Question

in the $f_2$ generation of mendels crosses,
all plants had the dominant trait.
half of the plants had the dominant allele.
all plants had both the dominant and recessive alleles.
one in four plants had two alleles for the recessive trait.

Explanation:

Mendel's monohybrid cross: When he crossed pure - breeding (homozygous) plants (e.g., \(TT\) - tall and \(tt\) - short), the \(F_1\) generation was all heterozygous (\(Tt\)). When \(F_1\) plants self - pollinated (\(Tt\times Tt\)), using the Punnett square:

Gametes of Parent 1	Gametes of Parent 2
\(T\)	\(TT\) (tall)	\(Tt\) (tall)
\(t\)	\(Tt\) (tall)	\(tt\) (short)

The genotypic ratio of \(F_2\) is \(TT:Tt:tt = 1:2:1\). The phenotypic ratio is \(3\) (tall) : \(1\) (short). Plants with two alleles for the recessive trait (\(tt\)) are \(\frac{1}{4}\) of the \(F_2\) generation.

• Option 1: In \(F_2\), not all plants have the dominant trait. There is a \(3:1\) phenotypic ratio (dominant : recessive).
• Option 2: Half of the plants (\(Tt\)) have the dominant allele, but this is not the most accurate description of the key result regarding recessive - trait - bearing plants.
• Option 3: Not all plants have both dominant and recessive alleles (\(TT\) has two dominant alleles, \(tt\) has two recessive alleles).
• Option 4: According to the \(F_2\) genotypic ratio (\(1TT:2Tt:1tt\)), one - fourth (\(tt\)) of the plants have two alleles for the recessive trait.

Answer:

one in four plants had two alleles for the recessive trait.