 USMLE Forum Step 1 Step 2 CK Step 2 CS Matching & Residency Step 3 Classifieds

 <<   < *  Step 1   * >   >>

 * can someone post hardyweinbergs for me #147598 calmer - 12/23/06 04:03 i just dont get this....please post Report Abuse

 * Re:can someone post hardyweinbergs for me #596302 iced - 12/23/06 06:25 In a certain population, the frequency of color-blind males is 1 in 100. Assuming that the population is in Hardy-Weinberg equilibrium at this locus, the frequency of color-blind females is approximately A. 0.0001 B. 0.0005 C. 0.01 D. 0.02 Explanation: The correct answer is A. Color blindness is an X-linked recessive trait. A male is hemizygous for the X chromosome, and thus has only one copy of each trait. The frequency of an X-linked recessive in males is thus equal to the frequency of the allele in the population. From this, we know that q = 0.01 and p = 0.99. A female has two copies of each gene on the X chromosome, so the equation for Hardy-Weinberg equilibrium is the same as for the autosomal traits. In this case, a homozygous recessive female would occur at a frequency of q2 or 0.0001. Choice B, 0.0005, is incorrect. If you remembered that color blindness was more frequent in males, but did not know how to use the equations to get the true estimate, you might have guessed this answer. Choice C, 0.01, makes the assumption that the trait is autosomal, and so the frequencies of affected males and affected females are equal. Choice D, 0.02, assumes that q = 0.01, and then calculates the frequency of carrier females (2pq). Choice E, 0.025 is also incorrect; it is a distracter. E. 0.025 Report Abuse

 * Re:can someone post hardyweinbergs for me #596393 calmer - 12/23/06 11:50 thank u still tried still couldnt get it, Report Abuse

 Page 1 of 1

 [<]   [Last >>]  Web USMLEforum.com

 Step 1 Step 2 CK Step 2 CS Matching & Residency Step 3 Classifieds