Principle of inheritance and variation practice · Mixed difficulty

Option B (Two type) is correct because in a monohybrid cross, selfing involves two types of gametes, each carrying one of the two alleles for the trait.

Option C (4) is correct because the cross ABcDE × AbcdE results in the genotype AABbcDdEE. Only the heterozygous loci (Bb and Dd) contribute to the number of gamete types. The number of gamete types is 2^n, where n is the number of heterozygous loci. In this case, n=2, so 2² = 4.

Option B ($1 / 16$) is correct because the probability of getting rryy from a self-crossed RrYy plant is the product of the probabilities of getting rr and yy, which are both 1/4.

Option C (Four) is correct because with two unlinked heterozygous genes (PpQq), the possible gametes are PQ, Pq, pQ, and pq.

Option B (Recessive gene, that is not expressed) is correct because a carrier organism carries a recessive gene that is not expressed in its phenotype due to the presence of a dominant allele.

Option D (No antigens A and B on RBCs) is correct because the 'O' blood type signifies the absence of both A and B antigens on the surface of red blood cells.

Option D (1 and 3) is correct because the unit of a linkage map is both a map unit and a centimorgan, which are equivalent.

Option A (A-8000-1000 BC, B-sexual, C-variations) is correct because humans recognized the role of sexual reproduction in causing variations as early as 8000-1000 BC.

Option A (Both (A) & (R) are true and the (R) is the correct explanation of the (A)) is correct because incomplete dominance allows for the distinction between heterozygotes and homozygotes due to the intermediate phenotype, as explained by the reason.

Option D (Both (1) and (3) are correct) is correct because a gene can be considered one pair of alleles and is present in allelic form on homologous chromosomes.

Option C (Mutated genes are frequently recessive) is correct because recessive mutations are less likely to be immediately expressed in heterozygotes, allowing them to persist and potentially affect future generations.

Option B (Alleles shows complete dominance) is correct because Mendel's laws are based on the principle of complete dominance, where one allele masks the expression of the other.

Option B ($1: 1: 1: 1$) is correct because a dihybrid test cross involves crossing a dihybrid individual (e.g., AaBb) with a double recessive individual (aabb), resulting in a 1:1:1:1 phenotypic ratio.

Option D (gynadromorph) is correct because a gynandromorph is an organism that contains both male and female characteristics.

Option B (Colour blindness) is correct because color blindness is a sex-linked recessive disorder, meaning the gene responsible is located on the X chromosome.

Option D ($X^C Y, X^C X^C$) is correct because for a colorblind female to be born, she must inherit one colorblind X chromosome from each parent, meaning both parents must carry the colorblind allele.

The correct answer is B ($1:2:1$) because in the F2 generation of a monohybrid cross, the genotypic ratio is 1 homozygous dominant : 2 heterozygous : 1 homozygous recessive.

Option B (22 pairs) is correct because human females have 46 chromosomes in total, with 2 sex chromosomes (XX) and 44 autosomes, which is 22 pairs of autosomes.

Option A (Male heterogamety) is correct because in both XO and XY systems, males produce two different types of gametes (X or O, and X or Y, respectively).

Option B (4) is correct because in a dihybrid cross with independent assortment, there are typically 4 different phenotypes.

Option A (Antigen-A and antibody-b) is correct because a person with blood group A has antigen A on their red blood cells and antibody b in their plasma.

Option D ($\mathrm{RrYy} \times \mathrm{rryy}$) is correct because this cross represents a test cross, where a dihybrid heterozygous individual (RrYy) is crossed with a homozygous recessive individual (rryy), resulting in a 1:1:1:1 phenotypic ratio.

Option A (Gene for eye colour and gene for wing size) is correct because genes that show a higher percentage of recombination are located farther apart on the chromosome.

Option C (a - linkage, b -recombination) is correct because linkage refers to the physical association of genes, while recombination refers to the generation of non-parental gene combinations.

Option B is correct because a color-blind girl needs to inherit the color-blind allele from both parents, meaning her father must be color-blind and her mother must carry the allele, which she inherited from her maternal grandfather.

Option B is correct because chromatids separate during mitosis, allele pairs replicate and are passed to daughter cells, maintaining the genetic composition.

Option A (Always expressed in male) is correct because males have only one X chromosome, so any recessive gene on that X chromosome will be expressed.

Option C (One of the two ${X}$-chromosomes in somatic cells of female) is correct because a Barr body represents an inactivated X chromosome in female somatic cells.

Option A (Trisomy of chromosome 21) is correct because the described symptoms are characteristic of Down syndrome, which is caused by trisomy 21.

Option D ($7 :9$) is correct because the cross involves two genes (tall/dwarf and yellow/green pods), and the parental combinations are homozygous dominant and recessive. The F2 generation will have a 9:3:3:1 ratio, and the parental combinations will be less frequent than the new combinations, resulting in a 7:9 ratio.