Let us imagine that as a result of the Human Genome Project, compulsive channel
ID: 34158 • Letter: L
Question
Let us imagine that as a result of the Human Genome Project, compulsive channel switching (i.e. channel surfing) is found to be caused by a recessive (but highly prevalent) gene on the X chromosome (cs). Also located on the X chromosome are two other recessive genes - one that causes an addiction to watching football on TV (fba), and one that causes excessive burping in public (bp). The location of these genes on the X chromosome are 25 (cs/ cs+), 43 (fba/fba+) and 76 (bp/bp+) map units. The level of interference over this linkage group (i.e. the Coefficient of Interference) is 0.15. Consider a normal female with the genotype cs+fba bp+ / cs fba+bp and her husband who is a channel surfing, burping, football addict (genotype cs bp fba / Y).
What is the observed DCO frequency for this linkage group? Hint: Start with the Coefficient of Interference and work backwards!
Consider the genotype provided (cs+fba bp+ / cs fba+bp ) for the female. What is the probability that she will produce a gamete that contains all three recessive alleles?
What is the probability that they could have a son who displays compulsive channel switching only? Hint: You will need to work backwards!
What is the probability that the woman will produce a parental type (or non-recombinant type) gamete?
Explanation / Answer
Answer 1
Interference = 0.15
So the coefficient of coincidence = 1 – 0.15 = 0.85
We know that, Observed DCO/ Expected DCO = coefficient of coincidence
According to question distance between cs and fba = 43 – 25 = 18 mu
distance between fba and bp = 76 – 43 = 33 mu
So, Expected rate of DCO = (18/100)*(33/100) = 0.0594 = 0.06 i.e. 60 in 1000
Therefore,
Observed DCO/ Expected DCO = 0.85
Observed DCO/ 60 = 0.85
Observed DCO = 60 x 0.85 = 51 in 1000 i.e. 5 in 100
Answer 2
To produce a gamete with all three recessive alleles there should be a DCO,
probability of crossing over between cs and fba = 18% = 0.18
and, probability of crossing over between fba and bp = 33% = 0.33
So the total expected probability of all recessive allele would be 0.18 x 0.33 = 0.059
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