Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
expand_more
expand_more
format_list_bulleted
Concept explainers
Textbook Question
Chapter 5, Problem 10P
In a particular human family, John and his mother both have brachydactyly (a rare autosomal dominant allele causing short fingers). John’s father has Huntington disease (another rare autosomal dominant allele). John’s wife is
a. | What are the genotypes of John’s parents? |
b. | What are the possible genotypes for John? How likely is John to have each of these genotypes? |
c. | What is the probability the child will express both brachydactyly and Huntington disease by age 50 if the two genes are unlinked? |
d. | How will your answer to part (c) change if instead these two loci are 20 m.u. apart? |
Expert Solution & Answer
Trending nowThis is a popular solution!
Students have asked these similar questions
Color blindness in humans is controlled by an X-linked completely recessive allele (Xc), while breast cancer is controlled by an autosomal completely dominant allele, B. A color blind male, who is a heterozygote carrier for breast cancer has three children/n with a normal eyed female (whose mother was color blind), who is homozygote recessive for the breast cancer allele. What is the probability that out of three children, 2 will be color blind males, and not show breast cancer, and one will be a color blind female, who shows breast cancer?
In human beings, the gene for red‑green colorblindness (r) is sex‑linked and recessive to its allele for normal vision (R), while the gene for freckles (F) is autosomal and dominant over its allele for nonfreckled (f). A nonfreckled, normal‑visioned woman whose father was freckled and colorblind, marries a freckled, colorblind man whose mother was nonfreckled.
What is the genotype of the woman's father?
What is the probability that the couple's first child will be a non-freckled, normal visioned girl?
What is the probability that the first two children born to the couple will be freckled and colorblind girls?
In humans, ABO blood types refer to glycoproteins in the membranes of red blood cells. There are three alleles for this autosomal gene: IA, IB, and i. The IA allele codes for the A sugar, The IB allele codes for the B sugar, and the i allele doesn't code for any sugar. IA and IB are codominant, and i is recessive to both IA and IB. If an individual with type AB blood has a child with an individual with type O blood, what blood types could their children possibly have?
Chapter 5 Solutions
Genetics: From Genes to Genomes
Ch. 5 - Choose the phrase from the right column that best...Ch. 5 - a. A Drosophila male from a true-breeding stock...Ch. 5 - Prob. 3PCh. 5 - The Punnett square in Fig. 5.4 shows how Mendels...Ch. 5 - In mice, the dominant allele Gs of the X-linked...Ch. 5 - In Drosophila, males from a true-breeding stock...Ch. 5 - If the a and b loci are 20 m.u. apart in humans...Ch. 5 - CCDD and ccdd individuals were crossed to each...Ch. 5 - In mice, the autosomal locus coding for the...Ch. 5 - In a particular human family, John and his mother...
