Question

dihybrid genetics problems
name madison maroon

1. in garden peas, tall (t) vine is dominant over short (t) vine, and round pea pod (r) is dominant over wrinkled pea pod (r). cross a heterozygous tall, round pea plant with a homozygous short, wrinkled pea plant. show the genotypes and phenotypes of the f₁ offspring, and the probability of each.

______ genotype of the homozygous tall, round pea plant.
______ genotype of the homozygous short, wrinkled pea plant.

1. in garden peas, tall (t) vine is dominant over short (t) vine. yellow (y) seed color is dominant over green (y) seed color. cross a heterozygous tall, yellow seeded plant with a heterozygous tall, yellow seeded plant.

______ genotypes of the parent plants

______ what is the probability for a tall offspring?
______ what is the probability for a short offspring?
______ what is the probability for a yellow offspring?
______ what is the probability for a green offspring?
______ what is the probability for a tall yellow offspring?
______ what is the probability for a tall green offspring?
______ what is the probability for a short yellow offspring?
______ what is the probability for a short green offspring?

Explanation:

Step1: Identify Parent Genotypes

Both parent plants are heterozygous tall (Tt) and heterozygous yellow (Yy), so their genotype is \( TtYy \).

Step2: Determine Gametes

Each parent produces gametes by independent assortment: \( TY, Ty, tY, ty \).

Step3: Create Punnett Square

A dihybrid Punnett square has 16 cells. The possible combinations from \( TtYy \times TtYy \) are:

• \( TTYY, TTYy, TtYY, TtYy, TTYy, TTyy, TtYy, Ttyy, TtYY, TtYy, ttYY, ttYy, TtYy, Ttyy, ttYy, ttyy \)

Step4: Analyze Tall (T_) Offspring

Tall genotypes: \( TT \) or \( Tt \). Count: 12 out of 16. Probability: \( \frac{12}{16} = \frac{3}{4} \).

Step5: Analyze Short (tt) Offspring

Short genotype: \( tt \). Count: 4 out of 16. Probability: \( \frac{4}{16} = \frac{1}{4} \).

Step6: Analyze Yellow (Y_) Offspring

Yellow genotypes: \( YY \) or \( Yy \). Count: 12 out of 16. Probability: \( \frac{12}{16} = \frac{3}{4} \).

Step7: Analyze Green (yy) Offspring

Green genotype: \( yy \). Count: 4 out of 16. Probability: \( \frac{4}{16} = \frac{1}{4} \).

Step8: Analyze Tall Yellow (T_Y_)

Count: 9 out of 16. Probability: \( \frac{9}{16} \).

Step9: Analyze Tall Green (T_yy)

Count: 3 out of 16. Probability: \( \frac{3}{16} \).

Step10: Analyze Short Yellow (ttY_)

Count: 3 out of 16. Probability: \( \frac{3}{16} \).

Step11: Analyze Short Green (ttyy)

Count: 1 out of 16. Probability: \( \frac{1}{16} \).

Answer:

• Genotypes of parent plants: \( TtYy \) and \( TtYy \)
• Probability for tall offspring: \( \frac{3}{4} \)
• Probability for short offspring: \( \frac{1}{4} \)
• Probability for yellow offspring: \( \frac{3}{4} \)
• Probability for green offspring: \( \frac{1}{4} \)
• Probability for tall yellow offspring: \( \frac{9}{16} \)
• Probability for tall green offspring: \( \frac{3}{16} \)
• Probability for short yellow offspring: \( \frac{3}{16} \)
• Probability for short green offspring: \( \frac{1}{16} \)