#4: Ch. 20- Hardy-Weinberg The ApoE gene has three common alleles: E2, E3 and E4. A population survey of Reno reveals the following frequencies: 0.10 for E2, 0.15 for E3, 0.75 for E4. Assume random mating is occurring. If the E4 allele confers resistance to hepatitis C (when 1 or 2 copies are present), what proportion of the population is NOT resistant to hepatitis C?

#4: Ch. 20- Hardy-Weinberg The ApoE gene has three common alleles: E2, E3 and E4. A population survey of Reno reveals the following frequencies: 0.10 for E2, 0.15 for E3, 0.75 for E4. Assume random mating is occurring. If the E4 allele confers resistance to hepatitis C (when 1 or 2 copies are present), what proportion of the population is NOT resistant to hepatitis C?

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter21: Microevolution: Genetic Changes Within Populations

Section: Chapter Questions

Problem 8TYK: If the genotype frequencies in a population are 0.60 AA, 0.20 Aa, and 0.20 aa, and if the...

See similar textbooks

Similar questions

If the genotype frequencies in a population are 0.60 AA, 0.20 Aa, and 0.20 aa, and if the requirements of the HardyWeinberg principle apply, the genotype frequencies in the offspring generation will be: a. 0.60 AA, 0.20 Aa, 0.20 aa. b. 0.36 AA, 0.60 Aa, 0.04 aa. c. 0.49 AA, 0.42 Aa, 0.09 aa. d. 0.70 AA, 0.00 Aa, 0.30 aa. e. 0.64 AA, 0.32 Aa, 0.04 aa.
Activity 1 Write down the correct biological term for each of the following descriptions in the space provided. DESCRIPTION TERM 1.1 A cross involving a single characteristic 1.2 A cross involving two characteristics A representation that traces the transmission of a characteristic over many generations of a family Allele that influences the phenotype only when in the homozygous combination but not in the heterozygous condition 1.3 1.4 1.5 Allele that influences the phenotype whether in the homozygous or heterozygous combination 1.6 Alternative forms of a gene found at the same locus 1.7 Characteristics controlled by alleles of a gene that are located on the sex chromosomes Cross involving alleles of a gene that are equally dominant, leading to a third phenotype which is a combination of the other two phenotypes 1.8 1.9 Cross involving alleles that are neither dominant or recessive, leading to an intermediate phenotype 1.10 Cross involving characteristics controlled by dominant and…
Q5.1. Which of the following is FALSE? If a genetic disease reduces fertility and the allele that causes the disease offers no other advantage, the allele will likely eventually disappear, due to natural selection. Natural selection does not favor individuals who are homozygous for the sickle-cell allele, because these individuals typically die before they are old enough to reproduce. Individuals who are heterozygous HbA/ HbS are protected from malaria, and this is why sickle-cell disease persists in wetter, mosquito-prone regions in Africa. OIn regions where malaria does not occur, individuals who are heterozygous HBAIHBS have a fitness advantage over those who are homozygous for the normal hemoglobin allele (HbA). Submit Q5.2. AFTER malaria is cured, the frequency of the HbS allele should decrease in regions with lots of mosquitoes because: People will no longer die from sickle-cell disease in these regions. Having one copy of the HbS allele will no longer be advantageous in these…
1: The A and B antigens in humans may be found in water-soluble form in secretions, including saliva, of some individuals (Se/Se and Se/se) but not in others (se/se). The population thus contains “secretors” and “nonsecretors.” (a) Determine the proportion of various phenotypes (blood type and ability to secrete) in matings between individuals that are blood type AB and type O, both of whom are Se/se. (b) How will the results of such matings change if both parents are heterozygous for the gene controlling the synthesis of the H substance (Hh)?
27. Which of the following statements best explains the prevalence and geographic distribution of sickle cell anemia allele (HbS). The homozygotes for normal hemoglobin protein are always at an advantage in all environments. Homozygous recessive individuals are selected for in malarial environments. Heterozygous individuals are at a selective advantage in malarial environments. Sickle cell anemia is found in high frequencies only among West Africans and their descendants. none of the above.