Ch. 5 - Albino rabbits lacking pigment are homozygous for...Ch. 5 - In corn, the allele A allows the deposition of...Ch. 5 - If the a and b loci are 40 cM apart and an AA BB...Ch. 5 - Write the number of different kinds of phenotypes,...Ch. 5 - A DNA variant has been found linked to a rare...Ch. 5 - Figure 5.7a shows chromosomes during prophase of...Ch. 5 - Figure 5.7b shows bivalents in mouse primary...Ch. 5 - Cinnabar eyes cn and reduced bristles rd are...Ch. 5 - In Drosophila, the autosomal recessive dp allele...Ch. 5 - From a series of two-point crosses, the following...Ch. 5 - Map distances were determined for four different...Ch. 5 - In the tubular flowers of foxgloves, wild-type...Ch. 5 - In Drosophila, the recessive allele mb of one gene...Ch. 5 - A snapdragon with pink petals, black anthers, and...Ch. 5 - In Drosophila, three autosomal genes have the...Ch. 5 - Drosophila females heterozygous for each of three...Ch. 5 - Male Drosophila expressing the autosomal recessive...Ch. 5 - a. In Drosophila, crosses between F1 heterozygotes...Ch. 5 - A true-breeding strain of Virginia tobacco has...Ch. 5 - Prob. 30PCh. 5 - The following list of four Drosophila mutations...Ch. 5 - Do the data that Mendel obtained fit his...Ch. 5 - Two genes control color in corn snakes as follows:...Ch. 5 - A mouse from a true-breeding population with...Ch. 5 - Neurospora of genotype a c are crossed with...Ch. 5 - A cross was performed between one haploid strain...Ch. 5 - Prob. 40PCh. 5 - Prob. 41PCh. 5 - Indicate the percentage of tetrads that would have...Ch. 5 - Prob. 43PCh. 5 - This problem leads you through the derivation of a...Ch. 5 - a. In ordered tetrad analysis, what is the maximum...Ch. 5 - Prob. 46PCh. 5 - A single yeast cell placed on a solid agar will...Ch. 5 - Figure 5.29 shows mitotic recombination leading to...Ch. 5 - A diploid strain of yeast has a wild-type...Ch. 5 - In Drosophila, the yellow y gene is near the...Ch. 5 - Neurofibromas are tumors of the skin that can...Ch. 5 - Two important methods for understanding the...
Knowledge Booster
Learn more about
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
- A 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_forwardIn human beings, the gene for red‑green colorblindness (r) is sex‑linked and recessive to its allele for normal vision (R), while the gene for freckles (F) is autosomal and dominant over its allele for nonfreckled (f). A nonfreckled, normal‑visioned woman whose father was freckled and colorblind, marries a freckled, colorblind man whose mother was nonfreckled. What is the probability that the first child born to the couple will either be a freckled, colorblind boy or a non‑freckled, normal visioned girl or a non-freckled, normal visioned boy? What is the probability that the first four children born to the couple will be freckled and normal visioned girls?arrow_forwardCystic fibrosis (CF) is an autosomal recessive condition triggered by the overproduction of sticky mucus that clogs the lungs and pancreas. It is a life-threatening disease, but medical advances helped the afflicted to live through adulthood. The mother of Claudia died from cystic fibrosis, but her father was normal and never had any relative with CF. Her fiancé, Marcus, turned out to be a carrier of the CF allele. What are the genotypes of Claudia and Marcus? Claudia: ________________________ Marcus: _____________________ They planned to have four children. What is the probability that: a. all children will be normal b. at least two will be normal PLEASE SHOW COMPLETE SOLUTIONarrow_forward
- Male-limited precocious puberty results from a rare, sex-limited autosomal allele (P) that is dominant over the allele for normal puberty (p) and is expressed only in males. Bill undergoes precocious puberty, but his brother Jack and his sister Beth underwent puberty at the usual time, between the ages of 10 and 14. Although Bill's mother and father underwent normal puberty, his two maternal uncles (his mother's brothers) underwent precocious puberty. All of Bill's grandparents underwent normal puberty. Match the most likely genotypes for each relative in his family. If there is an equal likelihood of multiple genotypes for a relative, place each possible genotype. Relative Genotype Answer Bank Bill Pp PP PP Bill's brother Jack Bill's sister Beth Bill's father Bill's mother Bill's uncle Bill's grandfather Bill's grandmotherarrow_forwardAlbinism, lack of pigmentation in humans, results from an autosomal recessive gene designated a. Two parents with normal pigmentation have an albino child. What is the probability that their next child will be albino? What is the probability