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Q. No.What would be the number of types of gametes produced by a plant with genotype AaBbCc [assume independent assortment between all loci]?
1. 6
2. 8
3. 9
4. 16
Subtopic: Dihybrid Cross Analysis |
82%
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In a typical dihybrid cross \(A a B b \times A a B b\) [symbols have usual meanings; assume independent assortment], what proportion of progeny are expected to carry both dominant alleles?
1. ¼
2. 9/16
3. 3/16
4. ½
1. ¼
2. 9/16
3. 3/16
4. ½
Subtopic: Dihybrid Cross Analysis |
69%
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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
1. 280
2. 31
3. 94
4. 187
Subtopic: Dihybrid Cross Analysis |
61%
From NCERT
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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 |
81%
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The physical basis of independent assortment is:
1. Metaphase I of Meiosis I
2. Metaphase II of Meiosis II
3. Anaphase I of Meiosis I
4. Anaphase II of Meiosis II
1. Metaphase I of Meiosis I
2. Metaphase II of Meiosis II
3. Anaphase I of Meiosis I
4. Anaphase II of Meiosis II
Subtopic: Dihybrid Cross: General Consideration |
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The pedigree shown below, most likely, exhibits the inheritance of a/an:
1. Autosomal recessive trait
2. Autosomal dominant trait
3. X linked recessive trait
4. X linked dominant trait
1. Autosomal recessive trait
2. Autosomal dominant trait
3. X linked recessive trait
4. X linked dominant trait
Subtopic: Pedigree Analysis: Problem Solving |
53%
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When preparing genetic map using recombination data, it is assumed that 1% of crossing over equals:
1. one map unit.
2. two map units.
3. five two map units.
4. ten map units.
1. one map unit.
2. two map units.
3. five two map units.
4. ten map units.
Subtopic: Recombination & Gene Mapping |
83%
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If the strength of linkage between y and w is higher than w and m [y, w and m are three linked genes located on a chromosome]:
| 1. |
the number of recombinants between y and w will be higher than the number of recombinants between w and m.
|
| 2. |
the number of recombinants between y and w will be lower than the number of recombinants between w and m.
|
| 3. |
the number of recombinants between y and w will be equal to the number of recombinants between w and m.
|
| 4. |
no recombinants between y and w and between w and m will be formed as they are linked.
|
Subtopic: Understanding Linkage |
84%
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Identify the incorrectly matched pair:
| Organism | Mechanism of sex determination | |
| 1. | Honey bees | Haplo-diploidy |
| 2. | Grasshoppers | XX-XO female heterogamety |
| 3. | Birds | ZZ-ZW female heterogamety |
| 4. | Fruit fly | Genic balance |
Subtopic: Sex Determination: Further Considerations |
66%
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Down’s syndrome:
1. Only I and II are correct
2. Only III and IV are correct
3. Only I and III are correct
4. Only II and IV are correct
| I: | is an aneuploidy |
| II: | is trisomy 21 |
| III: | always runs in a family |
| IV: | does not affect the life span and fertility |
1. Only I and II are correct
2. Only III and IV are correct
3. Only I and III are correct
4. Only II and IV are correct
Subtopic: Sex Aneuploidy - Turner & Klinefelter Syndrome |
82%
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