Genetics MCQ 1 - Biotechnology Engineering (BT) MCQ

Pigmentation in moth is the example of the maternal effect in which the cytoplasm and genotype of the mother affect the phenotype of the progeny as mother genotype is a+/a that’s why all the progenies will be pigmented at larval stages

Only the cytoplasm of the mother is inherited by the progeny and father does not contribute any cytoplasm. If a trait is inherited only and only through the mother, the genes must, most likely be present in the DNA of the cytoplasmic organelles.

Albino is an autosomal recessive condition but in this example, albino parents can have normal children if they have different mutated allele variants of albinism leads to complementation results in wild type pigmentation of the child.

The ratio of X chromosomes to autosomes should be 1 : 2 for a Drosophila to be a male. Here, the ratio of X chromosomes to autosomes is also 1 : 2.

Since the trait skips a generation , it cannot be a dominant trait. Only the males are affected, so, it has to be an X-linked recessive inheritance.

Phenotypic classes are calculated as 2n+1, where n is the number of gene pairs controlling the trait.

In humans, the presence of a Y chromosome results in formation of testis. If the Y chromosome is absent, then ovaries are formed.

According to the Mendelian inheritance theory, during the gametogenesis, the assortment of the genes of one pair is independent to the other pair. Thus, each pair alleles behaves independently and does not influence the segregation of the other. As a result, each gamete receives one member of every pair of alleles in equal frequency with different sets of combinations. Hence, independent assortment results new combinations of character in the offspring showing lot of phenotypic differences in the siblings of the same parent.

Since both the mother and the father have genotype dd, the progeny would contain dd as well. Since the mother did not have a dominant allele, her gametes would not contain the protein required for dextral coiling of the shell in the cytoplasm although she herself had a dextrally coiling shell. Hence, they would all be sinistral.

I^A and I^B alleles in human encode glycosyl transferase enzyme, each causing different modification in the terminal sugar of the H-structure (carbohydrate group) present on the surface of RBC. Addition of terminal N- acetylgalactosamine to the H-structure by I^A allele results in A antigen and addition of terminal galactose by I^B forms B antigen in human RBC, which is absent in the O type individuals.

The above cross can be illustrated as follows

Non-disjunction of chromosome 21 can lead to trisomy of 21 after fertilization with a haploid gamete. This results in Down Syndrome. When a part of chromosome 21 cuts off and gets translocated on chromosome 14, it gives a trisomy like effect when this 14/21 chromosome gets segregated into a gamete and is fertilized. So, both translocation and non-disjunction result in Down Syndrome.

Since females contain two X chromosomes, one X chromosome gets inactivated to prevent over-expression of genes in females and to balance the expression in males and females. Since only females contain an extra X chromosome, this happens only in females.

When alleles show a dominant recessive relationship, then the heterozygous and homozygous dominant genotypes show same phenotype. When alleles are co-dominant or incompletely dominant, the heterozygous individuals show different phenotypes such as in ABO blood groups and Antirrhinum flowers.

Polyploids are sterile because of irregular segregation of chromosomes during meiosis. It may result in aneuploid gametes. The zygotes obtained always die and this inviability explains the sterility of polyploid organisms.

Deletion and duplication result in an unequal number of alleles on a homologue pair. This results in the formation of a compensation loop in both of these events.

This is solved using binomial expansion where p = ¾, q = ¼ , n = 7, x = 4. The answer would be 0.173

q² = 0.04. Then, q = 0.2. p = 1-q = 0.8.
Frequency of heterozygotes will be 2pq = 2(0.8)(0.2) = 0.32

The gene w is in the middle. Distance between vg and w will be = Number of SCOs (79+84) + number of DCOs(8) / 1000 = 17.1 cM

q² = 0.04. Then, q = 0.2. p = 1-q = 0.8.
Frequency of heterozygotes will be 2pq = 2(0.8)(0.2) = 0.32

q = 0.30
p = 1–q = 0.70
p² = 0.70 * 0.70 = 0.49
% of individuals with p2 = 0.49 * 100 = 49%