Biology 11 U
posted by Caitlin .
In the U.S. 1 out of 10 000 babies is born with phenylketonuria (PKU). The disease is caused by a recessive allele. The frequency of individuals in the U.S. population born with PKU corresponds to q squared.
What is the given information? (e.g. p squared = 99999/10000)
Respond to this Question
Similar Questions

Genetics
Phenylketonuria is a metabolic disease of humans that results from an autosomal recessive gene. If the frequency of phenylketonurics in the population is 9/10,000, what is the probability that two normal individuals will produce a … 
biology
1. A population which meets the hardy weinberg requirements: a. Evolves b. is small and usually isolated c. has allele frequency in equilibrium d. changes genotypic distribution from generation to generation e. always has 75% A and … 
Biology
Populations are frequently NOT in equilibrium due to selection or other evolutionary forces. Imagine a population of 100 individuals. Fortynine of the individuals are homozygous dominant for beadyeye gene, 9 are heterozygous and … 
Science
Phenylketonuria (PKU) is a human metabolic disease caused by the recessive allele k. If 2 heterozygous carriers of the allele marry and plan a family of five children: a) what is the chance that all their children will be unaffected? 
Biology
you are studying a singlegene locus with two alleles in a population that is in HardyWeinberg equilibrium. Examination of a large sample of individuals from the population reveals there are six times as many heterozygotes as there … 
Biology (Please Help: Urgent)
The HardyWeinberg formulas allow scientists to determine whether evolution has occurred. Any changes in the gene frequencies in the population over time can be detected. The law essentially states that if no evolution is occurring, … 
bio
The probability that their baby will be a carrier of the gene for PKU is (A) ___3/4_______. The probability that their baby will be affected by PKU is (B) __________. If both genotypes were Pp, the probability that their baby would … 
biology Please Help
The probability that their baby will be a carrier of the gene for PKU is (A) __________. The probability that their baby will be affected by PKU is (B) __________. If both genotypes were Pp, the probability that their baby would be … 
Math
1. Cystic fibrosis (CF), a lifethreatening lung disease, is caused from the inheritance of a recessive allele from both parents for the CFTR gene. In order for this disease to manifest itself, one needs to be homozygous recessive … 
Math. (Didn't understand please help!)
Suppose 20% of babies born are born early, 50% are born on time, and 30% are born late. A nurse uses a randomnumber table to find the experimental probability that of 5 births, at least 1 baby will be born early. The digits 0 and …