Principles of Inheritance and Variation Test - 87

Heterozygous plant having yellow seeds - Yy

Presence of multiples of complete haploid chromosomal sets is called as polyploidy. Wheat, Triticum, has basic chromosome number is n= 7; thus the hexaploid species (Triticum aestivum) have 6n= 42. Monosomy is absence of a single chromosome from a diploid cell (2n-1); Nullisomy is absence of one complete pair of homologous chromosomes (2n-2); haploidy is characterized by presence of one complete set of chromosomes as is found in gametes; trisomy is presence of one chromosome in triplicate (2n+1). Thus, for a hexaploid wheat; monosomy (2n-1) = 41; haploidy (2n/2) = 21; Nullisomy (2n-2) = 40; trisomy (2n+1) = 43. This makes option D correct.

• Mendel's second law, the law of independent assortment, says that alleles of different genes assort independently of one another during gamete formation. The second law describes the outcome of dihybrid (two characters) crosses, or hybrid crosses involving more than two traits.
•  A dihybrid is an individual that is a double heterozygote. The random alignment of homologous pairs of chromosomes along the metaphase 1 plate accounts for the law of independent assortment. 
• This model works for genes that are on separate (non-homologous) chromosomes, but not necessarily for genes that are on the same chromosome. When Mendel studied these characters together, he found that all the seven characters chosen were controlled by seven pairs of alleles and they were located on seven different pairs of homologous chromosomes. If any two different pairs of alleles were to present on the same pair of chromosomes his conclusion would have been different. 
• Hence, it can be deduced that the pea plant could have at least seven pairs of chromosomes.

Total number of types of gamete produce by an organism is 2$$^n$$, where n is the number of heterozygous genes present. As we know that monohybrid plant is heterozygous for one gene, thus total possible gametes by it = 2$$^1$$=2. During fertilization, any of the 2 types of male gamete can fuse with any of the 2 types of female gamete to produce 4 possible combinations of zygote. Similarly, dihybrid, trihybrid and tetrahybrid plants are heterozygous for two, three and four genes respectively, thus total possible gametes= 2$$^2$$, 2$$^3$$ and 2$$^4$$ = 4, 8 and 16 respectively. Thus total possible genotypes by dihybrid, trihybrid and tetrahybrid cross are 16, 64 and 256 respectively. The total number of boxes in a Punnett square equals the number of rows times the number of columns and represents the total number of genotypes formed by gametes. Thus, monohybrid, dihybrid, trihybrid and tetrahybrid cross have 4, 16, 64 and 256 boxes respectively. 

A trait governed by more than one gene where dominant allele of each gene expresses only a part of trait and the full trait is expressed only in the presence of all dominant alleles of all multiple genes, is called as polygenic inheritance. A cross between AABBCC (dark wheat grain) and aabbcc (light wheat grain) produces AaBbCc; F$$_1$$ hybrid. Since grain colour is determined by three pairs of polygene; the genotypes having the dominant allele of all genes will show the phenotype of pure-breeding parents; rest genotype will show intermediate phenotypes. The hybrid produce 2$$^3$$ = 8 gametes and selfing of gametes from two hybrids produce total 64 genotypes with 1/64 AABBCC and 1/64 aabbcc (the parental types). Thus, total number of parental types in F$$_2$$ generation= 2/64= 3.12, i.e., less than 5%. The correct option is C.

As we know that tallness (T) is dominant over dwarfism (t) and round seed (R) is dominant over wrinkled (r).

A polygenic trait is the one that is governed by more than one gene. Human skin colour is a polygenic trait governed by more than one gene where dominant allele of each gene express only a part of trait and the full trait is expressed only in the presence of dominant alleles of all multiple genes, is called polygenic inheritance. These genes control skin pigmentation only, no other phenotypic trait is affected by these genes. Likewise, human eye colour and hair pigment in the mouse are polygenic but tongue rolling in humans is not polygenic.
Human blood group inheritance is the example of codominance and multiple alleles. It is governed by three alleles namely $$I^A$$, $$I^B$$ and i. Flower color in Mirabilis jalapa is governed by two alleles of a gene that show incomplete dominance.
Sickle cell anaemia is governed by two alleles of a gene $$ Hb^b$$ and $$Hb^s$$.
Therefore, the correct answer is option D.

Sickle cell anaemia is caused by a missense mutation in the Hb$$^b$$ gene that codes for $$\beta$$ chain. Replacement of A by T at the 17th nucleotide of the Hb$$^b$$ gene changes the codon GAG (glutamic acid) to GTG (which encodes valine). The mutated allele Hb$$^s$$ encodes the abnormal haemoglobin molecules which stick to one another and cause stiffness and sickle shape of red blood cells. The sickle-shaped RBC block and damage the vital organs and tissue. The hemoglobin-Beta gene is located on chromosome 11 and the heterozygotes have normal RBCs which means that disease is governed by the recessive gene. Option A is incorrect. The homozygotes for Hb$$^s$$ do not survive which means that homozygous state is fatal; option C is incorrect. The heterozygotes, Hb$$^b$$Hb$$^s$$, show increased resistance to malaria and therefore, have the survival advantage in regions where malaria is a common disease. That’s why sickle cell anaemia has not been eliminated from African population who reside in malaria-prone area. This makes option D correct answer.