Consider the two statements:
Statement I: Thalassemia is a qualitative problem of synthesising an incorrectly functioning globin.
Statement II: Sickle cell anaemia is a quantitative problem of synthesising too few globin molecules while the latter.
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 correct
4. Statement I is incorrect; Statement II is incorrect

Subtopic:  Mendelian Disorders: Thalassemia |
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Turner’s Syndrome:
I:  is an aneuploidy
II: causes infertility in affected females
III: does not affect the development secondary sexual characters in affected females
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:  Sex Aneuploidy - Turner & Klinefelter Syndrome |
 83%
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The cause of Klinefelter's syndrome in humans is : 
1. Absence of Y-chromosome
2. Absence of X-chromosome
3. Extra copy of an autosome
4. Extra copy of an X-chromosome
 
Subtopic:  Sex Aneuploidy - Turner & Klinefelter Syndrome |
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Select the incorrect pair : 
1. Polygenic inheritance Haemophilia
2.  Linkage Drosophila 
3. Incomplete dominance  Antirrhinum
4. Pleiotropy Phenylketonuria
Subtopic:  Polygenic Inheritance & Pleiotropy | Chromosomal Basis of Inheritance: Introduction | Linkage | Mendelian Disorders: Hemophilia |
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According to Mendel, the nature of the unit factors that control the expression of traits were : 
1. Stable 
2. Blending
3. Stable and discrete 
4. Discrete
 
Subtopic:  Chromosomal Basis of Inheritance: Introduction | Chromosomal Basis of Inheritance: Further Considerations |
 86%
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Which of the following animals exhibit male heterogamety?
(i) Fruit fly 
(ii) Fowl
(iii) Human
(iv) Honey bee
1. (i) and (iii) 
2. (ii) and (iv)
3. (ii) and (iii) 
4. (i) and (iv) 
Subtopic:  Sex Determination |
 79%
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The probability of all possible genotypes of offsprings in a genetic cross can be obtained with the help of :
1. Test cross 
2. Back cross 
3. Punnett square 
4. Linkage cross
Subtopic:  Monohybrid Cross: 1 | Monohybrid Cross: Further Understanding |
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The number of different type of gametes that would be produced from a parent with genotype AABBCc is:
1. 1
2. 2
3. 3
4. 4
Subtopic:  Monohybrid Cross: 1 | Dihybrid Cross Analysis |
 70%
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Assertion (A): Accumulation of phenylalanine in the brain results in mental retardation in Phenylketonuria.
Reason (R): The affected person lacks phenylalanine which is therefore not converted to tyrosine. 

In light of the above statements choose answer from the options given below:
 
1. Both (A) and (R) are true and (R) is the correct explanation of (A).
2. Both (A) and (R) are true but (R) is not the correct explanation of (A).
3. (A) is true, but (R) is false 
4. (A) is false, but (R) are true. 


 

 
 
 
Subtopic:  Mendelian Disorders |
From NCERT
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In Pisum sativum, the flower position may be axial (allele A) or terminal (allela a). What would be the percentage of the offspring with respect to axial flower position, if a cross is made between parents \(Aa \times aa\) ? 
1. 25% 
2. 50%
3. 75% 
4. 100%
 
Subtopic:  Monohybrid Cross: Further Understanding |
 85%
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