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Rajavel-Problem Set 1 #1 and #2 1. A man with vitamin D-resistant rickets with a normal mate has no affected sons and no normal daughters. a. What type of inheritance is observed in this case? Type of inheritance is X-linked dominant trait (NLM, 2010). b. What is the inheritance pattern for both the male and female offspring of female carriers? And what percentage of the offspring will be unaffected? For x-linked affected mother crossed with normal father, their offspring are 1 normal boy: 1 affected girl: 1 affected boy: 1 normal girl. Since the mother carries two copies of the x-chromosome, 50% of the offspring will be normal. c. If you draw a pedigree pattern for this condition, it will be similar to what other type of inheritance pattern? And how would you distinguish one from the other? The inheritance pattern is similar to autosomal dominant conditions; dominate conditions are expressed when the individual has one mutant allele copy (UVM, 2012). Affected males and females have equal probability of passing on trait to offspring (UVM, 2012). Like X-linked dominant, each offspring will have a 50% chance of inheriting the mutant allele- having one normal copy and one mutant copy of the gene (NLM, 2010). In autosomal dominant inheritance, the proportion of affected males should be equal to the number of females affected; also male-to-male transmission is observed (e.g. from father to son) (UVM, 2012). In x-linked dominant inheritance, an affected male will pass the affected trait to all of his daughters and none of his sons (NLM, 2010). 2. Given the recombination frequencies between the three genes (X, Y, Z), which are located on the same chromosome, indicate the order of these three genes. Which of the two genes are closer together? Defend your response Recombination frequency of Gene X and Gene Y is 15% Recombination frequency of Gene X and Gene Z is 25% Recombination frequency of Gene Z and Gene Y is 5% The order of the genes is as follows: Z-Y-X. Evidence from the Morgan Columbia University group study, distances between the genes on the genetic map of a chromosome are measured in map units that directly correspond to the genes’ recombination frequency value (Watson et al., 2014). Thus, genes Z-Y at 5% recombination frequency are the genes that are the closest to each other. Since&recombination&va lue &for&Z‐Y&crossing&is&5%,&we&can&surmise&that&these&two&genes&are&the&clos est&to&each&other.&Next,&we&know&that&the&furthest&distance& is&25%&from&the&X‐Z&recombina nt&value.&We&also&know&that&one&of&the&X‐Z&genes&has&to&be&on&t he& end&since& Z‐Y&is&the&sh ort est&distance;&from&this&information,&Z&and&X&have&to&be&on&the&flanking&ends.&Lastly,&the&gene&in&the&middle


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Johns Hopkins AS 410 602 - Problem Set 1

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