Final answer:
Calculating the probability of at most 1 crossover in meiosis using the Poisson distribution with λ=0.66 gives a sum that is not listed in the provided answer options. None of the options match the calculated value, suggesting a possible miscalculation or error in the list of options.
Step-by-step explanation:
The question relates to the meiosis process in sexual reproduction, where a diploid cell divides to produce haploid gametes with genetic variation. According to the Haldane model for crossovers, which is described by the Poisson distribution, we need to calculate the probability of obtaining at most 1 crossover between two loci on the genome when the expected number (λ) is 0.66. Using the Poisson probability formula P(X) = (e-λλX)/X!, where X=0 and X=1 for this case, we find:
- Probability of 0 crossovers: P(0) = (e-0.660.660)/0! = e-0.66 ≈ 0.516
- Probability of 1 crossover: P(1) = (e-0.660.661)/1! = 0.66e-0.66 ≈ 0.340
The total probability of at most 1 crossover is the sum P(0) + P(1), which equals ≈ 0.856. This value is not listed in the answer options, therefore none of the provided options a) 0.283, b) 0.481, c) 0.715, d) 0.233 is correct. There may have been a miscalculation, or the correct answer is not provided. The student would be encouraged to recheck calculations or the values presented in the answer options.