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Q. No.Study the pedigree analysis of human given below and identify the type of inheritance along with an example :
1. Sex-linked recessive, Haemophilia
2. Sex-linked dominant, Vitamin D resistant rickets
3. Autosomal recessive, Sickle-cell anaemia
4. Autosomal dominant, Myotonic dystrophy
Subtopic: Pedigree Analysis: More Examples |
65%
From NCERT
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Possibility of the blood groups of the children in a family where the father is heterozygous for blood group 'A' and the mother is heterozygous for blood group 'B', would be :
1. Blood groups 'A','B'
2. Blood groups 'A','B','O'
3. Blood groups 'AB','O'
4. Blood groups 'A','B','AB','O'
1. Blood groups 'A','B'
2. Blood groups 'A','B','O'
3. Blood groups 'AB','O'
4. Blood groups 'A','B','AB','O'
Subtopic: Co-dominance |
83%
From NCERT
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The correct statement with respect to Thalassemia in humans is:
1. \(\alpha -\)Thalassemia is controlled by a single gene HBB.
2. The gene for \(\alpha -\)Thalassemia is located on chromosome-16.
3. \(\beta -\)Thalassemia is controlled by two closely linked genes HBA-1 and HBA-2.
4. In \(\beta-\)Thalassemia the production of \(\alpha -\)globin chain is affected.
1. \(\alpha -\)Thalassemia is controlled by a single gene HBB.
2. The gene for \(\alpha -\)Thalassemia is located on chromosome-16.
3. \(\beta -\)Thalassemia is controlled by two closely linked genes HBA-1 and HBA-2.
4. In \(\beta-\)Thalassemia the production of \(\alpha -\)globin chain is affected.
Subtopic: Mendelian Disorders: Thalassemia |
80%
From NCERT
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Given below are the pairs of contrasting traits in Pisum sativum as studied by Mendel. Identify the incorrect Pair traits :
| Character | Dominant | Recessive | |
| 1. | Stem height | Tall | Dwarf |
| 2. | Seed shape | Round | Wrinkled |
| 3. | Pod colour | Yellow | Green |
| 4. | Flower position | Axial | Terminal |
Subtopic: Introduction to Genetics: 1 | Introduction to Genetics: 2 |
75%
From NCERT
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A plant breeder crossed a pure bred tall plant having white flowers with a pure bred dwarf plant having blue flowers. He obtained 202 \(F_1\) progeny and found that they are all tall having blue flowers. Upon selfing these \(F_1\) plants he obtained a progeny of 2160 plants. Approximately how many of these likely to be short having blue flowers?
1. 1215
2. 405
3. 540
4. 135
1. 1215
2. 405
3. 540
4. 135
Subtopic: Dihybrid Cross Analysis |
From NCERT
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Given below is a karyotype of a human foetus obtained for screening to find any probable genetic disorder :
Based on the karyotype, the chromosomal disorder detected in unborn foetus and the consequent symptoms the child may suffer from are :
1. Turner's syndrome : Sterile ovaries, short stature
2. Down's syndrome : Gynecomastia, overall masculine stature
3. Turner's syndrome : Small round head, flat back of head
4. Down's syndrome : Furrowed tongue, short stature
Based on the karyotype, the chromosomal disorder detected in unborn foetus and the consequent symptoms the child may suffer from are :
1. Turner's syndrome : Sterile ovaries, short stature
2. Down's syndrome : Gynecomastia, overall masculine stature
3. Turner's syndrome : Small round head, flat back of head
4. Down's syndrome : Furrowed tongue, short stature
Subtopic: Non - Disjunction & Aneuploidy | Sex Aneuploidy - Turner & Klinefelter Syndrome |
73%
From NCERT
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In the dihybrid cross that was conducted by Morgan involving mating between parental generation for genes yellow bodied, white eyed female Drosophila and wild type male Drosophila, upto \(F_2\) generation is given below :
Study the result obtained of the \(F_2\) progeny. Select the correct option from the given choice for the \(F_2\) progeny.
1. Parental type, \(1.3 \%\) : Strength of linkage high
2. Recombinant types, \(1.3 \%\) : Strength of linkage high
3. Parental type \(98.7 \%\) : Strength of linkage high
4. Recombinant types, \(98.7 \%\) : Strength of linkage low
Study the result obtained of the \(F_2\) progeny. Select the correct option from the given choice for the \(F_2\) progeny.
1. Parental type, \(1.3 \%\) : Strength of linkage high
2. Recombinant types, \(1.3 \%\) : Strength of linkage high
3. Parental type \(98.7 \%\) : Strength of linkage high
4. Recombinant types, \(98.7 \%\) : Strength of linkage low
Subtopic: Chromosomal Basis of Inheritance: Introduction | Chromosomal Basis of Inheritance: Further Considerations |
61%
From NCERT
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Chromosomal aberrations are caused by:
1. Change in the structure of the gene
2. Change in the number of chromosomes
3. Change in the arrangement or position of genes
4. Change in the number or arrangement of genes in the chromosomes
1. Change in the structure of the gene
2. Change in the number of chromosomes
3. Change in the arrangement or position of genes
4. Change in the number or arrangement of genes in the chromosomes
Subtopic: Non - Disjunction & Aneuploidy |
57%
From NCERT
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1:2:1 phenotypic and genotypic ratio is found in:
1. Complementary genes.
2. Pseudo alleles.
3. Blending inheritance
4. Multiple alleles.
1. Complementary genes.
2. Pseudo alleles.
3. Blending inheritance
4. Multiple alleles.
Subtopic: Monohybrid Cross: Further Understanding |
53%
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Possibilities of the blood groups of the children in a family where the father is heterozygous for blood group A and the mother is heterozygous for blood group B would be:
1. Blood Group A and B
2. Blood Group A, B and O
3. Blood Group AB and O
4. Blood Group A, B, AB and O
1. Blood Group A and B
2. Blood Group A, B and O
3. Blood Group AB and O
4. Blood Group A, B, AB and O
Subtopic: Co-dominance |
84%
From NCERT
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