Question

The genotype of a plant is \(AaBbDd\). The genes \(A\) and \(B\) are linked in coupling phase with a map distance of \(10\) cM. The gene \(D\) is present in different chromosome. Under such situation, the frequency of \(AbD\) gametes would be?

• A. \(1/8\)
• B. \(9/40\)
• C. \(9/20\)
• D. \(1/40\)

Hint:

For linked genes in coupling phase \(AB/ab\), remember that the parental gametes are \(AB\) and \(ab\), while the recombinant gametes are \(Ab\) and \(aB\).
If the map distance is \(10\) cM, total recombinant frequency is \(10\%\), so each recombinant gamete has a frequency of \(5\%\).
For a gene located on a different chromosome in a heterozygous condition such as \(Dd\), multiply by \(1/2\) because of independent assortment.
Thus, for \(AbD\): \(5\%\times 1/2=2.5\%=1/40\).

Answer 1

The Correct Option is D

•

Step 1: Understanding the Question:
The plant has genotype \(AaBbDd\).
The genes \(A\) and \(B\) are linked in coupling phase, meaning that the parental chromosome arrangement is \(AB/ab\).
The map distance between \(A\) and \(B\) is \(10\) cM, which represents a recombination frequency of \(10\%\).
The gene \(D\) is located on a different chromosome, so it undergoes independent assortment with respect to genes \(A\) and \(B\).
The required gamete is \(AbD\).

•

Step 2: Key Formula or Approach:
For two linked genes, the recombination frequency is equal to the percentage of recombinant gametes.
\[ \text{Recombination frequency}=10\%=0.10 \] Since the genes are in coupling phase \(AB/ab\), the parental gametes are \(AB\) and \(ab\), while the recombinant gametes are \(Ab\) and \(aB\).
Each recombinant type occurs with half of the total recombinant frequency:
\[ \text{Frequency of }Ab=\frac{10\%}{2}=5\% \] Since \(D\) is on a different chromosome and the genotype is \(Dd\), the probability of obtaining \(D\) in a gamete is:
\[ P(D)=\frac{1}{2} \] Therefore,
\[ P(AbD)=P(Ab)\times P(D) \]
• Detailed Explanation:

• The coupling arrangement is \(AB/ab\).

• The map distance of \(10\) cM means that \(10\%\) of the gametes are recombinant.

• The two recombinant gamete types are \(Ab\) and \(aB\).

• Therefore, the frequency of each recombinant gamete is:
\[ \frac{10}{2}=5\% \]
• Thus, the frequency of the \(Ab\) gamete is \(0.05\).

• Gene \(D\) is located on a different chromosome and therefore assort independently of \(A\) and \(B\).

• Because the genotype at the \(D\) locus is \(Dd\), half of the gametes receive \(D\), while the other half receive \(d\).
\[ P(D)=\frac{1}{2} \]
• Hence, the frequency of the required \(AbD\) gamete is:
\[ P(AbD)=\frac{10\%}{2}\times\frac{1}{2} \] \[ =\frac{10}{100}\times\frac{1}{2}\times\frac{1}{2} \] \[ =\frac{10}{400} \] \[ =\frac{1}{40} \]
• Therefore, the frequency of \(AbD\) gametes is \(1/40\).

• Final Answer:
The frequency of the \(AbD\) gamete is:
\[ \boxed{\frac{1}{40}} \] Hence, the correct option is (D) \(1/40\).