My answer key says its 7.5 percent but I don\'t think that considers recombinati

Question

My answer key says its 7.5 percent but I don't think that considers recombination if she is a carrier.

2. The human gene for hemophilia is on the X chromosome. Below is a pedigree from a family afflicted with hemophilia. Blackened symbols indicate that the person has hemophilia. To help with genetic diagnosis, a probe that detects an RFLP (restriction fragment length polymorphism) on the X chromosome is used. This probe detects either a 7 kb restriction enzyme fragment or 3 kb and 4 kb restriction enzyme fragments. The RFLP pattern for all the members of the pedigree is shown. The recombination distance between the RFLP and the hemophilia locus is 15%. What information could be given to the woman designated with the arrow as to the likelihood of her first son having hemophilia? Show your work and circle your answer. 7 kb 4 kb 3 kb

Explanation / Answer

As it can be seen in the figure, the affected individual sample generates 2 fragments (4kb and 3 kb ), while the unaffected healthy individual sample generates one 7 kb fragment.

However, sample from the designated woman shows all three (7kb, 4kb, 3kb) fragments. Hence, this indicates that she is the carrier.If the female silently carries the deficient gene, her son will have a 50% chance of inheriting that gene from her , hence; there is 50% chance of her producing an affected son. Since the recombination distance is 15%, if she mates with a healthy male, her first son may have 7.5%, (15/2=7.5 %) chance of having haemophilia.

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