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Q7(b):
When a cross in made between tall plant with yellow seeds (\(TtYy\)) and tall plant with green seed (\(Ttyy\)), what proportions of phenotype in the offspring could be expected to be (b) dwarf and green.

Solution :

Step 1: Identifying Parental Genotypes and Gametic Potential

The cross involves two dihybrid parents:

  • Parent 1: Tall with yellow seeds ($TtYy$)
  • Parent 2: Tall with green seeds ($Ttyy$)
[Per the Law of Independent Assortment, alleles for height and seed color segregate independently during gamete formation.]

Gametes produced by Parent 1 ($TtYy$): The possible combinations are $TY$, $Ty$, $tY$, and $ty$. Each occurs with a probability of $0.25$.

Gametes produced by Parent 2 ($Ttyy$): Since the plant is homozygous recessive for color, the only allele for color is $y$. The possible combinations are $Ty$ and $ty$. Each occurs with a probability of $0.5$.

Step 2: Punnett Square Construction for Phenotypic Probability

Gametes $Ty$ ($0.5$) $ty$ ($0.5$)
$TY$ ($0.25$) $TtYy$ (Tall, Yellow) $TtYy$ (Tall, Yellow)
$Ty$ ($0.25$) $Ttyy$ (Tall, Green) $Ttyy$ (Tall, Green)
$tY$ ($0.25$) $ttYy$ (Dwarf, Yellow) $ttYy$ (Dwarf, Yellow)
$ty$ ($0.25$) $ttyy$ (Dwarf, Green) $ttyy$ (Dwarf, Green)

Step 3: Calculating the Proportion of Dwarf and Green Offspring

The phenotype "dwarf and green" requires the homozygous recessive genotype ($ttyy$). From the Punnett Square calculation:

Mathematical derivation:

  • Probability of $tt$ (dwarf) from $Tt \times Tt$: $1/4$
  • Probability of $yy$ (green) from $Yy \times yy$: $1/2$
  • Combined probability (Product Rule of Probability): $P(tt) \times P(yy) = 1/4 \times 1/2 = 1/8$
[Per the Product Rule of Probability for independent events.]

Visual Representation of Genetic Segregation

Parent 1 (TtYy) Gametes: TY, Ty, tY, ty Parent 2 (Ttyy) Gametes: Ty, ty Target Offspring: ttyy (Dwarf/Green) = 1/8

Final Solution: The proportion of offspring expected to be dwarf and green is $1/8$ (or $0.125$, which is $12.5\%$).


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