Concept explainers
To analyze:
The presence of cheek and chin dimples is dominant to the lack of dimples, and the ability to taste the compound PTC is dominant to the failure to taste the compound, in humans. Both these qualities are autosomal and unlinked. The allele frequencies for dimples are
Calculate the frequency of genotypes for each gene and the frequency of each
Estimate the expected frequencies of the four probable phenotype combination: Dimpled nontasters, dimpled tasters, undimpled tasters, and undimpled nontasters.
Introduction:
Hardy-Weinberg explains that allelic frequency in a population can be maintained but population should fulfill some criteria i.e. there should be random mating, no mutation, large
Hardy-Weinberg mathematical equation to calculate the allelic and genotypic frequency is
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Genetic Analysis: An Integrated Approach (2nd Edition)
- In Ayrshire cattle, the spotting of the animals can be either red and white or mahogany and white. The mahogany and white phenotype is caused by the allele SM. The red and white phenotype is controlled by the allele SR. The table below shows the relationship between genotype and phenotype for males and females:arrow_forwardMale-pattern baldness is an example of a sex-influenced trait. The baldness allele, which causes hair loss, is influenced by the hormones testosterone and dihydrotestosterone, but only when levels of the two hormones are high. In general, males have much higher levels of these hormones than females, so the baldness allele has a stronger effect in males than in females. (B-baldness; N-full or normal) hair) A normal man marries a normal woman. What can be the possible genotypes of the parents respectively? * a. BN x BN b. NN x BN c. NN x bb d. NN x BB e. BN x bb f. BB x BBarrow_forwardA pedigree analysis was performed on the family of a man with schizophrenia. Based on the known concordance statistics, would his MZ twin be at high risk for the disease? Would the twins risk decrease if he were raised in an environment different from that of his schizophrenic brother?arrow_forward
- Using the HardyWeinberg Law in Human Genetics Suppose you are monitoring the allelic and genotypic frequencies of the MN blood group locus (see Question 2 for a description of the MN blood group) in a small human population. You find that for 1-year-old children, the genotypic frequencies are MM = 0.25, MN = 0.5, and NN = 0.25, whereas the genotypic frequencies for adults are MM = 0.3, MN = 0.4, and NN = 0.3. a. Compute the M and N allele frequencies for 1-year-olds and adults. b. Are the allele frequencies in equilibrium in this population? c. Are the genotypic frequencies in equilibrium?arrow_forwardUsing the two Hardy-Weinberg equations, calculate the allelic and genotypic frequencies. Tongue rollers [R_-]= 840, Non rollers [rr] = 160, Total = 1000 1. What are the phenotypic frequencies for tongue rollers? 0.84 2. What are the phenotypic frequencies for non-tongue rollers? 0.16 3. What is the allelic frequency for r? 0.4 4. What is the allelic frequency for R? 0.6 5. What is the homozygous dominant frequency? 6. What is the heterozygous frequency? 7. What is the homozygous recessive frequency?arrow_forwardAlbinos produce very little of the pigment melanin in their skin and hair. Albinism is inherited as a homozygous recessive trait. In North America, about 1 in 20,000 people are albinos. a. What is the frequency of the dominant allele for albinism?b. What is the frequency of albinos?c. What is the frequency of heterozygotes?arrow_forward
- PTC, or phenylthiocarbamide, is a chemical that, to humans, either tastes bitter or has no taste at all. The ability to taste the bitterness of PTC is controlled by a single gene that codes for a bitter taste receptor on the tongue. Tasters, or individuals that can taste the bit- terness of PTC, have at least one copy of the dominant allele, T. Nontasters, or individuals who cannot taste the bitterness of PTC, must have two copies of the recessive allele, t. The numbers of tasters and nontasters in the human population are shown in the table below. A B Phenotype с Tasters Select all that apply - There are 2 correct options. D Nontasters Total If the population is in Hardy-Weinberg equilibrium, which of the following identifies the correct genotype frequencies for the population? Number of Individuals 1,762 338 2,100 A total of 2.56 percent of the population are homozygous dominant. A total of 13.44 percent of the population are heterozygous. A total of 48 percent of the population are…arrow_forwardAchondroplasia is a hereditary condition caused by a dominant allele in humans (dominant allele A). This disorder affects bone growth specifically in long bones of the upper and lower limbs by preventing the ossification of bones from cartilage. Determine the genotypes of the parents and offspring for the following family scenarios in a and b below. One parent with the Achondroplasia phenotype and a normal parent have 2 children. Both children have the Achondroplasia phenotype NOTE: You must draw a Punnet square to determine the possible genotypes of te children. When two alternative genotypes are possible for an individual, indicate both.arrow_forwardPlease choose the correct answer. If narrow sense heritability is equal to 1, which of the following is most likely responsible for phenotypic variations? a. additive variance b. dominance variance c. environmental variance d. interactive variancearrow_forward
- Suppose a geneticist is using a three-point test cross to map three linked rabbit morphology and behavioral mutations called si, sf, and H. The gene si is associated with the silky fur phenotype, and sf is associated with the short-footed phenotype. Both si and sf are recessive mutations with respect to wild type. H is a dominant mutation that confers the hyper phenotype. The geneticist first crosses true-breeding hyper rabbits to true-breeding silky fur, short-footed rabbits. Next, the geneticist backcrosses the F₁ progeny to the silky fur, short-footed parents, and obtains the results reported in the table. Phenotype hyper silky fur, short-footed short-footed silky fur, hyper silky fur short-footed, hyper silky fur, short-footed, hyper wild type Place the genes in the correct order in the chromosome. LLIIN HEL H Number 815 807 175 169 4 5 27 29 Answer Bank si sfarrow_forwardSuppose a geneticist is using a three-point test cross to map three linked rabbit morphology and behavioral mutations called si, It, and Le. The gene si is associated with the silky fur phenotype, and It is associated with the long-tailed phenotype. Both si and It are recessive mutations with respect to wild type. Le is a dominant mutation that confers the lethargic phenotype. The geneticist first crosses true-breeding lethargic rabbits to true-breeding silky fur, long-tailed rabbits. Next, the geneticist backcrosses the Fj progeny to the silky fur, long-tailed parents, and obtains the results reported in the table. Phenotype Number lethargic 815 silky fur, long-tailed 807 long-tailed 177 silky fur, lethargic 179 silky fur 7 long-tailed, lethargic 5 silky fur, long-tailed, lethargic 29 wild type 29 Place the genes in the correct order in the chromosome. Answer Bank si It Learrow_forwardSome heterozygotes express a phenotype that is intermediate between the dominant and recessive phenotype. For example, in 4 o’clock flowers the gene for red pigmentation is dominant and the gene for white pigmentation is recessive. However, heterozygotes are pink. The dominant allele does not completely mask expression of the recessive allele: it is incompletely dominant. a. By observing flower color in 4 o’clock flowers, is it possible to unambiguously determine the genotype? YES/NO. Explain your answer. b. Is the same true for flower color in snow peas? YES/NO Why or why not?arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning