Human Heredity: Principles and Issues (MindTap Course List)
11th Edition
ISBN: 9781305251052
Author: Michael Cummings
Publisher: Cengage Learning
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Chapter 10, Problem 17QP
Knowing that individuals who are homozygous for the GD allele show no symptoms of galactosemia, is it surprising that galactosemia is a recessive disease? Why?
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Take the example of B-thalassemia, an autosomal recessive genetic disease that particularly affects people from around the Mediterranean. This disease is associated with an anomaly of hemoglobin, a protein essential for the transport of oxygen, which is composed of four chains: two alpha (a) and two beta (B). In case of B-thalassemia, the ẞ chains are produced in insufficient or no quantity in an individual homozygous recessive resulting in insufficient production of overall hemoglobin leading to anemia and other physiological challenges.
The gene that controls the synthesis of the ẞ chains is located on chromosome 11. Here is part of the coding portion of this gene (which controls a total of 146 amino acids and of which you only see the portion 36 to 41) and one of the targeted mutations:
1. Give the sequence of amino acids from the template and mutated strands. 2. What type of point mutation is it?
3. Using the principles of the theory of evolution, explain briefly and generally why…
In humans the allele coding for normal metabolism of the amino acid phenylalanine (P) is dominant over the recessive autosomal allele (p) that causes the disease phenylketonuria, or PKU. Without treatment, this disease is characterized by brain damage due to the failure of the gene to code correctly for the enzyme phenylalanine hydroxylase. If a man marries a woman and both are heterozygous for hemophilia and heterozygous for PKU, what is the chance that they will have a child afflicted by both diseases? Your response MUST include:(a): list the traits involved (example: Color: Blue- B (dominate) Green - b (recessive) -Do not use this example as this is not part of the problem(b): identify if its a monohybrid or dihybrid cross and why (use prompt above) (c): create a punnet square that represents the problem above
Consider two hypothetical recessive autosomal genes a and b, where a heterozygote is testcrossed to a double- homozygous mutant. Predict the phenotypic ratios under the following conditions: (a) a and b are located on separate autosomes. (b) a and b are linked on the same autosome but are so far apart that a crossover always occurs between them. (c) a and b are linked on the same autosome but are so close together that a crossover almost never occurs.
Chapter 10 Solutions
Human Heredity: Principles and Issues (MindTap Course List)
Ch. 10.4 - Prob. 1GRCh. 10.4 - Prob. 2GRCh. 10.7 - Prob. 1EGCh. 10.7 - Prob. 2EGCh. 10 - A couple was referred for genetic counseling...Ch. 10 - A couple was referred for genetic counseling...Ch. 10 - A couple was referred for genetic counseling...Ch. 10 - Many individuals with metabolic diseases are...Ch. 10 - Prob. 2QPCh. 10 - Enzymes have all the following characteristics...
Ch. 10 - Questions 4 through 6 refer to the following...Ch. 10 - Questions 4 through 6 refer to the following...Ch. 10 - Prob. 6QPCh. 10 - Prob. 7QPCh. 10 - Prob. 8QPCh. 10 - a. Compounds A, B, C, and D are known to be...Ch. 10 - b. Compounds A, B, C, and D are known to be...Ch. 10 - a. If an individual who is homozygous for the...Ch. 10 - Prob. 12QPCh. 10 - Suppose that in the formation of phenylalanine...Ch. 10 - If phenylalanine was not an essential amino acid,...Ch. 10 - Phenylketonuria and alkaptonuria are both...Ch. 10 - The normal enzyme required for converting sugars...Ch. 10 - Knowing that individuals who are homozygous for...Ch. 10 - Prob. 18QPCh. 10 - A person was found to have very low levels of...Ch. 10 - If an extra nucleotide is inserted in the first...Ch. 10 - Transcriptional regulators are proteins that bind...Ch. 10 - Prob. 22QPCh. 10 - Prob. 23QP
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- Maple syrup urine disease (MSUD) is an autosomal recessive metabolic disorder that presents in newborns, typically within the first 48 hours after birth. As the name suggests, a key indicator of the disease is the presence of a sweet odor in the urine that smells like maple syrup. Left untreated, MSUD can result in failure of central neurological function and the respiratory system and can be fatal. MSUD is caused by mutations in components of the branched-chain alpha-keto acid dehydrogenase complex (BCKDC). These mutations result in the inability for cells to break down branched-chain amino acids (BCAAs). BCAAs and their byproducts accumulate and are excreted in the urine, giving rise to the maple syrup scent. QUESTIONS: Draw the structures of the three branched-chain amino