Question

multiple choice 1 point
in garden peas, a single gene controls stem length. the recessive allele (t) produces short stems when homozygous. the dominant allele (t) produces long stems. a short - stemmed plant is crossed with a heterozygous long - stemmed plant. which of the following represents the expected phenotypes of the offspring and the ratio in which they will occur?
short - stemmed plants only
1 long - stemmed plant: 1 short - stemmed plant
long - stemmed plants only
1 long - stemmed plant: 3 short - stemmed plants
3 long - stemmed plants: 1 short - stemmed plant

Explanation:

Step1: Determine Genotypes

The short - stemmed plant is homozygous recessive, so its genotype is \(tt\). The heterozygous long - stemmed plant has the genotype \(Tt\) (since \(T\) is dominant for long stems and \(t\) is recessive for short stems).

Step2: Set Up Punnett Square

When we cross \(tt\) (short - stemmed) with \(Tt\) (heterozygous long - stemmed), the Punnett square will have the following combinations:

• From the \(tt\) parent, we get only \(t\) gametes.
• From the \(Tt\) parent, we get \(T\) and \(t\) gametes.

The Punnett square is:

	\(T\)	\(t\)
\(t\)	\(Tt\)	\(tt\)

Step3: Determine Phenotypes and Ratio

The genotype \(Tt\) results in a long - stemmed phenotype (because \(T\) is dominant), and the genotype \(tt\) results in a short - stemmed phenotype.
From the Punnett square, we have 2 \(Tt\) (long - stemmed) and 2 \(tt\) (short - stemmed)? Wait, no, wait. Wait, the \(Tt\) parent: gametes are \(T\) and \(t\), and \(tt\) parent: gametes are \(t\) and \(t\). So the cross is:
First row: \(t\) (from \(tt\)) and \(T\) (from \(Tt\)) gives \(Tt\) (long - stemmed); \(t\) (from \(tt\)) and \(t\) (from \(Tt\)) gives \(tt\) (short - stemmed).
Second row: \(t\) (from \(tt\)) and \(T\) (from \(Tt\)) gives \(Tt\) (long - stemmed); \(t\) (from \(tt\)) and \(t\) (from \(Tt\)) gives \(tt\) (short - stemmed). Wait, no, actually, the \(Tt\) parent has two gametes: \(T\) and \(t\), and \(tt\) parent has two gametes: \(t\) and \(t\). So the possible offspring genotypes are \(Tt\), \(tt\), \(Tt\), \(tt\). So the number of long - stemmed (\(Tt\)) is 2 and short - stemmed (\(tt\)) is 2? Wait, no, the question's options: let's re - check. Wait, maybe I made a mistake. Wait, the heterozygous long - stemmed is \(Tt\), and short - stemmed is \(tt\). So the cross is \(Tt\times tt\). The Punnett square:
Gametes of \(Tt\): \(T\), \(t\); Gametes of \(tt\): \(t\), \(t\).
So the offspring:

• \(T\times t=Tt\) (long)
• \(t\times t = tt\) (short)
• \(T\times t=Tt\) (long)
• \(t\times t = tt\) (short)

Wait, so that's 2 long - stemmed (\(Tt\)) and 2 short - stemmed (\(tt\))? But the options are: 1 long - stemmed: 1 short - stemmed? Wait, wait, no, maybe I messed up the parent. Wait, the problem says "a short - stemmed plant is crossed with a heterozygous long - stemmed plant". Short - stemmed is \(tt\), heterozygous long - stemmed is \(Tt\). So the Punnett square has two \(Tt\) and two \(tt\), which is a ratio of 1:1 (2 long - stemmed and 2 short - stemmed is equivalent to 1:1 when simplified? Wait, no, 2:2 simplifies to 1:1. Wait, the options: "1 long - stemmed plant: 1 short - stemmed plant" is one of the options. Wait, let's count the number of each phenotype. \(Tt\) is long - stemmed, and there are 2 of them, \(tt\) is short - stemmed, 2 of them. So the ratio of long - stemmed to short - stemmed is 2:2 = 1:1. So the expected phenotypes are 1 long - stemmed: 1 short - stemmed.

Answer:

1 long - stemmed plant: 1 short - stemmed plant (the option with "1 long - stemmed plant: 1 short - stemmed plant")