Remark: XO type is male heterogamety (sperm with X or O) seen in grasshoppers. XY type is male heterogamety (sperm with X or Y) seen in humans. ZW type is female heterogamety seen in birds. The Haplo-diploid system in honey bees determines sex based on chromosome sets.
Match List-I (Mendelian Law/Concept) with List-II (Core Principle):
List-I
List-II
A. Law of Dominance
I. Explains the 3:1 proportion obtained in F2
B. Law of Segregation
II. Based on the fact that alleles do not show any blending
C. Factors
III. Discrete units that control characters
D. Gamete formation
IV. Alleles of a pair separate from each other
Choose the correct answer from the options given below:
Remark: The Law of Dominance explains the F1 expression and the 3:1 ratio in F2. The Law of Segregation is based on the non-blending of alleles. Factors are discrete units controlling characters. During gamete formation, alleles of a pair segregate.
Match List-I (Monohybrid Cross Result) with List-II (Ratio/Proportion):
List-I
List-II
A. F2 Phenotypic ratio
I. 3:1
B. F2 Genotypic ratio
II. 1:2:1
C. Proportion of homozygous dominant (TT) in F2
III. 1/4th
D. Resulting phenotypic ratio of a Test Cross (Tt x tt)
IV. 1:1
Choose the correct answer from the options given below:
Remark: The F2 phenotypic ratio is 3:1. The F2 genotypic ratio is 1:2:1. Homozygous dominant (TT) is 1/4th (or 25%) of the F2 generation. A test cross (Tt x tt) yields a 1:1 phenotypic and genotypic ratio.
Match List-I (Pleiotropy/Gene Action) with List-II (Affected Trait/Pathway):
List-I
List-II
A. Pleiotropic gene
I. Single gene exhibiting multiple phenotypic expression
B. Starch synthesis gene (Pea)
II. Influences both seed shape and starch grain size
C. Phenylketonuria gene
III. Codes for enzyme phenyl alanine hydroxylase
D. Underlying pleiotropy mechanism
IV. Effect of gene on metabolic pathways
Choose the correct answer from the options given below:
Remark: A pleiotropic gene exhibits multiple phenotypic expressions. The pea starch synthesis gene influences both seed shape (dominance) and grain size (incomplete dominance). Phenylketonuria is caused by mutation in the gene coding for phenyl alanine hydroxylase. Pleiotropy works by affecting metabolic pathways.
Match List-I (Drosophila Suitability) with List-II (Reason):
List-I
List-II
A. Life Cycle Time
I. Complete life cycle in about two weeks
B. Progeny
II. Large number of offspring from a single mating
C. Sexes
III. Clear differentiation and easily distinguishable male and female
D. Hereditary Variation
IV. Many types visible with low power microscopes
Choose the correct answer from the options given below:
Remark: Drosophila completes its life cycle in about two weeks. A single mating yields a large number of progeny. Male and female flies are easily distinguishable. They possess many hereditary variations visible under a low power microscope.
Match List-I (Mutation Type) with List-II (Defining Feature/Result):
List-I
List-II
A. Mutation
I. Alteration of DNA sequences
B. Point Mutation
II. Change in a single base pair of DNA (e.g., Sickle-cell anaemia)
C. Chromosomal Aberrations
III. Loss or gain of a segment of DNA (commonly seen in cancer cells)
D. Frame-shift mutations
IV. Deletions and insertions of base pairs of DNA
Choose the correct answer from the options given below:
Remark: Mutation is the alteration of DNA sequences. Point mutation is a change in a single base pair. Chromosomal aberrations involve the loss or gain of a DNA segment. Frame-shift mutations are caused by deletions and insertions of base pairs.
Match List-I (Gene/Chromosome Location) with List-II (Position):
List-I
List-II
A. Genes
I. Located on DNA present in the chromosome
B. Alleles of a gene pair
II. Located on homologous sites on homologous chromosomes
C. Sex-linked genes
III. Located on sex chromosomes (e.g., X chromosome)
D. Autosomes
IV. Chromosomes other than sex chromosomes
Choose the correct answer from the options given below:
Remark: Genes are units of inheritance located on DNA. Alleles are located on homologous sites on homologous chromosomes. Sex-linked genes are located on sex chromosomes. Autosomes are the non-sex chromosomes.
Match List-I (Syndrome) with List-II (Associated Symptom):
List-I
List-II
A. Down’s Syndrome
I. Short statured with small round head, furrowed tongue
B. Klinefelter’s Syndrome
II. Overall masculine development with development of breast (Gynaecomastia)
C. Turner’s Syndrome
III. Ovaries rudimentary and lack of secondary sexual characters
D. Phenylketonuria
IV. Mental retardation and excretion of phenylpyruvic acid in urine
Choose the correct answer from the options given below:
Remark: Down’s Syndrome symptoms include short stature, small round head, and furrowed tongue. Klinefelter’s Syndrome features masculine development along with gynaecomastia. Turner’s Syndrome features rudimentary ovaries and lack of secondary sexual characteristics in females. Phenylketonuria leads to mental retardation due to the accumulation of phenylpyruvic acid in the brain.
Match List-I (Reason for Mendel’s Work Neglect) with List-II (Core Issue):
List-I
List-II
A. Communication difficulties
I. Work could not be widely publicised
B. Concept of Factors (Genes)
II. Not accepted due to the continuous variation seen in nature
C. Use of Mathematical Logic
III. Totally new and unacceptable to biologists
D. Lack of Physical Proof
IV. Could not explain the existence or nature of factors
Choose the correct answer from the options given below:
Remark: Mendel’s work was not widely published due to lack of communication. His concept of discrete factors was countered by the observation of continuous variation. His application of mathematics was new and unacceptable. He could not provide physical proof for the existence of factors.
Match List-I (Allele in ABO Group) with List-II (Sugar Production):
List-I
List-II
A. Allele $I^A$
I. Produces a slightly different form of sugar (A sugar)
B. Allele $I^B$
II. Produces a slightly different form of sugar (B sugar)
C. Allele $i$
III. Does not produce any sugar
D. Alleles $I^A$ and $I^B$ together
IV. Both express their own types of sugars (Co-dominance)
Choose the correct answer from the options given below:
Remark: Alleles $I^A$ and $I^B$ produce slightly different sugars. Allele $i$ does not produce any sugar. When $I^A$ and $I^B$ are present together, they both express their sugars (Co-dominance).
