Question
hi this a hardy weinberg problem (H-W) that i need help understanding, please help me know how to solve it!
3. 1 in 1700 US Caucasian newboms have cystic fibrosis. C is the normal allele, dominant over the recessive c. Individuals mu
0 0
Add a comment Improve this question Transcribed image text
Answer #1

Answer 3)

It is given that 1 out of 1700 people new borns have cystic fibrosis.

Percentage of cystic fibrosis in the population is (1/1700)×100= 0.0588235294% or 0.059 % (round off up to three decimal places)

The genotypic frequency of cystic fibrosis (cc) individual is 1/1700=0.000588235294

Allelic frequency of c is

​​​​​√ (1/1700)=0.0242535625 or 0.024 (round off up to 3 decimal places.)

Allelic frequency of C is 1-c = 1-0.024= 0.976

please rate.

Add a comment
Know the answer?
Add Answer to:
hi this a hardy weinberg problem (H-W) that i need help understanding, please help me know...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • 2.3 Problem 3 The Hardy-Weinberg equation is useful for predicting the percent of a hu- man...

    2.3 Problem 3 The Hardy-Weinberg equation is useful for predicting the percent of a hu- man population that may be heterozygous carriers of recessive alleles for certain genetic diseases. Phenylketonuria (PKU) is a human metabolic dis- order that results in mental retardation if it is untreated in infancy. In the United States, one out of approximately 10.000 babies is born with the disor- der. Approximately what percent of the population are heterozygous carriers of the recessive PKU allele? If you...

  • The Hardy-Weinberg principle and its equations predict that frequencies of alleles and genotypes remain constant from generation to generation in populations that are not evolving

    .1. The Hardy-Weinberg principle and its equations predict that frequencies of alleles and genotypes remain constant from generation to generation in populations that are not evolving. What five conditions does this prediction assume to be true about such a population? a._______  b._______  c._______  d._______  e._______  2. Before beginning the activity, answer the following general Hardy-Weinberg problems for practice (assume that the population is at Hardy-Weinberg equilibrium).a. If the frequency of a recessive allele is 0.3, what is the frequency of the dominant...

  • Q3 Question 3 1 Point In a Hardy-Weinberg situation, suppose that cystic fibrosis (a rare genetic...

    Q3 Question 3 1 Point In a Hardy-Weinberg situation, suppose that cystic fibrosis (a rare genetic disorder caused by having two recessive alleles) occurs in approximately 1 out of 1600 births in Canada. What is the estimated proportion of the population that would be carriers for the disorder? Q4 Question 4 2 Points For some genetically linked disorders, the survival of the heterozygous individual may be less than that of a homozygous dominant individual (while homozygous recessive individuals still have...

  • Hardy-Weinberg Practice Problems: You need to list equations used and provide steps of problem solving. Providing...

    Hardy-Weinberg Practice Problems: You need to list equations used and provide steps of problem solving. Providing answer itself is not enough for full grade. 1. You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Calculate the frequency of the heterozygous genotype, homozygous dominant genotype and homozygous recessive genotype. 2. You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 49%....

  • EXERCISE 6 HARDY-WEINBERG EQUILIBRIUM Work in a small group or alone to complete this exercise. In...

    EXERCISE 6 HARDY-WEINBERG EQUILIBRIUM Work in a small group or alone to complete this exercise. In human population X consider the simple Mendelian trait for freckles. F is the dominant allele and fis the recessive allele. Individuals who are homozygous dominant (FF) or heterozygous (F) for the trait express freckles. Individuals who are homozygous recessive (ff) for the trait do not express freckles. In this population, 30% (0.3) of the alleles are recessive (1) and 70% (0.7) are dominant (F)....

  • please click on the photo to see all of it The basic equations of Hardy-Weinberg Equilibrium...

    please click on the photo to see all of it The basic equations of Hardy-Weinberg Equilibrium p² + 2pq + q2 = 1 p+q=1 p= frequency of the dominant allele in the population 9 = frequency of the recessive allele in the population př= percentage of homozygous dominant individuals q* = percentage of homozygous recessive individuals 2pq - percentage of heterozygous individuals 1. You have sampled a population in which you know that the percentage of the homozygous recessive genotype...

  • Cystic fibrosis is an autosomal recessive disorder. In one population, the frequency of affected individuals (A...

    Cystic fibrosis is an autosomal recessive disorder. In one population, the frequency of affected individuals (A 2 A 2) is 0.0004 Assuming that this population is under Hardy- Weinberg equilibrium, calculate all allele frequencies and genotype frequencies. Enter your numerical answers in the boxes below. Express each answer to four decimal places. Frequency of A1: Frequency of A2: Frequency of homozygous dominant: Frequency of homozygous recessive: Frequency of heterozygous

  • The occurrence of the NN blood group genotype in the US population is 1 in 400,...

    The occurrence of the NN blood group genotype in the US population is 1 in 400, consider NN as the homozygous recessive genotype in this population. You sample 1,000 individuals from a large population for the MN blood group, which can easily be measured since co-dominance is involved (i.e., you can detect the heterozygotes). They are typed accordingly: BLOOD TYPE GENOTYPE NUMBER OF INDIVIDUALS RESULTING FREQUENCY M MM 490 0.49 MN MN 420 0.42 N NN 90 0.09 Using the...

  • 2. Hardy-Weinberg Equilibrium; chi-square test Sickle cell anemia is a recessive disorder caused by a recessive...

    2. Hardy-Weinberg Equilibrium; chi-square test Sickle cell anemia is a recessive disorder caused by a recessive mutation (S) in the b-hemoglobin gene. 80% of affected SS individuals die before reproducing.   Heterozygotes (AS) and homozygous dominant (AA) individuals do not have sickle cell anemia. The table below shows the number of people of each genotype in a population of 100 people in population of Cameroon. Observed # individuals in a Cameroon population AA AS SS 62 32 6 What are the...

  • Hardy Weinberg assignment P + Q = 1 In which P represents frequency of dominant allele...

    Hardy Weinberg assignment P + Q = 1 In which P represents frequency of dominant allele and Q represents frequency of the recessive allele P2 + 2PQ + Q2 =1 P2 represents frequency of homozygous dominant 2PQ represents frequency of heterozygous Q2 represents frequency of homozygous recessive Consider a population of beetles on an island. There are 1000 beetles and they have different colored wings. Black wings are dominant over silver wings. Calculate the allele and the genotypic frequencies in...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT