MET D7S8
Answer
MET: cf allele (diseased) is associated with the upper, purple band. Wild-type cf +allele is associated with lower, green band.
D7S8: cf allele (diseased) is associated with the lower, purple band. Wild-type cf +allele is associated with upper, green band.
MET D7S8
Create a genetic map of cf and the RFLP markers MET and D7S8.
Answer
Answer
There are two possible ways to diagram this linkage:
How can we determine which map is correct?
Answer
Comparing MET and D7S8 inheritance
Answer
Therefore, only this map can be correct:
You are the genetic counselor, what do you tell these parents when considering the probability of child #6 having CF?
Answer
It looks like child #6 is homozygous cf since it is homozygous for the band that is linked to the cf allele. However, there is a 6% chance that recombination occurred in either parent, so you cannot simply say that you are 94% sure the child will have CF. If one parent OR the other had recombination between the CF locus and the D7S8 locus, then the child will be a healthy carrier.
One way to look at this is what is the probability the child WILL have CF: 0.94 (no recombination) X (and) 0.94 (no recombination) = 0.8836.
Alternatively, you could ask what is the chance the child WILL NOT have CF. If one parent does have recombination AND the other does not (0.94 X 0.06) OR recombination happens in the other parent (0.94 X 0.06) OR it happens in both parents (0.06 X 0.06) = 0.0564 + (or) 0.0564 + 0.0036 = 0.1164 or 11.64 % chance the child WILL NOT have CF.
You are the genetic counselor, what do you tell these parents when considering the probability of child #6 having CF?
RFLP analysis using D7S8
Answer
In this case, for child #6 to develop CF, the upper band would have to be linked to the cf alllele AND the lower band would have to be linked to the cf allele. The odds that the upper band is linked to cf is 6% (requires recombination) while the odds of the lower band being linked to cf is 94% (does not require recombination). Therefore,the odds of #6 developing the disease is 0.06 X 0.94 = 0.0564 or 5.64%.
© Copyright 2007 Department of Biology, Davidson
College, Davidson, NC 28036
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