypes, totaling 10/16
D.
All four phenotype classes are recombinant because four classes appear
Correct Answer:
A_B_ and aabb are recombinant phenotypes, totaling 10/16
Explanation:
The parental cross contributed the phenotype combinations A_bb and aaB_. Therefore, A_B_ and aabb are the nonparental recombinant classes, with frequencies 9/16 and 1/16, respectively, totaling 10/16.
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