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Q. No.A typical dihybrid cross results in the formation of 500 F2 progeny. What is the expected number of recombinants among these progeny?
1. 280
2. 31
3. 94
4. 187
Subtopic: Dihybrid Cross Analysis |
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Level 2: 60%+
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In a monohybrid cross, ¼ of the random fertilisations lead to:
1. Only I and II are correct
2. Only I and III are correct
3. Only II and III are correct
4. I, II and III are correct
| I: | Homozygous dominant progeny |
| II: | Homozygous recessive progeny |
| III: | Heterozygous progeny |
1. Only I and II are correct
2. Only I and III are correct
3. Only II and III are correct
4. I, II and III are correct
Subtopic: Monohybrid Cross: Further Understanding |
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Level 2: 60%+
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Consider the given two statements:
| Assertion (A): | Mendel self-pollinated the F2 plants and found that dwarf F2 plants continued to generate dwarf plants in F3 and F4 generations. |
| Reason (R): | The genotype of the dwarfs was homozygous. |
| 1. | Both (A) and (R) are True but (R) does not correctly explain (A). |
| 2. | Both (A) and (R) are True and (R) correctly explains Assertion. |
| 3. | Both (A) and (R) are False. |
| 4. | (A) is True but (R) is False. |
Subtopic: Monohybrid Cross: Further Understanding |
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In Antirrhinum majus, how many of the given cross will produce all progeny exhibiting similar phenotypes [symbols carry usual meanings]?
1. 1
2. 2
3. 3
4. 4
| I: | \(Rr \times Rr\) | II: | \(Rr \times rr\) |
| III: | \(RR \times rr\) | IV: | \(Rr \times RR\) |
1. 1
2. 2
3. 3
4. 4
Subtopic: Monohybrid Cross: Further Understanding |
Level 3: 35%-60%
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Consider the given two statements:
| Assertion (A): | Female must have the colour blindness gene on both of their X chromosomes to be colour blind. |
| Reason (R): | Colour blindness is a sex linked dominant trait. |
| 1. | Both (A) and (R) are True but (R) does not correctly explain (A). |
| 2. | Both (A) and (R) are True and (R) correctly explains (A). |
| 3. | Both (A) and (R) are False. |
| 4. | (A) is True but (R) is False. |
Subtopic: Pedigree Analysis: More Examples |
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Level 2: 60%+
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Genes that are located on the same chromosome are called as:
1. Duplicate genes
2. Pseudo genes
3. Linked genes
4. Codominant genes
1. Duplicate genes
2. Pseudo genes
3. Linked genes
4. Codominant genes
Subtopic: Understanding Linkage |
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The karyotype shown below is of a person affected with:
1. Klinefelter’s syndrome
2. Turner’s syndrome
3. Down’s syndrome
4. Edward’s syndrome
1. Klinefelter’s syndrome
2. Turner’s syndrome
3. Down’s syndrome
4. Edward’s syndrome
Subtopic: Sex Aneuploidy - Turner & Klinefelter Syndrome |
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How many of the given statements are correct?
1. Only I
2. Only II and III
3. None
4. I, II and III
| I: | F1 resembled either of the two parents in co-dominance |
| II: | F1 was in-between in complete dominance. |
| III: | F1 generation resembles both parents in incomplete dominance. |
2. Only II and III
3. None
4. I, II and III
Subtopic: Co-dominance |
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Level 2: 60%+
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Regarding ABO blood group genetics:
1. Only I and II are correct
2. Only I and III are correct
3. Only II and III are correct
4. I, II and III are correct
| I: | Three alleles control blood type: A, B, and O. |
| II: | A and B are codominant, while O is recessive. |
| III: | Six genotypes and four phenotypes are possible. |
2. Only I and III are correct
3. Only II and III are correct
4. I, II and III are correct
Subtopic: Co-dominance |
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Level 2: 60%+
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Starch synthesis in pea seeds is controlled by one gene. What will be true?
| Statement I: | BB homozygotes produce large starch grains, Bb heterozygotes produce large starch grains and bb homozygotes produce smaller starch grains. |
| Statement II: | BB homozygotes produce round seeds, Bb heterozygotes produce oval seeds and bb homozygotes produce wrinkled seeds. |
| 1. | Statement I is correct; Statement II is correct |
| 2. | Statement I is correct; Statement II is incorrect |
| 3. | Statement I is incorrect; Statement II is incorrect |
| 4. | Statement I is incorrect; Statement II is correct |
Subtopic: Co-dominance |
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Level 3: 35%-60%
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