acids (BCAA). Draw the structures of the three BCAA products released by functional BCKDC. Indicate the original amino acid.arrow_forwardGalactosemia is a metabolic condition, caused by a recessive allele, which prevents affected patients from breaking down certain sugars. In infants galactosemia can lead to a toxic build-up of the by- product galactol, but negative effects can be minimized if the condition is identified early and the child is kept on a diet free from lactose or galactose. John and Maggie both know they are carriers of the galactosemia allele. They have just been married and hope to start a large family. If they have four children, what is the chance that none of them will be affected by the disease? (4 points)arrow_forwardIs the variation associated with lactose tolerance found always on one chromosome, bothchromosomes, or either chromosome in individuals with the trait?arrow_forward
- Shown below is a pedigree for Phenylketonuria (PKU), an autosomal recessive metabolic disorder. The characteristic feature of PKU is severe mental retardation A) What is the probability that individual II-1 is heterozygous for this gene? B) What is the probability that individual III-4 is heterozygous for this gene? C) If individuals III-3 and III-4 were to marry, what is the probability that their child would express PKU?arrow_forwardWhich of the following best explains how individuals who inherit phenylketonuria alleles can avoid the symptoms of this disease (mental impairment, foul smelling urine)?arrow_forwardRegarding Mendelian inheritance in diploid individuals, (Read each statement carefully. Select all of the statements below that are true (that you agree with). Leave any statements that are false (that you do not agree with) un- selected.) a diploid individual receives two copies of every autosome from the previous generation. for every autosomal gene inherited by an individual, both copies can come from one parent. a diploid individual gives two copies of every autosome to a child in the next generation. to be diploid means that two independent genes are specified in the individual's genotype.arrow_forward
- Salim and Sara are contemplating having children, but Salim’s brother has galactosemia and Sara has a sister who has three children, one of whom is affected. Sara dad has no history in his family of any sign of the diease and it is assumed to be homozygous normal. What is the probability that salim and saras first child will have galactosemia?arrow_forwardProduce a Punnett square to illustrate the dihybrid cross described below: There are two common alleles for the TAS2R38 gene on Chromosome 7. This gene encodes a seven-transmembrane G-protein coupled receptor. This receptor controls the ability to taste glucosinolates. Phenylthiocarbamide (PTC) is a synthetic glucosinolate. The recessive TAS2R38 allele produces a non-functional receptor. The father in this dihybrid cross is heterozygous for these alleles, meaning that he can taste PTC. The mother is homozygous recessive, meaning that she cannot taste PTC The father has X-Linked Protoporphyria which means that he is very sensitive to sunlight exposure, he is hemizygous for the dominant causative mutation. The mother is homozygous wild type at the same locus. Add a file here showing your diagram.arrow_forwardSalim and Sara are contemplating having children, but Salim’s brother has galactosemia and Sara has a sister who has three children, none of whom is affected. Sara dad has no history in his family of any sign of the diease and it is assumed to be homozygous normal. What is the probability that salim and saras first child will have galactosemia?arrow_forward
- a. If an individual who is homozygous for the mutation found in individual 2 and heterozygous for the mutation found in individual 4 marries an individual who is homozygous for the mutation found in individual 4 and heterozygous for the mutation found in individual 2, what will be the phenotype of their children? b. List the intermediate that would build up in each of the types of children who could not produce protein E.arrow_forwardPhenylketonuria and alkaptonuria are both autosomal recessive diseases. If a person with PKU marries a person with AKU, what will the phenotype of their children be?arrow_forwardFamilial retinoblastoma, a rare autosomal dominant defect, arose in a large family that had no prior history of the disease. Consider the following pedigree (the darkly colored symbols represent affected individuals): a. Circle the individual(s) in which the mutation most likely occurred. b. Is the person who is the source of the mutation affected by retinoblastoma? Justify your answer. c. Assuming that the mutant allele is fully penetrant, what is the chance that an affected individual will have an affected child?arrow_forward
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