Question

Assume that a marker M1 is present 5 cM away on one side of a gene “X” (a desirable allele of the gene), while marker M2 is present 10 cM away on the other side of the gene. The donor’s genotype is M1M1XXM2M2, while the recipient has m1m1xxm2m2 genotype. A cross was made between these two individuals. The F1 is crossed to the recipient. The progeny of this cross had 1000 plants. How many plants from this progeny will have both the markers (M1 & M2) present while the desired gene is absent? (Assume no interference)

1. 0

2. 5

3. 10

4. 15

Detailed Explanation

To solve this problem, we need to apply the concepts of genetic distance, crossovers, and recombination based on the given distances between the markers and the gene of interest. Let’s break down the information:

Key Information:

• Marker M1 is 5 cM away from gene “X”.

• Marker M2 is 10 cM away from gene “X”.

• Genotype of the donor: M1M1XXM2M2

• Genotype of the recipient: m1m1xxm2m2

• The F1 generation will have the genotype M1m1XxM2m2.

• Cross between F1 and the recipient results in a progeny of 1000 plants.

Genetic Distance and Recombination Frequency:

• The recombination frequency (RF) is equal to the genetic distance between two loci, given in centimorgans (cM).

• For the M1 and X loci (5 cM apart), the recombination frequency is 5%. This means there is a 5% chance that there will be a crossover between M1 and X.

• For the X and M2 loci (10 cM apart), the recombination frequency is 10%, meaning there is a 10% chance of crossover between X and M2.

F1 Progeny Genotypes:

• The F1 progeny from the initial cross (M1M1XXM2M2 × m1m1xxm2m2) will have the genotype M1m1XxM2m2.

Progeny Cross and Expected Outcomes:

• In the F1 cross (M1m1XxM2m2 × m1m1xxm2m2), we can calculate the probability of no crossover and recombination events that lead to the absence of gene “X” (xx) and the presence of both markers (M1 and M2).

  The crossovers for M1-X and X-M2 can occur independently. So, the probability of no crossover for M1-X is 95% (1 – 0.05), and the probability of no crossover for X-M2 is 90% (1 – 0.10). Thus, the probability that both markers (M1 and M2) are present but the desirable gene “X” is absent (i.e., the genotype is M1m1xxm2m2) can be calculated as:

  $$\text{Probability of the desired genotype}=(\text{Probability of no crossover for M1-X})\times (\text{Probability of no crossover for X-M2})\times \text{Frequency of m1m1xxm2m2}$$

  The probability of getting M1m1xxm2m2 will involve calculating the combined crossing over rates:

  For both M1 and M2, we find that 5% of progeny are recombinants for M1, and 10% for M2.

Final Answer:

After calculating the recombination frequencies, the number of progeny showing both markers M1 and M2 but without the desirable gene would be approximately 5 plants in a progeny of 1000 plants.

Conclusion

Thus, the number of progeny with both markers but absent gene X is 5.

The correct answer is:

2. 5