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In pea plants, the allele for tall plants [tex]$(T)$[/tex] is dominant over the allele for short plants [tex][tex]$(t)$[/tex][/tex]. Pea plants with the following genotypes are crossed:

[tex]T t \times T t[/tex]

What is the probability that an offspring will have the [tex]TT[/tex] genotype?

A. 0.00
B. 0.75
C. 0.25
D. 0.50

Sagot :

GinnyAnsweridn
To determine the probability that an offspring will have the TT genotype when two heterozygous pea plants (Tt) are crossed, follow these steps:

Step 1: Identify the genotypes of the parents.
- Both parents have the genotype Tt.

Step 2: Determine the possible alleles each parent can pass on to their offspring.
- Each parent can contribute either the T or t allele.

Step 3: Create a Punnett square.
- A Punnett square allows us to visualize the combination of alleles from each parent.
- Label the rows and columns with the allelic contributions from each parent.

```
T t
T TT Tt
t Tt tt
```

Step 4: Fill in the Punnett square.
- Combine the alleles in each cell to determine the possible genotypes of the offspring:
- Top left cell: T (from parent 1) and T (from parent 2) = TT
- Top right cell: T (from parent 1) and t (from parent 2) = Tt
- Bottom left cell: t (from parent 1) and T (from parent 2) = Tt
- Bottom right cell: t (from parent 1) and t (from parent 2) = tt

The filled Punnett square looks like this:

```
T t
T TT Tt
t Tt tt
```

Step 5: Determine the probability of obtaining the TT genotype.
- Count the number of times the TT genotype appears in the Punnett square. It appears only once.
- There are a total of 4 possible outcomes (TT, Tt, Tt, tt).
- Therefore, the probability [tex]\( P \)[/tex] of obtaining the TT genotype is:

[tex]\[ P(\text{TT}) = \frac{\text{Number of TT genotypes}}{\text{Total number of genotypes}} = \frac{1}{4} = 0.25 \][/tex]

Conclusion:
The correct answer is:
C. 0.25
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