The human DNA probe L9 recognizes an RFLP locus as indicated below. This locus w
ID: 226289 • Letter: T
Question
The human DNA probe L9 recognizes an RFLP locus as indicated below. This locus which has two alleles (a1 and a2) shows linkage with a locus affecting disease R (allele Rd causes the disease and is fully recessive to the wild-type allele, R+)
a) Different Southern blots of EcoR I digests of DNA from three individuals, a1a1, a2a2, and a1a2 were prepared. Show the expected autoradiographic pattern of each of the individual’s blots when probed with the 32P labelled L9 DNA
b) Linkage studies between the L9 locus and the R locus show a distance of 7 cM. The following cross is made:
a1 R+ a2Rd
X
a1 R+ a1R+
i)What is the probability of having an a1a1R+Rd child?
ii) If an offspring of a1a2 genotype is born, what is the probability of it also being a carrier of the disease allele?
EcoRI EcoR I L9 4 Kbp EcoRI 10 Kbp 6 Kbp EcoRI EcoRI L9 probe a allele a alleleExplanation / Answer
1. There is 7 cM distance between locus L9 and R, therefore there percentage of recombination can calculate by following formula.
Recombination= (Distance/ 50) x 100= (7/ 50) x 100= 7x 2 =14%
Percentage of linkage= 100-14 = 86%
The probability of recombination = 14/100=0.14
In the cross between a1 R+ and a2Rd
Genotype a2Rd has linked together with 86% or with probability 0.86 (86/100 =0.86). Therefore, there genotype would found in 86% offspring. The remaining 14% possible recombination will dissociate in two gametes a1 and R+. The possible chances of having an a1a1R+Rd = 14/2 = 7% or 0.07p.
2. If a1a2 born there would 100% possibility of having carrier of the disease allele:
a1a1 R+R+ x a2 a2 Rd Rd
Gametes
a2 Rd
a1 R+
a1a2R+Rd
Gametes
a2 Rd
a1 R+
a1a2R+Rd
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