that the next child will be an albino girl? If the child is normal, what is the probability that it will be a carrier (heterozygous) for the albino gene?arrow_forwardIn addition to the allelic pair determining pattern baldness in man (B,b), consider early baldness to be due to another autosomal allele (E) on a different pair of chromosomes and also dominant in males but recessive in females. The phenotype for ee may be late or nonbaldness depending on sex and the genotype for B, b alleles. Two doubly heterozygous persons marry. What is the phenotype of the male parent? What is the phenotype of the female parent? Give the phenotypic ratio expected among male children of couples such as this one. Show corresponding genotypes for each phenotype mentioned in your phenotypic ratio. Give the phenotypic ratio expected among female children of couples such as this one. Show corresponding genotypes for each phenotype mentioned in your phenotypic ratio.arrow_forward
- —Hereditary canine spinal muscular atrophy (HCSMA) is a motor neuron disease in Brittany Spaniels. Breeding studuies within a kindred of more than 125 dogs (Brittany spaniel and beagle-Brittany outcrosses) have established an autosomal dominant inheritance for HCSMA. Which of the following genotypes would cause the dog to NOT HAVE HCSMA? Hh HH Hh A and C B and C– —(see question above for genotype) is a spaniel who is homozygous recessive for the HCSMA trait had a puppy with a spaniel who was heterozygous for this trait, what is the probability that the puppy would be normal? 0% 25% 50% 75% 100%arrow_forwardTake 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…arrow_forwardIn the snail Cepaea nemoralis, an autosomal allele causing a banded shell (BB) is recessive to the allele for an unbanded shell (BO). Genes at a different locus determine the background color of the shell; here, yellow (CY) is recessive to brown (CBw). A banded, yellow snail is crossed with a homozygous brown, unbanded snail. The F1 are then crossed with banded, yellow snails (a testcross). a. What will the results of the testcross be if the loci that control banding and color are linked with no crossing over? b. What will the results of the testcross be if the loci assort independently? c. What will the results of the testcross be if the loci are linked and 20 m.u. apart?arrow_forward
- In the snail Cepaea nemoralis, an autosomal allele causing a banded shell (BB) is recessive to the allele for an unbanded shell (BO). Genes at a different locus determine the background color of the shell; here, yellow (CY) is recessive to brown (CBw). A banded, yellow snail is crossed with a homozygous brown, unbanded snail. The F1 are then crossed with banded, yellow snails (a testcross). Q.What will the results of the testcross be if the loci are linked and 20 m.u. apart?arrow_forwardIn the snail Cepaea nemoralis, an autosomal allele causing a banded shell (BB) is recessive to the allele for an unbanded shell (BO). Genes at a different locus determine the background color of the shell; here, yellow (CY) is recessive to brown (CBw). A banded, yellow snail is crossed with a homozygous brown, unbanded snail. The F1 are then crossed with banded, yellow snails (a testcross). Q.What will the results of the testcross be if the loci assort independently?arrow_forwardIn the snail Cepaea nemoralis, an autosomal allele causing a banded shell (BB) is recessive to the allele for an unbanded shell (BO). Genes at a different locus determine the background color of the shell; here, yellow (CY) is recessive to brown (CBw). A banded, yellow snail is crossed with a homozygous brown, unbanded snail. The F1 are then crossed with banded, yellow snails (a testcross). Q.What will the results of the testcross be if the loci that control banding and color are linked with no crossing over?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Human Biology (MindTap Course List)BiologyISBN:9781305112100Author:Cecie Starr, Beverly McMillanPublisher:Cengage LearningHuman Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning
Human Biology (MindTap Course List)
Biology
ISBN:9781305112100
Author:Cecie Starr, Beverly McMillan
Publisher:Cengage Learning
Human Heredity: Principles and Issues (MindTap Co...
Biology
ISBN:9781305251052
Author:Michael Cummings
Publisher:Cengage Learning
How to solve genetics probability problems; Author: Shomu's Biology;https://www.youtube.com/watch?v=R0yjfb1ooUs;License: Standard YouTube License, CC-BY
Beyond Mendelian Genetics: Complex Patterns of Inheritance; Author: Professor Dave Explains;https://www.youtube.com/watch?v=-EmvmBuK-B8;License: Standard YouTube License, CC-BY