Genetic Analysis: An Integrated Approach (2nd Edition)
2nd Edition
ISBN: 9780321948908
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 4, Problem 6P
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If a man with dark skin whose genotype is AaBB reproduces with a woman who has light skin (aaBb), what are the possible skin colors that their children will have? Develop a Punnett square and list the possible genotypes & phenotypes. For your convenience, a table is presented below that shows several possible genotypes & phenotypes:
Genotypes Phenotypes
AABB Very dark skin
AABb or AaBB Dark skin
AaBb, AAbb, or aaBB Medium brown skin
Aabb or aaBb Light skin
aabb Very light skin
Achondroplasia 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.
Two parents who have the Achondroplasia phenotype have 4 children where 1 is normal
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.
Given the following pedigree:
Is the trait autosomal or sex-linked?
Is the trait dominant or recessive?
Based only on the information given, what is the probability that I-2 is heterozygous?
Give the genotypes of individuals II-3, II-4.
What is the probability that individual III-1 is purebreeding?
Chapter 4 Solutions
Genetic Analysis: An Integrated Approach (2nd Edition)
Ch. 4 - 1. Define and distinguish incomplete penetrance...Ch. 4 -
2. Define and distinguish epistasis and...Ch. 4 - When working on barley plants, two researchers...Ch. 4 - Fifteen bacterial colonies growing on a complete...Ch. 4 - 5. In a type of parakeet known as a “budgie,”...Ch. 4 - 6. The and blood groups are given below for four...Ch. 4 - The wild-type color of horned beetles is black,...Ch. 4 - 8. Two genes interact to produce various...Ch. 4 - Prob. 9PCh. 4 - 10. In rats, gene produces black coat color if the...
Ch. 4 - 11. In the rats identified in Problem, a third...Ch. 4 - Using the information provided in Problems 10 and...Ch. 4 - 13. Total cholesterol in blood is reported as the...Ch. 4 - 14. Flower color in snapdragons results from the...Ch. 4 - 5. A plant line with reduced fertility comes to...Ch. 4 - Prob. 16PCh. 4 - The coat color in mink is controlled by two...Ch. 4 - Prob. 18PCh. 4 - 19. Feather color in parakeets is produced by the...Ch. 4 - Brachydactyly type D is a human autosomal dominant...Ch. 4 - 21. A male and a female mouse are each from...Ch. 4 - Xerodermapigmentosum (XP) is an autosomal...Ch. 4 - 23. Three strains of green-seeded lentil plants...Ch. 4 - Blue flower color is produced in a species of...Ch. 4 - 25. The following crosses are performed between...Ch. 4 - Two pure-breeding strains of summer squash...Ch. 4 - Marfan syndrome is an autosomal dominant disorder...Ch. 4 - 28. Yeast are single-celled eukaryotic organisms...Ch. 4 - Prob. 29PCh. 4 - Dr. Ara B. Dopsis and Dr. C. Ellie Gans are...Ch. 4 - Human ABO blood type is determined by three...Ch. 4 - In rabbits, albinism is an autosomal recessive...Ch. 4 - Dr. O. Sophila, a close friend of Dr. Ara B....Ch. 4 - In a breed of domestic cattle, horns can appear on...
