Principles of Inheritance and Variation — NCERT Solutions

Advantages of pea plant for Mendel’s experiments:

• It had many clear contrasting traits such as tall/dwarf, round/wrinkled, yellow/green, etc.
• It was a true-breeding plant, so stable traits were available for experiments.
• It is naturally self-pollinating, but can also be cross-pollinated easily by Mendel.
• It had a short life cycle, so successive generations could be studied quickly.
• It produced a large number of offspring, giving reliable statistical results.

Dominance: In a heterozygous condition, one allele expresses itself and masks the other. The expressed allele is the dominant one.

Recessive: The allele whose effect is masked in a heterozygote and is expressed only in the homozygous condition is recessive.

For example, in pea plant height, T (tall) is dominant over t (dwarf).

Homozygous: A condition in which the two alleles of a gene are identical. Example: TT or tt.

Heterozygous: A condition in which the two alleles of a gene are dissimilar. Example: Tt.

Monohybrid cross: A cross involving one pair of contrasting traits.

Dihybrid cross: A cross involving two pairs of contrasting traits.

Example: TT × tt is a monohybrid cross, while RRYY × rryy is a dihybrid cross.

A heterozygous locus produces 2 kinds of gametes. For 4 heterozygous loci, the number of gamete types is:

$2^4 = 16$

So, 16 types of gametes can be produced.

Mendel explained law of dominance by a monohybrid cross between true-breeding tall pea plant and true-breeding dwarf pea plant.

• Parental cross: TT × tt
• Gametes formed: T and t
• F$_1$ genotype: Tt
• F$_1$ phenotype: all tall

This shows that in a heterozygous pair, one factor dominates the other. Here T is dominant over t. When F$_1$ plants are selfed:

• F$_2$ genotypes: 1 TT : 2 Tt : 1 tt
• F$_2$ phenotypes: 3 tall : 1 dwarf

Thus, the recessive trait appears only when it becomes homozygous, while the dominant trait appears in the heterozygous condition.

A test-cross is a cross between an individual showing a dominant phenotype but whose genotype is unknown, and a homozygous recessive individual.

For example, a tall plant of unknown genotype may be crossed with a dwarf plant (tt) to determine whether the tall plant is TT or Tt.

Design:

• If the unknown tall plant is TT: cross TT × tt gives all tall offspring.
• If the unknown tall plant is Tt: cross Tt × tt gives 1 tall : 1 dwarf offspring.

So, by examining the offspring, the genotype of the dominant individual can be identified.

Let the homozygous female be recessive and the heterozygous male be dominant, as in a single-locus cross.

Example: tt × Tt

Gametes:

• Female produces only t
• Male produces T and t

Punnett square gives offspring:

• Tt = tall
• tt = dwarf

So the phenotypic distribution is:

• 1 tall : 1 dwarf
• i.e., 50% tall and 50% dwarf

If the dominant allele is not specified, this is the standard expected distribution for a test cross.

Cross: TtYy × Ttyy

For height:

• Tt × Tt gives 3 tall : 1 dwarf

For seed colour:

• Yy × yy gives 1 yellow : 1 green

Combine the probabilities independently:

(a) Tall and green

• Tall = $3/4$
• Green = $1/2$
• So, tall and green = $\frac{3}{4} \times \frac{1}{2} = \frac{3}{8}$

(b) Dwarf and green

• Dwarf = $1/4$
• Green = $1/2$
• So, dwarf and green = $\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$

If two heterozygous parents are crossed and the two loci are linked, the genes tend to be inherited together because they are on the same chromosome.

So, unlike an unlinked dihybrid cross which gives 9:3:3:1, linkage produces a deviation from independent assortment.

The offspring show more parental combinations and fewer recombinants. If linkage is very strong, the parental types may greatly outnumber the non-parental types.

Therefore, the distribution in F$_1$ progeny of such a dihybrid cross would not be 9:3:3:1; it would be dominated by the parental phenotypes.