20-21. If the litter resulting from the mationg of 2 short-tailed cats contains 3 kittens without tails, 3 with long tails, and 6 with short tails: A. What would be the simplest way explaining the inheritance of tail length in these cats? B. What are the genotypes of the kittens (1) without tails: _____ (2) with long tails: (3) with short tails: ? A = Sex-Linked Inheritance | B = (1) = ss, (2) = SS, (3) = Ss A = Incomplete Dominance | B = (1) = ss, (2) = SS, (3) = Ss A = Mendelian Dominant | B = (1) = ss, (2) = SS, (3) = Ss A = Incomplete Dominace | B = (1) = ss, (2) = ss, (3) = ss

20-21. If the litter resulting from the mationg of 2 short-tailed cats contains 3 kittens without tails, 3 with long tails, and 6 with short tails: A. What would be the simplest way explaining the inheritance of tail length in these cats? B. What are the genotypes of the kittens (1) without tails: _____ (2) with long tails: (3) with short tails: ? A = Sex-Linked Inheritance | B = (1) = ss, (2) = SS, (3) = Ss A = Incomplete Dominance | B = (1) = ss, (2) = SS, (3) = Ss A = Mendelian Dominant | B = (1) = ss, (2) = SS, (3) = Ss A = Incomplete Dominace | B = (1) = ss, (2) = ss, (3) = ss

Name: 20-21. If the litter resulting from the mationg of 2 short-tailed cat
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Description: 20-21. If the litter resulting from the mationg of 2 short-tailed cats contains 3 kittenswithout tails, 3 with long tails, and 6 with short tails:A. What would be the simplest way explaining the inheritance of tail length in thesecats?B. What are th

Biology 2e

2nd Edition

ISBN:9781947172517

Author:Matthew Douglas, Jung Choi, Mary Ann Clark

Publisher:Matthew Douglas, Jung Choi, Mary Ann Clark

Chapter12: Mendel's Experiments And Heredity

Section: Chapter Questions

Problem 34CTQ: A heterozygous pea plant produces violet flowers and yellow, round seeds. Describe the expected...

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Ch. 13-6 Suppose you identify a new gene in mice. One of its alleles specifies white fur, another specifies brown. You want to see if these alleles are inherited in a Mendelian pattern, or with incomplete dominance. What crosses would give you the answer?
Ch. 14-4 Duchenne muscular dystrophy, which is inherited in an X-linked recessive pattern, nearly always occurs in males. Explain why.
A heterozygous pea plant produces violet flowers and yellow, round seeds. Describe the expected genotypes of the gametes produced by Mendelian inheritance. If all three genes are found on the same arm of one chromosome should a scientist predict that inheritance patterns will follow Mendelian genetics?
In a mating between two individuals that are heterozygous for a recessive lethal allele that is expressed in utero, what genotypic ratio (homozygous dominant: heterozygous:homozygous recessive) would you expect to observe in the offspring? a. 1:2:1 b. 3:1:1 c. 1:2:0 d. 0:2:1
Does the phenotype indicated by the red circles and squares in this pedigree show an inheritance pattern that is autosomal dominant, autosomal recessive, or X-linked?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What if the couple wanted prenatal testing so that a normal fetus could be aborted?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What is the chance that this couple will have a child with two copies of the dominant mutant gene? What is the chance that the child will have normal height?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?
Having malformed hands with shortened fingers is a dominant trait controlled by a single gene; people who are homozygous for the recessive allele have normal hands and fingers. Having woolly hair is a dominant trait controlled by a different gene; homozygous recessive individuals have normal, nonwoolly hair. Suppose a woman with normal hands and nonwoolly hair marries a man who has malformed hands and woolly hair. Their first child has normal hands and nonwoolly hair. What are the genotypes of the mother, the father, and the child? If this couple has a second child, what is the probability that it will have normal hands and woolly hair?
Ch. 13-8 In sweet pea plants, an allele for purple flowers (P) is dominant when paired with a recessive allele for red flowers (p). An allele for long pollen grains (L) is dominant when paired with a recessive allele for round pollen grains (l). Bateson and Punnett crossed a plant having purple flowers and long pollen grains with one having white flowers and round pollen grains.All F1 offspring have purple flowers and long pollen grains. Among the F2 generation, the researchers observed the following phenotypes:296 purple flowers/long pollen grains19 purple flowers/round pollen grains27 red flowers/long pollen grains85 red flowers/round pollen grainsWhat is the best explanation for these results?
7-5. In corn a pair of genes determines leaf shape and another pair determines pollen shape. A ragged-leafed plant with round pollen was crossed to a ragged-leafed plant with angular-pollen, and the resultant progeny were classified as follows: Class Phenotypes 186 ragged-leaf round-pollen 11 IV 174 ragged-leaf angular-pollen 57 smooth-leaf round-pollen Total 63/480 smooth-leaf angular-pollen (a) Using alphabetical letters of your choice, designate the genes for the different leaf and pollen characters. (b) On the basis of the symbols given in (a) 1 provide genotypes two parents. (c) (forth)/(nbers) According to your hypothesis what would you have expected for each of the four classes of progeny? (d) After re viewing Chapter 8, use the chi-square method (p. 132) to test your hypothesis and indicate whether you accept or reject it.
Ch. 14-9 Mutations in the genes for clotting factor VIII and IX cause hemophilia A and B, respectively. A woman may be heterozygous for mutations in both genes, with a mutated factor VIII allele on one X chromosome, and a mutated factor IX allele on the other. All of her sons should have either hemophilia A or B. However, on rare occasions, one of these women gives birth to a son who does not have hemophilia, and his one X chromosome does not have either mutated allele. Explain.