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Author13 Posts
  #1

If an X-linked dominant disorder affects 1/100 males in a population, what is the gene frequency for the disorder in the population?





  #2

how geroo? can u solve it for me pls.





  #3

Because of X-linked dominant, so the affect on males and female are the same (except X inactivation in females). therefore frequency in the whole population is 2/100.

Any idea?



  #4

I agree with it : 2pq=1/100+ 1/100=2/200, then q=1/100


  #5

ok, I am confused with this one...why would you add the 2?

Wouldn't the frequency in the whole population simply be 2pq= 1/100?

Huh??confused



  #6

ya doc2378, agree with u, q2=q in this case, as x linked dominant condition, hence 1/100


  #7

If it was an X-linked recessive the gene frequency of the population would be the same as the gene frequency of the male, that is 1/100. But the Q says it is an X-linked Dominant. shdnt the answer be different, now that the females are also involved?


  #8

let me explain. if males have it 1 in 100 that is males with one allele(having one x is 1 in 100)they cannot have another x. so all males are covered. now females having one x diseased(carriers) is same frequency as males have one diseased x. but they are not diseased so should not count.. now females homozygous is chance on one x diseased multiply chance of another x diseased. 1 / 10000.very small.so the occurence of disease is 1/100. please correct me.


  #9

thanks geroo.

ok same Q, anthr scenario, if you substitute the words 'X-linked dominant' with 'autosomal dominant' ...wht wud the answer be?

msyamp, if iam not mistaken u seem to be describing x-linked recessive, no? i mean if the female gets one X its enuff for her to be affected, right? there wudnt be a 'carrier female' in an x-linked dominant, i think.



  #10

If an autosomal dominant disorder affects 1/100 males in a population, what is the gene frequency for the disorder in the population? <?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />

In autosomal dominant: prevalence of disease is calculated through: P2+2pq, (because AA and Aa are all phenotypically diseased) only q2 (whose genotype is aa) is disease free.



  #11

well in autosomal dominant too, the gene frequency would still be 1/100. this is because the pt with autosomal dominant disease has only one gene, and he expresses the disease, and soo here the mutation fre=gene fre

same for x linked dominant, u just need one gene to express the disease, hence the gene frequency, would be same as mutation fre.

but for autosomal recessive diseases, gene frequency would be q2 and carrier frequency would be 2pq


by the way as prof, lionel raymon in kaplan audios, said, u will only be asked hardy weinberg law in autosomal recessive and x linked recessive, so chillwink



  #12

Well guys ,It is nothing but the gene frequency of X chromosome.men has one x chromosome

hence gene frequency is 1/100 it self





  #13

X-LINKED DOMINENT,,,CHRO.X IS DOMINENT=EXPRESS SYMPTOMS=AT THE SAME TIME IT IS ACTS AS CARRIER(PREVILENCE =CARRIER CASES IN POPULATION IN GENERAL TERM )SO FREQUENCY IN X-LINK. DOMINANT =CARRIERS=1/100
NOTE;YOU CAN NOT ADD 1/100 OF FEMAL COZ WE DEAL WITH X-DOMINANT ,,CHROMOS.. CASES,,,





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