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- As it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women. During his tenure as guard, he had numerous clandestine affairs. In each case, children resulted. Subsequently, some of the childrenwho had no way of knowing that they were relatedmarried and had children of their own. Assume that two pairs of genes determine height. The genotype of the 7-foot-tall Potsdam Guard was A9A9B9B9, and the genotype of all of his 5-foot clandestine lovers was AABB. An A9 or B9 allele in the offspring each adds 6 inches to the base height of 5 feet conferred by the AABB genotype. a. What were the genotypes and phenotypes of all the F1 children? b. Diagram the cross between the F1 offspring, and give all possible genotypes and phenotypes of the F2 progenyarrow_forwardThe young woman shown at right has albinismvery pale skin, white hair, and pale blue eyes. This phenotype is due to the absence of melanin, which imparts color to the skin, hair, and eyes. It typically is caused by a recessive allele. In the following situations, what are the probable genotypes of the father, the mother, and their children? a. Both parents have normal phenotypes; some of their children are albino and others are not. b. Both parents and all their children are albino. c. The mother is not albino, the father is albino, and one of their four children is albino.arrow_forwardGiven the karyotype shown at right, is this a male or a female? Normal or abnormal? What would the phenotype of this individual be?arrow_forward
- Along with the trait in the pedigree, individual IV-6 and the woman are also both heterozygous for the autosomal dominant allele causing Huntington's disease. If they have a child, what is the probability that it will be affected by at least one of these traits? Remember to include both the trait in the pedigree and Huntington's disease in your calculations. Enter your answer to two decimal places (e.g., 0.55).arrow_forwardAlbinism is a recessive disorder where there is a lack of melanin. Andrea and her husband Claude both have normal skin pigmentation. Andrea’s mother has the albino phenotype, but her father and her brother do not (normal pigmentation). Claude’s parents are both normal, but he has a sister who has the condition (is albino). Answer the following questions If Andrea and Claude are carriers for the albino allele, what is the probability that they have an albino child? If Andrea and Claude have a second child, what is the probability this child be normal (non-albino)? NOTE: Draw a punnet square or show your work.arrow_forwardThe allele for albinism is recessive to the allele for normal skin pigmentation. All individuals who are homozygous for this recessive allele (m) are unable to produce the enzyme needed for melanin production and are referred to as albinos. In the following statements, determine the correct genotypes: An albino male (genotype A) whose parents both have normal skin pigmentation (genotype B) marries a woman who does not have albinism (genotype C). This woman’s father is an albino (genotype D). The married couple has three children, two with normal skin pigmentation (genotype E) and one exhibiting albinism (genotype F)arrow_forward
- Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. Individuals with PKU have two recessive alleles and have very low levels of an enzyme that is needed to properly break down proteins. If a woman and her husband are both carriers and have three children, what is the probability of each of the following? Show your math. (Hint: You can represent your probabilities as fractions or decimals, but probabilities are always between 0 and 1) a. All three children are of normal phenotype. b. One or more of the three children have the disease. c.All three children have the disease.arrow_forwardBlood type is determined by multiple codominant alleles of the I gene, IA, IB, and i, where ii results in type O. The genotypes of a husband and wife are IAIB x IAi. Among the blood types of their children, how many different genotypes and phenotypes are possible? Group of answer choices 2 genotypes, 3 phenotypes 4 genotypes, 3 phenotypes 3 genotypes, 3 phenotypes 3 genotypes, 4 phenotypesarrow_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_forward
- A study of SNPs associated with familial hypercholesterolemia has revealed that 7 or more contributing alleles at 6 loci can lead to elevated levels of cholesterol in the blood. What is the probability that the following parents, AaBbccDDeeFF x AAbbCCDdEeFF, will have children that are susceptible to hypercholesterolemia?arrow_forwardConsider three genes L, U, and W, for which the count of F2 phenotypes after a 3-point cross is as follows: Phenotype F2 count: L U w 19 L u W 1 l u W 21 L U W 33 l U W 274 l u w 41 l U w 2 L u w 259 Which of the following statements about genes L, U, and W are TRUE? (may be more than one correct ans) A. L, U, and W are each on a different chromosome B. Only U and L are on the same chromosome C. Only U and W are on the same chromosome D. Only W and L are on the same chromosome E. L, U, and W are all on the same chromosomearrow_forwardIn pedigrees, individuals are usually specified by using a Roman numeral for their generation in the chart and an Arabic number for their position (reading left to right) within that generation. If we use the letter c for the allele that causes cystic fibrosis, what are the genotypes of individuals III-3 and III-4 (the third and fourth individuals shown in generation III) in the pedigree that shows this disease?arrow_forward
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