Phenotype to Population 1. Describe how the phenotype of individuals with sickle-cell disease influences how common the Hbs allele is in the population: Do individuals with sickle cell disease typically have many offspring? What effect does the fact that individuals with sickle-cell disease do not produce many offspring have on the frequency of the Hbs allele in the population over time? Given this, do you expect the Hbs allele to be common or rare in populations?
Q5.1. Which of the following is FALSE? O If a genetic disease reduces fertility and the allele that causes the disease offers no other advantage, the allele will likely eventually disappear, due to natural selection. Natural selection does not favor individuals who are homozygous for the sickle-cell allele, because these individuals typically die before they are old enough to reproduce. Individuals who are heterozygous HbAlHbS are protected from malaria, and this is why sickle-cell disease persists in wetter, mosquito-prone regions in Africa. O In regions where malaria does not occur, individuals who are heterozygous HbAIHbS have a fitness advantage over those who are homozygous for the normal hemoglobin allele (HbA).
2. 235O 4) Q7. Haemophiliacs possess a non-functional form of the gene responsible for the production of blood clotting factors. Shown below is the occurrence of haemophilia in one family. = male = female = male haemophiliac 7. 8. 5. 9. 3. 11 12 Usingthe following symbols: H = dominant allele h = recessive allele 1) State the genotypes of the following individuals. Individual Genotype 1. 6. 2) On the basis of the information provided, is the inheritance of haemophilia: (i) autosomal or sex-linked? (ii) dominant or recessive? 3) State the probability of individual 8 being a carrier of haemophilia. 4) Explain why only females can be carriers of haemophilia.
l Metro by T-Mobile 1 2:29 PM • 16% I In flies, long wings are a dominant trait, and short wings are a recessive trait. Medium wings are the heterozygous trait. Based on this information, if a homozygous long-winged fly is crossed with another a heterozygous'fly, their offspring will have which percentages for long, medium, and short wings? Assume random chromosome segregation. 17) A) 25% long, 75% medium, 0% short B) 50% long, 25% medium, 25% short 50% long, 50% medium, 0% short D) 100% medium, 0% long, 0% short
2) The year is 2050. The climate has continued to warm. Areas in northern Canada and Alaska are without snow or ice; the average annual temperature has risen and previously snow-covered areas are now shrub land and some temperate forests. What effect would such a climatic change have on the gene pool of the rabbits? All BUT ONE could apply. A) The allelic frequency has most likely shifted to 85% GG and 15% gg. B) The dominant allele in the gene pool would become G, the previously rare allele. C) The allele for white, g, would be scarce and perhaps nonexistent in the population. D) White rabbits would be at a disadvantage in terms of camouflage and would likely fall prey to predators.
A.) Assuming there is random mating, find the gene frequencies, genotypic frequencies, and the proportion of individuals with the different blood types (Type A, B, AB and O) if the population is composed of the following: Туре А - 80 Туре АВ - 35 Туре В - 90 Type o - 195 Type O = 195/400 = 0.488 Type A = 80/400 = 0.200 Type AB = 35/400 = 0.087 Type B = 90/400 = 0.225 Gene frequency: Genotypic frequency: f(0)= r=r2 =0.4882 f(AA) = p2=0.22 f(AA)=0.04 f(0)=r= 0.488 f(AO) =2pr =2(0.2)(0.488) =0.195 E(A)=p=1-(g+r) f(BB) = q2 =0.4882=0.238 =1-(0.225+0.488) f(BO) =2qr=2(0.225)(0.488)=0.220 f(A)=p =0.287 f(AB)=2pq=2(0.200)(0.488)=0.195 f(o0)=r2=0.4882=0.488 f(B)=q=1-(p+r) f(B)=q=1-(0.200+0.488) f(B)=q=0.312 Proportion: 1.) TYPE A = AA+AO 4.) TYPE O = 00 (p2+2pr) x total=(0.04+0.195) x 400 = 94 (r2) x total=(0.488) x 400 = 195.2 2.) TYPE B = BB+BO (q2+2qr) x total=(0.238+0.220) x 400 = 183.2 3.) TYPE AB = AB (2pq) x total=(0.195) x 400 = 78
3 (b). The S-s antigen system in human is controlled by two codominant alleles, S and s. In a group of 3146 individuals, the following genotypic frequencies were found: 188 SS, 717 Ss, and 2241 ss. Calculate the frequency of s. O q = 0.8263 O q = 0.9402 O q = 0.4701 O q = 0.1438