Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN: 9780134580999
Author: Elaine N. Marieb, Katja N. Hoehn
Publisher: PEARSON
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- Sickle cell anemia is caused by a recessive allele at a single gene. As we discussed in class, being a homozygote for the sickle cell allele is almost always lethal, but heterozygotes tend to be resistant against malaria although they have a mild form of anemia. Because of this heterozygote advantage, the allele for sickle cell anemia has a frequency of more than 10% in some human populations. How would present allele frequencies of the sickle cell allele change, if there was no heterozygote advantage or disadvantage (that is, that heterozygotes would be identical to ‘normal’ homozygotes – no malaria resistance, no anemia)? How would the change in sickle cell allele frequencies compare to scenario a (extirpation of malaria)arrow_forwardPhenylthiocarbamide (PTC) is an organic compound that either tastes extremely bitter or is tasteless, depending on the genotype of the taster. The tasting allele (T) is dominant over the non-tasting allele (t). Normal skin pigmentation (N) is dominant over albinism (n). (It is important to note that this example of PTC tasting cannot be explained with simple genetics, but for the purpose of this question, assume that it can.) A normally-pigmented woman who cannot taste PTC has an albino father. The woman married an albino man who can taste PTC and whose mother cannot taste PTC. For the married couple, 1. correctly identify their genotypes, 2. complete a digital Punnett square showing the expected offspring of the couple, and 3. identify the genotypic and phenotypic ratios of their offspringarrow_forwardTay-Sachs disease is a recessive genetic disease. Individuals with this disease rarely survive past the age of four. In the general population, approximately 1 person in 300 carries the allele for this disease. However, in some populations, including the Irish Americans, the Ashkenazi Jews, and the Cajuns from Louisiana, the proportion of Tay-Sachs carriers is much higher (1 in 27 to 1 in 50) than in other populations. Such high frequency of an otherwise rare allele is expected when Question 24 options: populations experienced disruptive selection populations were founded by a small number of settlers the allele is advantageous at the heterozygous state populations have higher than average mutation rates populations experienced stabilizing selectionarrow_forward
- Consider the following scenario: A man without freckles (freckles are a dominant trait, determined by the dominant allele “F”) is a carrier of cystic fibrosis (recall that CF is a recessive trait, determined by the recessive allele "a"), mates with a woman whose genotype is heterozygous for freckles and is also a carrier of CF. Assume the two genes in question are in different chromosomes and, therefore, assort independently. Complete the following Punnett square to generate the offspring probabilities from this couple, by entering the genotypes of the parents, the gametes, and the offspring for the two traits described above. (4) Father’s genotype ffAa Mother’s genotypeFfAa ● Sperm: fA ● Sperm: fa ● Sperm: fA ● Sperm: fa ● Egg: FA ● Egg: fA ● Egg: Fa ● Egg: fa What is the probability for this couple to have a child with freckles? What is the…arrow_forwardThere may be a number of possible alleles for a given gene within a population. In a multiple allele system the dominance relationships between the various alleles must be considered. One of the more familiar examples of a multiple allelic system is that of the human ABO blood group. The gene involved codes for a protein located on the outside of red blood cell membranes. Three alleles (IA, IB and i) determine whether the protein is present or absent and which form of the protein (if any) is present. The A and B alleles code for the A and B forms of the protein and are co-dominant with each other. The O allele (i) codes for no protein and is recessive to both A and B alleles. This means there are four possible phenotypes (blood types: A, B, AB, and O). This also means there are 6 possible genotypes: IAIA, IBIB, IAi, IBi, IAIB and ii. If you have the letter ‘O’ anywhere in any of your genotypes, you are doing it wrong. 6- A couple with the following blood types: the man has type AB and…arrow_forwardFor practice, let's work with blood groups. Three alleles exist for human blood types. They were not numbered, but instead marked with the letters for A, B and O. The labeling convention is : IA, IB, and Ii. IA and IB are codominant alleles, whereas Ii is the recessive allele for type O. Humans are diploid; every individual carries only two alleles. For the following question list all 6 possible Blood group genotypes. The first has been done for you below:arrow_forward
- In humans, there is a gene that influences the ability to smell the thioesters present in urine after asparagus is eaten. The ability to smell these compounds is an autosomal dominant trait. A is dominant and confers the ability to smell these thioesters a is recessive and does NOT confer the ability to smell these thioesters p = freq (A) = frequency of allele A in the population = number of A alleles/ total number of alleles %3D q = freq (a) = frequency of allele a in the population where p + q = 1 and, therefore, p? + 2pq + q? = 1 and %3D %D p? = freq (AA) = frequency of genotype AA = number of AA individuals/ total number of individuals %3D 2pq = freq (Aa) = frequency of genotype Aa q? = freq (aa) = frequency of genotype aa Phenotype Number of People Able to smell thioesters 378 Unable to smell thioesters 1323 Total 1701 1. Calculate p and q for this population. 2. Calculate p?, 2pq, and q? for this population.arrow_forwardSimilar or matching HLA alleles are important for organ/stem cell donation How would a mismatch between donor and recipient affect the recipient differently if it was a hematopoietic stem cell (bone marrow) transplant than if it was an organ, such as a kidney, that was transplanted. Why?arrow_forwardPlease help with the question belowarrow_forward
- Goatee A Bearding in goats is inherited as an autosomal trait determined by two alleles, B1 and B2. Interestingly, the bearded trait is recessive in females, while it is dominant in males. In both sexes, homozygotes for the B1 allele are beardless. Which of the following is true? (more than one correct answer possible) Males with the genotype B1B1 are beardless Females with the genotype B1B2 are beardrd Males with the genotype B1B2 are bearded Females with the genotype B1B1 are bearded Females with the genotype B2B2 are beardless. Males with the genotype B2B2 are beardedarrow_forwardAn allelic series determines coat color in rabbits: C (full color), cch (chinchilla, gray color), c* (Himalayan, white with black extremities), and c (albino, all white). The C allele is dominant over all others, cch is dominant over c and c, ch is dominant over c, and c is recessive to all other alleles. This dominance hierarchy can be summarized as C> cch > c^ > c. The rabbits in the table are crossed and produce the progeny shown. Cross Parents Offspring А full color x albino 50% full color, 50% albino В Himalayan x albino 50% Himalayan, 50% albino C full color x albino 50% full color, 50% chinchilla D full color × Himalayan 50% full color, 25% Himalayan, 25% albino E full color x full color 75% full color, 25% albino Match the parental genotypes to the letter corresponding to the appropriate cross listed in the table. A В C D E Answer Bank Ccch x cc Cc x cc СС х сс hch x cc Сс х сс Сс х Сс chc x ccarrow_forwardSuppose we define the Alzheimer's disease phenotype as being diagnosed with the disease by age 75 years. In the human population, there are three alleles of the ApoE gene: e2, e3, e4. They form an allelic series such that: 70% of 75 year olds with the e4/e4 genotype have the Alzheimer's phenotype 60% of 75 year olds with the e3/e4 genotype have the Alzheimer's phenotype 40% of 75 year olds with the e3/e3 genotype have the Alzheimer's phenotype 30% of 75 year olds with the e2/e4 genotype have the Alzheimer's phenotype 10% of 75 year olds with the e2/e3 genotype have the Alzheimer's phenotype If I have the e4/e4 genotype and my wife has the e2/e3 genotype, what is the probability that our child will have Alzheimer's by age 75. Explain your reasoning.arrow_forward
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