1. This pedigree tracks the inheritance of freckles, an autosomal dominant trait, for 3 generations.IlIII车A. Determine the genotypes of all individual in the chart.III-1III-2III-3III-4III-51-1Il-11-2Il-2Il-3Il-4Il-5III-6B. If II-6 marries someone without freckles, is it possible for them to have a non-freckled child?Why or why not?

Name: 1. This pedigree tracks the inheritance of freckles, an autosomal dom
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Description: 1. This pedigree tracks the inheritance of freckles, an autosomal dominant trait, for 3 generations.IlIII车A. Determine the genotypes of all individual in the chart.III-1III-2III-3III-4III-51-1Il-11-2Il-2Il-3Il-4Il-5III-6B. If II-6 marries someone wi

In this question, we are shown a pedigree of autosomal dominant trait which is showing inheritance…

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1. This pedigree tracks the inheritance of freckles, an autosomal dominant trait, for 3 generations. II II A. Determine the genotypes of all individual in the chart. Il-1 Il-2 Il-3 III-1 1-1 1-2 III-2 III-3 III-4 III-5 Il-4 Il-5 III-6 B. If III-6 marries someone without freckles, is it possible for them to have a non-freckled child? Why or why not?

1. This pedigree tracks the inheritance of freckles, an autosomal dominant trait, for 3 generations. II II A. Determine the genotypes of all individual in the chart. Il-1 Il-2 Il-3 III-1 1-1 1-2 III-2 III-3 III-4 III-5 Il-4 Il-5 III-6 B. If III-6 marries someone without freckles, is it possible for them to have a non-freckled child? Why or why not?

BIOLOGY:CONCEPTS+APPL.(LOOSELEAF)

10th Edition

ISBN:9781305967359

Author:STARR

Publisher:STARR

Chapter14: Human Inheritance

Section: Chapter Questions

Problem 9SA: Alleles for Tay-Sachs disease are inherited in an autosomal recessive pattern. Why would two parents...

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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?
Achondroplasia is a rare dominant autosomal defect resulting in dwarfism. The unaffected brother of an individual with achondroplasia is seeking counsel on the likelihood of his being a carrier of the mutant allele. What is the probability that the unaffected client is carrying the achondroplasia allele?
A heterozygous individual has a _______ for a trait being studied. a. pair of identical alleles b. pair of nonidentical alleles c. haploid condition, in genetic terms
Examine the following pedigrees. Which is the most likely mode of inheritance of each disorder? (a) autosomal recessive (b) autosomal dominant (c) X-linked recessive (d) a, b, or c (e) a or c 10.
Alleles for Tay-Sachs disease are inherited in an autosomal recessive pattern. Why would two parents with a normal 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-Sachs allele. c. New mutations gave rise to Tay-Sachs in the child. d. b or c
In Section 12.3, ''Laws of Inheritance," an example of epistasis was given for the summer squash. Cross white WAvYy heterozygotes to prove the phenotypic ratio of 12 white:3 yellow:l green that was given in the text.
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1. A human female "carrier" who is heterozygous for the recessive, sex-linked trait causing red-green color blindness (or alternatively, hemophilia), marries a normal male. What proportion of their male progeny will have red-green color blindness (or alternatively, will be hemophiliac)? * a. 100% b. 75% c. 50% d. 25% e. 0% 2. A human female "carrier" who is heterozygous for the recessive, sex-linked trait red color blindness, marries a normal male.What proportion of their female progeny will show the trait? * a. All b. ½ c. ¼ d. 0 e. 3/4 3. Women have sex chromosomes of XX, and men have sex chromosomes of XY. Which of a man's grandparents could not be the source of any of the genes on his Y-chromosome? * a. Father's Mother b. Mother's Father c. Father's Father d. Mother's Mother, Mother's Father, and Father's Mother e. Mother's Mother 4. Male-pattern baldness is an example of a sex-influenced trait. The baldness allele, which causes hair loss, is influenced by the hormones…
A man is brachydactylous (very short fingers; rare autosomal dominant), and his wife is not. Both can taste the chemical phenylthiocarbamide (autosomal dominant; common allele), but their mothers could not.a. Give the genotypes of the couple.If the genes assort independently and the couple has four children, what is the probability ofb. all of them being brachydactylous?c. none being brachydactylous?d. all of them being tasters?e. all of them being nontasters?f. all of them being brachydactylous tasters?g. none being brachydactylous tasters?h. at least one being a brachydactylous taster?
Hi, I'm having trouble with my study guide for my upcoming genetics exam. If someone could please help with work shown and an explanation it would help so much! Thank you!! 2a. The pedigree below represents inheritance of rare condition. What pattern of inheritance is most consistent with the data? Assign alleles to all individuals to support your answer. If an allele is unknown, assign it a ? symbol. NOTE: Individuals whose phenotype or genotype cannot be determined are assumed to be unaffected and homozygous, unless otherwise indicated. 2b. In addition to the alleles you’ve indicated, describe 2 overall features of the pedigree that make it consistent with your chosen form of inheritance. 2c. Based on your mode of inheritance, what is the probability that the child of couple IV-4 x IV-5 will be affected? Show your work. attached is the pedigree
Pedigree attached shows an autosomal recessive genetic disease. G is the normal allele and g is the disease-causing allele. Individual 1’s father is heterozygous (*) and his mother is homozygous dominant. Other individuals in the pedigree may be carriers, but are not marked. The question mark (?) indicates that you do not yet know anything about this individual’s phenotype with regard to the disease. part a) What is the probability that individuals 1 and 2 will have a child (5) who is a boy with the disease (the child is unborn and the sex is not yet known)? a)1/8 b)1/4 c)0 d)1/16 part b) What is the probability that the daughter (6) that individual 3 and 4 just had will have the disease? a)1/8 b)1/6 c)1/4 d)1/12