Answer :
Let's analyze the given problem step-by-step to determine the phenotype ratio for this cross.
First, we interpret the Punnett square provided:
[tex]\[ \begin{array}{c|c|c} & L & I \\ \hline L & LL & LI \\ \hline I & LI & II \\ \hline \end{array} \][/tex]
Each cell in the Punnett square represents a possible combination of alleles from the parent cacti.
1. From the top row and left column, we obtain these combinations:
- LL: Long needles
- LI: Long needles
- LI: Long needles (note that this appears twice, once in the top row, once in the second row)
- II: Short needles
Now, let's count the phenotypes:
- LL results in long needles.
- LI also results in long needles.
- II results in short needles.
So, the counts are:
- Long needles: LL, LI, LI (total 3 times)
- Short needles: II (total 1 time)
This gives us the phenotypes as follows:
- 3 cacti with long needles
- 1 cactus with short needles
Therefore, the phenotype ratio is:
3 long needles : 1 short needle
So, the correct answer is:
C. Three long, one short
First, we interpret the Punnett square provided:
[tex]\[ \begin{array}{c|c|c} & L & I \\ \hline L & LL & LI \\ \hline I & LI & II \\ \hline \end{array} \][/tex]
Each cell in the Punnett square represents a possible combination of alleles from the parent cacti.
1. From the top row and left column, we obtain these combinations:
- LL: Long needles
- LI: Long needles
- LI: Long needles (note that this appears twice, once in the top row, once in the second row)
- II: Short needles
Now, let's count the phenotypes:
- LL results in long needles.
- LI also results in long needles.
- II results in short needles.
So, the counts are:
- Long needles: LL, LI, LI (total 3 times)
- Short needles: II (total 1 time)
This gives us the phenotypes as follows:
- 3 cacti with long needles
- 1 cactus with short needles
Therefore, the phenotype ratio is:
3 long needles : 1 short needle
So, the correct answer is:
C. Three long, one short