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Perform a dihybrid cross between parents with the genotypes NnTt and NnTt (N = messy, n...

Perform a dihybrid cross between parents with the genotypes NnTt and NnTt (N = messy, n = neat, T = relaxed, t = tense). What is the probability that an F1 generation plant will be homozygous dominant for both traits, knowing that it is messy and relaxed?

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Answer #1

A Dihybrid Punnett square has 16 coloumns , hencforth the probability of obtaning homozygous dominant for messy and relaxed that is NNTT would 1 out of 16. So by the empirical probability the occurance of NNTT would be

Number of times it appeared in the square divided Total number of the squares :

Therefore the probability would be 1/16 = 0.0625 rounded off to 0.13 .

Since both the parents exhibit similar genotype henceforth this would be considered as a selfing test cross and the punnett square of F1 progeny would be either be similar to that of F2 progeny as well as the probability of it's occurance or the probability would be zero if a reverse punnett square is constructed.

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