Question

Neurofibromatosis is an autosomal dominant disease. The frequency of the recessive allele is 0.98. In population...

Neurofibromatosis is an autosomal dominant disease. The frequency of the recessive allele is 0.98. In population of 2250 how many individuals would you expect to not have the disease?

Homework Answers

Answer #1

If we consider p as frequency of dominant allele and q as frequency of receccive allele. Then it has given that q=0.98.

According to hardy Weinberg equilibrium, p+q=1 ,so p=1-0.98 =0.02

We have p2+2pq+q2=1 where p2= frequency of homozygous dominant individuals, q2=frequency of homozygous recessive individuals ,2pq=frequency of heterozygous individuals. Since it is autosomal dominant disorder, homozygous recessive and heterozygous individuals will be affected.and homozygous recessive individuals will be unaffected.

Here frequency of unaffected homozygous recessive individuals =q2=0.982=0.9604.

So number of unaffected individuals = 0.9604×2250 =2160.9 that is approximately 2161 individuals.

If the factors affecting hardy Weinberg equilibrium are not present.practically human populations don't meet its conditions and so the answer will be incorrect.

Know the answer?
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
Similar Questions
4 In a population of peas, the frequency of the dominant allele for a yellow seed...
4 In a population of peas, the frequency of the dominant allele for a yellow seed is 0.29 and the frequency of the recessive allele for a green seed is 0.71. What would the genotypic frequencies be if the population is in equilibrium? _____ = Frequency of homozygous dominant individuals _____ = Frequency of heterozygous individuals _____ = Frequency of homozygous recessive individuals How many individuals would you expect of each genotype in a population of 350 peas? _____ =...
In humans, the widow’s peak allele is dominant over the straight hairline allele. In a population...
In humans, the widow’s peak allele is dominant over the straight hairline allele. In a population of humans, the frequency of the allele for a widow’s peak is 0.61 and the frequency of the allele for a straight hairline is 0.39. What would the genotypic frequencies be if the population is in equilibrium? _____ = Frequency of homozygous dominant individuals _____ = Frequency of heterozygous individuals _____ = Frequency of homozygous recessive individuals How many individuals would you expect of...
3) In general, p = allele frequency of the dominant allele and q = allele frequency...
3) In general, p = allele frequency of the dominant allele and q = allele frequency of the recessive allele. Using the product and/or sum rule(s), use these variables (p and q) to determine the probabilities of an individual in the next generation being: a) Homozygous dominant (A dominant allele (A) AND another dominant allele (A))? b) Homozygous recessive (A recessive allele (a) AND another recessive allele (a))? c) Heterozygote (Aa OR aA)? 4a) In a population in Hardy-Weinberg equilibrium,...
5. Tay-Sach disease is due to an autosomal recessive allele t. The disease results in death...
5. Tay-Sach disease is due to an autosomal recessive allele t. The disease results in death of tt individuals before reproductive age. The relative fitnesses of the genotypes are TT=1, Tt=1, tt=0. a. What type of selection is occurring on this locus, and what do you expect will happen to the f(t) over time? b. Explain in general terms how Mutation-Selection Balance can keep the t allele in the population at a constant (equilibrium) frequency.
Consider a gene with a dominant and recessive allele in a population conforming to the Hardy-Weinberg...
Consider a gene with a dominant and recessive allele in a population conforming to the Hardy-Weinberg conditions. Assuming the frequency of the recessive allele in the entire population is q, show that the frequency of the recessive allele in individuals with the dominant phenotype is q/(1+q).
Consider a gene with a dominant and recessive allele in a population conforming to the H-W...
Consider a gene with a dominant and recessive allele in a population conforming to the H-W conditions. Assuming the frequency of the recessive allele in the entire population is q, show that the frequency of the recessive allele in individuals with the dominant phenotype i q/(1 +q). Be sure to show your work.
In mice, a mahogany coat color allele is dominant over the tan coat color allele. In...
In mice, a mahogany coat color allele is dominant over the tan coat color allele. In a population of mice, the frequency of the allele for mahogany coat color is 0.19 and the frequency of the allele for tan coat color is 0.81. What would the genotypic frequencies be if the population is in equilibrium? ___=Frequency of homozygous dominant individuals ___=Frequency of heterozygous individuals ___=Frequency of homozygous recessive individuals How many individuals would you expect of each genotype in a...
Among U.S. Caucasians, the frequency of the CF- allele causing the recessive autosomal disease cystic fibrosis...
Among U.S. Caucasians, the frequency of the CF- allele causing the recessive autosomal disease cystic fibrosis is 0.02. What is the Δq for the CF allele after one generation of random mating and selection?
The frequency of a dominant allele A in a population is P = 0.4, and the...
The frequency of a dominant allele A in a population is P = 0.4, and the rate of a recessive allele a at the same locus is q = 0.6. Assuming that allele a is fatal in double dose (individuals with genotype aa die in the fetus). a) What happens with the frequencey to the a-allel in the long term? b) Why will a never completely be gone if the population has infinite size?
If the allele frequency of a dominant allele is 0.6, then the genotype frequency of individuals...
If the allele frequency of a dominant allele is 0.6, then the genotype frequency of individuals that are homozygous for the recessive trait is: 60%. 48%. 40%. 36%. 16%.
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT