biology : class12 Linkage Versus Independent Assortment : Dihybrid Cross, Independent Assortment : Law of Independent Assortment MCQ Questions and Answers
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Linkage Versus Independent Assortment
MCQ questions and answers.
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MCQ Questions
1. In a testcross of a dihybrid with linked genes in coupling phase (AB/ab), the progeny are: AB = 412, ab = 398, Ab = 96, and aB = 94. What is the best estimate of the recombination frequency between the two genes?
A.
9.5%
B.
19.0%
C.
41.0%
D.
50.0%
Correct Answer:
19.0%
Explanation:
The recombinant classes are Ab and aB, totaling 190 progeny out of 1000. Therefore, recombination frequency = 190/1000 × 100 = 19.0%. The excess of parental over recombinant types indicates linkage rather than independent assortment.
2. A dihybrid in repulsion phase has the arrangement Ab/aB. If the genes are 12 map units apart, which gamete proportions are expected from this individual, assuming no complicating factors?
A.
AB = 44%, ab = 44%, Ab = 6%, aB = 6%
B.
Ab = 44%, aB = 44%, AB = 6%, ab = 6%
C.
AB = 25%, Ab = 25%, aB = 25%, ab = 25%
D.
Ab = 50%, aB = 50%, AB = 0%, ab = 0%
Correct Answer:
Ab = 44%, aB = 44%, AB = 6%, ab = 6%
Explanation:
In repulsion phase, Ab and aB are parental gametes. A recombination frequency of 12% gives 6% for each recombinant class, AB and ab, while the two parental classes each occur at 44%.
3. A testcross gives progeny in the ratio 46% AB, 44% ab, 5% Ab, and 5% aB. Which conclusion is most appropriate?
A.
The genes assort independently because all four gametes are present
B.
The genes are completely linked because parental classes are more frequent
C.
The genes are linked with an estimated map distance of 10 map units
D.
The genes are linked with an estimated map distance of 90 map units
Correct Answer:
The genes are linked with an estimated map distance of 10 map units
Explanation:
Ab and aB are recombinant classes and together constitute 10% of the progeny. Thus, recombination frequency is 10%, corresponding approximately to 10 map units. The unequal frequencies also reject independent assortment.
4. Under independent assortment, a testcross of an individual heterozygous for two genes is expected to produce four phenotypic classes in equal numbers. If the observed classes are 252, 248, 251, and 249, which interpretation is most justified?
A.
The genes show complete linkage
B.
The genes show strong linkage with 25% recombination
C.
The data are consistent with independent assortment
D.
The genes must be in repulsion phase with 50% parental gametes
Correct Answer:
The data are consistent with independent assortment
Explanation:
The four classes are approximately 1:1:1:1, the expected testcross ratio when the genes assort independently. At 50% recombination, linkage cannot be distinguished from independent assortment using this testcross pattern.
5. An individual with genotype AB/ab is testcrossed, and only AB and ab offspring are recovered in nearly equal numbers. Which statement best explains this result?
A.
The genes independently assort, but recombinant gametes are lethal in every case
B.
The genes are completely linked, so only parental gametes are effectively produced
C.
The individual is homozygous for both genes
D.
The genes are linked in repulsion phase
Correct Answer:
The genes are completely linked, so only parental gametes are effectively produced
Explanation:
For an AB/ab individual, AB and ab are parental gametes. Their exclusive recovery in a testcross indicates complete linkage or an effectively zero recombination frequency, whereas independent assortment would produce all four classes equally.
6. A testcross of an AB/ab individual produces 360 AB, 340 ab, 150 Ab, and 150 aB offspring. Which statement correctly compares this result with independent assortment?
A.
The result supports independent assortment because recombinant offspring exceed parental offspring
B.
The result indicates 30% recombination and therefore linkage
C.
The result indicates 15% recombination and therefore complete linkage
D.
The result indicates 70% recombination and therefore independent assortment
Correct Answer:
The result indicates 30% recombination and therefore linkage
Explanation:
The recombinant classes are Ab and aB, totaling 300 of 1000 offspring, giving 30% recombination. Since the parental classes are more frequent than the recombinant classes, the genes are linked; independent assortment would give 50% recombination and equal classes.
7. Two genes are 8 map units apart. In a testcross involving a heterozygote in coupling phase, which progeny distribution is theoretically closest to expectation?
A.
AB = 46%, ab = 46%, Ab = 4%, aB = 4%
B.
AB = 4%, ab = 4%, Ab = 46%, aB = 46%
C.
AB = 25%, ab = 25%, Ab = 25%, aB = 25%
D.
AB = 50%, ab = 50%, Ab = 0%, aB = 0%
Correct Answer:
AB = 46%, ab = 46%, Ab = 4%, aB = 4%
Explanation:
For AB/ab, AB and ab are parental. An 8% recombination frequency assigns 4% to each recombinant gamete, Ab and aB, leaving 92% divided equally between the parental gametes at 46% each.
8. A researcher observes 1:1:1:1 progeny in a dihybrid testcross. Which additional result would most strongly support independent assortment rather than loose linkage?
A.
A recombination frequency close to 50% in a much larger sample
B.
A slight excess of the two parental classes
C.
The two most frequent classes corresponding to parental combinations
D.
A recombination frequency of 12% in a second testcross
Correct Answer:
A recombination frequency close to 50% in a much larger sample
Explanation:
Independent assortment produces a recombination frequency of 50% and equal frequencies of all four gamete classes. A large sample with recombination near 50% provides stronger support than a small sample that happens to approximate a 1:1:1:1 ratio.
9. A heterozygote has genotype Ab/aB, and the genes are 20 map units apart. In a testcross, which pair should be the most frequent and which pair should be the least frequent?
A.
AB and ab most frequent; Ab and aB least frequent
B.
Ab and aB most frequent; AB and ab least frequent
C.
All four classes equally frequent
D.
AB and aB most frequent; Ab and ab least frequent
Correct Answer:
Ab and aB most frequent; AB and ab least frequent
Explanation:
Because the heterozygote is in repulsion phase, Ab and aB are parental and therefore most frequent. With 20% recombination, AB and ab are recombinant classes and are expected at lower frequencies.
10. In a testcross, the two parental classes together constitute 84% of the offspring, while the two recombinant classes together constitute 16%. If the heterozygote was AB/ab, what are the expected frequencies of the four classes assuming reciprocal recombination?
A.
AB = 42%, ab = 42%, Ab = 8%, aB = 8%
B.
AB = 8%, ab = 8%, Ab = 42%, aB = 42%
C.
AB = 21%, ab = 21%, Ab = 29%, aB = 29%
D.
AB = 84%, ab = 0%, Ab = 8%, aB = 8%
Correct Answer:
AB = 42%, ab = 42%, Ab = 8%, aB = 8%
Explanation:
For AB/ab, AB and ab are the parental classes. The 84% parental total is divided equally into 42% each, and the 16% recombinant total is divided equally into 8% each. This 16% recombination frequency demonstrates linkage.
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