Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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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).
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.
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.
The pedigree below shows the phenotypes of the ABO blood groups and Rhesus factors [positive (+) and negative (-)] for several members of a family.
I
(B+
AB-
1
2
3
4
II
O-
A+
В-
B-
AB+
A+
1
2
4
5 6
a. What are the ABO blood group genotypes of individuals I-1 and I-2?
b. Which child/ren of individual I-4 can donate blood to him?
c. Which individual in the pedigree can donate blood to all the other individuals in the pedigree?
Chapter 4 Solutions
Genetic Analysis: An Integrated Approach (3rd 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...Ch. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - 37. Epistatic gene interaction results in a...Ch. 4 - 38. Draw a pedigree containing two parents and...
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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_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. 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.arrow_forwardThe blood serum from one individual (let’s call this person individual 1) is known to agglutinate the red blood cells from a secondindividual (individual 2). List the pairwise combinations of possible genotypes that individuals 1 and 2 could have. If individual 1is the parent of individual 2, what are his or her possiblegenotypes?arrow_forward
- Please calculate the Chi-square valuearrow_forwardConsider the following dihybrid testcross: B/b • E/e × b/b • e/e For the progeny from this testcross, determine the relative proportions (from 0% to 100%) of each genotype if the two genes: a) are linked (dominant alleles in cis conformation) with no crossing over: Be/be: be/be: BE/be: bE/be: b) assort independently. B/b; E/e: B/b; e/e: b/b; E/e: b/b; e/e: c) are linked (dominant alleles in cis conformation) and 20 map units apart. Be/be: be/be: BE/be: bE/be:arrow_forwardPurple flowers are dominant to white flowers. Identify the phenotypefor the following genotype Ff, FF, ff and determine if the genotype is heterozygous or homozygous. * For each row, you should select two columns. Purple flowers White flowers Heterozygous Homozygous Ff FF ff Brown eyes are dominant to blue eyes. Identify the phenotype for the following genotype BB, Bb, bb and determine if the genotype is heterozygous or homozygous. * 口 ロ口arrow_forward
- In 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_forwardIn humans, four different blood types (A, B, AB, and O) are encoded by three alleles 1, 1, and i Individuals with both I and I alleles have blood type AB (red blood cells with both A and B antigens). Two copies of the i allele are required for an individual to have blood type O (red blood cells with no antigens). Which of the following correctly indicates the relationship between the I and / alleles for the blood type gene? Select one: OA. I is dominant to / OB. I is recessive to i OC. I and I are co-dominant OD. I and/exemplify incomplete dominancearrow_forwardSeveral genes in humans in addition to the ABO gene(I) give rise to recognizable antigens on the surface ofred blood cells. The MN and Rh genes are two examples. The Rh locus can contain either a positive or anegative allele, with positive being dominant to negative. M and N are codominant alleles of the MN gene.The following chart shows several mothers and theirchildren. For each mother-child pair, choose the fatherof the child from among the males in the right column, assuming one child per male.Mother Child Malesa. O M Rh(pos) B MN Rh(neg) O M Rh(neg)b. B MN Rh(neg) O N Rh(neg) A M Rh(pos)c. O M Rh(pos) A M Rh(neg) O MN Rh(pos)d. AB N Rh(neg) B MN Rh(neg) B MN Rh(pos)arrow_forward
- A prospective father has two dominant traits dependent on single autosomal genes, cataract (an eye abnormality), which he inherited from his mother, and polydactyly (extra fingers and/or toes), which he inherited from his father. If the loci for these two traits are very closely linked, which of the following possibilities would the man's child be more likely to have (assume his partner is unaffected by either condition): Select one: either cataract or polydactyly both cataract and polydactyly neither traitarrow_forwardFor the following problem: Identify the gametes for each parent, build a Punnett Square of the probability of offspring. Show the resulting genotype ratios and the resulting phenotype ratios. The submission file on Canvas will typically have a multiple-choice component or may be a fill-in-the-blank question related to genotype and/or phenotype outcomes. Be prepared to address all of these details in the questions. a )In Pansy flowers, a gene for follower color is “incompletely dominant”. All flowers resulting from the cross of a homozygous red (R) flowering plant with a homozygous white (r) flowering plant were pink. If a pink-flowering plant is bred with another pink-flowering plant, determine the probable genotype and phenotype ratios of the offspring. b) The ratio of genotypes is ________________ .c) The ratio of phenotypes (red:pink:white) is ______________.arrow_forwardClassical hemophilia is a sex-linked disease caused by a recessive gene on the X chromosome. (Hemophilia refers to diseases that cause delays in blood clotting.) If a woman who is acarrierof classical hemophilia has children with a normal male, give the ratios of the possible offspring with respect to classical hemophilia. Be sure to state both the genotypes and the phenotypes of each offspring. For genotypes, use X for a normal X chromosome, Xh for an X chromosome with the hemophilia gene, and Y for a normal Y chromosome. For phenotypes, if the offspring is female, be sure to state if homozygous normal, a carrier, or has the disease. If the offspring is a male, be sure to state if normal or has the disease.arrow_forward
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