Human Biology (MindTap Course List)
11th Edition
ISBN: 9781305112100
Author: Cecie Starr, Beverly McMillan
Publisher: Cengage Learning
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Textbook Question
Chapter 19, Problem 1CT
One gene has alleles A and a. Another has alleles B and b. For each genotype listed, what type(s) of gametes can be produced? (Assume independent assortment occurs.)
- a. AABB
- b. AaBB
- c. Aabb
- d. AaBb
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Questions a to e are answerable by yes or no. Indicate the possible parental genotypes if your answer is yes.a. Can a man with hairy ears have a hairy-eared daughter?b. Can two normal parents produce a colorblind son?c. Can two normal parents produce a colorblind daughter?d. Can a colorblind woman have a normal son?e. Can a bald man have a nonbald daughter?
Hemophilia is a disease caused by a gene found on the X
chromosome. Therefore, it is referred to as a sex-linked disease.
The recessive allele causes the disease. A man with hemophilia
(xhy) marries a woman who is homozygous dominant (XHXH.
A. Illustrate using a Punnett square the probability that their children
will have the disease.
B. Will any of their children have the disease?
C. Predict the probabilities of their children having the disease.
Using the same information listed above for hair and eye color, (E = dominant brown eyes with e = recessive blue eyes; R = dominant brown hair and r = recessive blond hair), if the mom was heterozygous for both eye color and hair color what would her genotype be?
a. EERR
b. eerr
c. EeRr
What would be the 4 possible gametes the heterozygous mom could contribute?
a. RE, Re, rE, re
b. Rr, Ee, RE, re
c. RR, rr, EE, ee
Chapter 19 Solutions
Human Biology (MindTap Course List)
Ch. 19 - Define the difference between (a) gene and allele,...Ch. 19 - Prob. 2RQCh. 19 - What is probability, and how is it applied in...Ch. 19 - What is independent assortment? Does independent...Ch. 19 - Alleles are ___________. a. alternate forms of a...Ch. 19 - A heterozygote has _____. a. only one of the...Ch. 19 - Prob. 3SQCh. 19 - Offspring of a cross AA aa are ______. a. all AA...Ch. 19 - Prob. 5SQCh. 19 - Prob. 6SQ
Ch. 19 - Which statement best fits the principle of...Ch. 19 - Prob. 8SQCh. 19 - Prob. 9SQCh. 19 - Prob. 10SQCh. 19 - One gene has alleles A and a. Another has alleles...Ch. 19 - Still referring to Problem 1, what will be the...Ch. 19 - Go back to Problem 1, and assume you now study a...Ch. 19 - The young woman shown at right has albinismvery...Ch. 19 - When you decide to breed your Labrador retriever...Ch. 19 - The ABO blood system has been used to settle cases...Ch. 19 - Prob. 7CTCh. 19 - A man is homozygous dominant for ten different...Ch. 19 - Prob. 9CTCh. 19 - Bill and Marie each have flat feet, long...Ch. 19 - You decide to breed a pair of guinea pigs, one...
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- 17. Parents Jacob and Emma have had two children, Samuel and Matthew. Baby Samuel died at the age of nine days. When baby Matthew has trouble feeding, his parents take him in to the doctor. He is diagnosed with Maple Syrup Urine Disease (MSUD), a life-threatening condition in which the patient is not able to break down proteins from 19 their food. As Matthew receives treatment, Jacob and Emma are referred to a genetic counselor. The genetic counselor collects information about their family. Neither of Jacob's parents were affected by MSUD. Jacob had one brother and three sisters. One of these sisters died shortly after birth. Similarly, neither of Emma's parents were affected by MSUD. Emma had four brothers, and two of these died shortly after birth. a. Given what you know about Jacob and Emma's family, construct a pedigree that includes all three generations.arrow_forwardThe condition phenylketonuria is caused by a recessive allele. There are two carriers who have progeny.a. Give the gene notation. b. Give the expected genotypic and phenotypic ratios. c. What is the probability that their child will be heterozygous if they have a normal child?d. What is the probability of having two affected children and one normal child if they have three children?arrow_forwardRalnh has type B blood and his wife Rachel has type A blood. They are very shocked to hear that their baby has type O blood, and think that a switch might have been made at the hospital. Can this baby be theirs? O A. Yes, if both parents are homozygous recessive O B. Yes, if Ralph is homozygous and Rachel is heterozygous O C. No, this baby is not theirs. D. Yes, if both parents are heterozygous for their blood typesarrow_forward
- Tay–Sachs disease is caused by recessive alleles on anautosome. In which case(s) could two parents with anormal phenotype have a child with Tay–Sachs?a. Both parents are homozygous for a Tay–Sachs allele.b. Both parents are heterozygous for a Tay–Sachsallele.c. One parent is homozygous for a Tay–Sachs allele,and the other is heterozygous.arrow_forwardA woman with a rare autosomal recessive disorder was told that it was unlikely that her children would have the disorderas her husband did not have it. However, her first child has the disorder. a. What is the most likely explanation? b. Diagram the cross between the woman and her husband using a Punnett square, give the genotypic ratio (GR) and phenotypic ratio (PR) from the Punnett square. c. Based on the Punnett square results, what is the chance that her next child will have the disorder?arrow_forwardA. heterozygous blue-eyed man is marrying a homozygous brown-ayed woman. Determine the ratio of the following from the offspring. B. Genotype C. Phenotypearrow_forward
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