Answer :
To determine the possible outcomes of a genetic cross where one parent has genotype \( Tt \) (one dominant allele \( T \) and one recessive allele \( t \)) and the other parent has the same genotype \( Tt \), we can use a Punnett square.
Here is the setup of the Punnett square:
[tex]\[ \begin{array}{|c|c|c|} \hline & T & t \\ \hline T & TT & Tt \\ \hline t & Tt & tt \\ \hline \end{array} \][/tex]
Now, let's fill in the Punnett square step by step:
1. The top row of the square represents the alleles from one parent: \( T \) and \( t \).
2. The first column represents the alleles from the second parent: \( T \) and \( t \).
To fill in the squares:
- The top-left square combines the \( T \) from the first parent with the \( T \) from the second parent, resulting in \( TT \).
- The top-right square combines the \( T \) from the first parent with the \( t \) from the second parent, resulting in \( Tt \).
- The bottom-left square combines the \( t \) from the first parent with the \( T \) from the second parent, resulting in \( Tt \).
- The bottom-right square combines the \( t \) from the first parent with the \( t \) from the second parent, resulting in \( tt \).
So, the filled Punnett square looks like this:
[tex]\[ \begin{array}{|c|c|c|} \hline & T & t \\ \hline T & TT & Tt \\ \hline t & Tt & tt \\ \hline \end{array} \][/tex]
From the Punnett square, we can see that the possible outcomes of the cross are:
- \( TT \)
- \( Tt \)
- \( Tt \) (which is the same as \( Tt \) above, but counted twice for clarity)
- \( tt \)
Since \( Tt \) results in the same genotype regardless of parent order, we can conclude the unique possible outcomes are:
- \( TT \)
- \( Tt \)
- \( tt \)
Now, let's match these outcomes with the given choices:
- Choice A: TT and tt only. This is incomplete because it does not include \( Tt \).
- Choice B: \( TT \), \( tt \), and \( Tt \). This is correct because it includes all the possible outcomes.
- Choice C: tt only. This is incorrect because it does not include \( TT \) and \( Tt \).
- Choice D: TT only. This is incorrect because it does not include \( tt \) and \( Tt \).
Thus, the correct answer is:
B. [tex]\( TT, tt, \)[/tex] and [tex]\( Tt \)[/tex]
Here is the setup of the Punnett square:
[tex]\[ \begin{array}{|c|c|c|} \hline & T & t \\ \hline T & TT & Tt \\ \hline t & Tt & tt \\ \hline \end{array} \][/tex]
Now, let's fill in the Punnett square step by step:
1. The top row of the square represents the alleles from one parent: \( T \) and \( t \).
2. The first column represents the alleles from the second parent: \( T \) and \( t \).
To fill in the squares:
- The top-left square combines the \( T \) from the first parent with the \( T \) from the second parent, resulting in \( TT \).
- The top-right square combines the \( T \) from the first parent with the \( t \) from the second parent, resulting in \( Tt \).
- The bottom-left square combines the \( t \) from the first parent with the \( T \) from the second parent, resulting in \( Tt \).
- The bottom-right square combines the \( t \) from the first parent with the \( t \) from the second parent, resulting in \( tt \).
So, the filled Punnett square looks like this:
[tex]\[ \begin{array}{|c|c|c|} \hline & T & t \\ \hline T & TT & Tt \\ \hline t & Tt & tt \\ \hline \end{array} \][/tex]
From the Punnett square, we can see that the possible outcomes of the cross are:
- \( TT \)
- \( Tt \)
- \( Tt \) (which is the same as \( Tt \) above, but counted twice for clarity)
- \( tt \)
Since \( Tt \) results in the same genotype regardless of parent order, we can conclude the unique possible outcomes are:
- \( TT \)
- \( Tt \)
- \( tt \)
Now, let's match these outcomes with the given choices:
- Choice A: TT and tt only. This is incomplete because it does not include \( Tt \).
- Choice B: \( TT \), \( tt \), and \( Tt \). This is correct because it includes all the possible outcomes.
- Choice C: tt only. This is incorrect because it does not include \( TT \) and \( Tt \).
- Choice D: TT only. This is incorrect because it does not include \( tt \) and \( Tt \).
Thus, the correct answer is:
B. [tex]\( TT, tt, \)[/tex] and [tex]\( Tt \)[/tex]
