The order of the genes on a plant chromosome is A, B, C, where A and B are located 10 cM apart and B and C are located 3 cM apart. What is the probability that the trihybrid ABC/abc will produce any kind of recombinant gamete

Explanation:

Available data:  

The genetic distance = recombination frequency x 100 expressed in map units (MU). One centiMorgan (cM) equals one map unit (MU).

We can use the provided map with the distances between genes to predict the occurrence of any kind of recombinant gamete.  

We know that there is a probability of a simple crossing over to occur between genes A and B, or between genes B and C. And we also know that there is a probability of double crossing-over occurrence.

We can get the probability of the double-crossing over by multiplying the recombination frequencies between A-B and B-C. So,

DCO prob = recombination frequency A-B x recombination frequency B-C

                 = 0.1  x  0.03 = 0.003 = 0.3%

DCO prob = 0.3%

According to this, we would expect to find

Now we need to know what are the probabilities of getting a simple crossing-over, SCO. We already have the genetic distances between genes. 10 cM equals 0.1 of recombination frequency, and 3cM equals 0.03 of recombination frequency. However, we can not estimate the probabilities of simple crossing over using only this information, because this data is including the probabilities of the double crossing-over occurrence. So we need to substrate this percentage.    

SCO prob A-B = 0.1 - 0.003 = 0.097 = 9.7%

SCO prob A-B = 9.7%

we would expect to find

SCO prob B-C = 0.03 - 0.003 = 0.027 = 2.7%

SCO prob B-C = 2.7%

we would expect to find

The total Crossing-over probability, TCO, is then the sum of all the crossing over probabilities,

TCO= SCO A-B + SCO B-C + DCO

TCO prob = 9.7% + 2.7% + 0.3%

TCO = 12.7%

i think the answer is g1

hope i

-lavira: )