The pedigree below is for a rare autosomal recessive trait with complete penetrance.IIIIIIIV1122IT?12233probability that IV-1 is a carrier?3410a. What does the double horizontal line indicate?A Using A and a alleles, what is the genotype of III-2?A/What is the

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Description: The pedigree below is for a rare autosomal recessive trait with complete penetrance.IIIIIIIV1122IT?12233probability that IV-1 is a carrier?3410a. What does the double horizontal line indicate?A Using A and a alleles, what is the genotype of III-2?A/

A pedigree analysis is used in genetic studies to find out the inheritance patterns of various…

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The pedigree below is for a rare autosomal recessive trait with complete penetrance. I II III IV 1 1 2 2 2 2 3 probability that IV-1 is a carrier? 4 3 4 5 3 4 a. What does the double horizontal line indicate? A Using A and a alleles, what is the genotype of III-2? A/ What is the

The pedigree below is for a rare autosomal recessive trait with complete penetrance. I II III IV 1 1 2 2 2 2 3 probability that IV-1 is a carrier? 4 3 4 5 3 4 a. What does the double horizontal line indicate? A Using A and a alleles, what is the genotype of III-2? A/ What is the

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter18: Genetics Of Behavior

Section: Chapter Questions

Problem 14QP: A pedigree analysis was performed on the family of a man with schizophrenia. Based on the known...

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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?
A pedigree analysis was performed on the family of a man with schizophrenia. Based on the known concordance statistics, would his MZ twin be at high risk for the disease? Would the twins risk decrease if he were raised in an environment different from that of his schizophrenic brother?
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 progeny
The following pedigree shows the pattern of inheritance of red-green color blindness in a family. Females are shown as circles and males as squares; the squares or circles of individuals affected by the trait are filled in black. What is the chance that a son of the third-generation female indicated by the arrow will be color blind if the father is not color blind? If he is color blind?
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?
Please consider the pedigree below. There are no cases of false paternity. I II III IV в 1 a. Which individual/s definitely has/have Bombay phenotype in the descendants of I-1 and I-2? b. What are the genotypes of individuals II-2 and III-2 at the AB0 and H loci? Please label your answers a and b, Il-2: and Ill-2:.
Please consider the pedigree below. There are no cases of false paternity. I B II A 2 3 III AB (A IV в 1 a. Which individual/s definitely has/have Bombay phenotype in the descendants of I-1 and I-2? b. What are the genotypes of individuals II-2 and III-2 at the ABO and H loci?
A- Normal xy Aa XX Carrier ху ху A- XX XX What inheritance pattern is Ад 43. illustrated in the following pedigree? a- Aa
|| AB ||| 1 AA 0₂ AA 0₁ In this pedigree A and B represent alleles at a marker locus very closely linked to the disease locus. Affected individuals are shown as shaded. The disease status in III 1 (a female of 10 years) is unknown as yet as the disease does not onset until teenage years or soon after. Which of the following is correct? 2 AB BB OA. The probable pattern of inheritance shown by the disease in this family is autosomal recessive. OB. If recombination does not occur the probability that III 1 will be affected if she has an AB marker genotype is 1. OC. If recombination does not occur, the probability that III 1 will be affected if she has a BB marker genotype is 1. OD. If the recombination fraction between the disease and marker loci equals 0.04, the probability that III 1 will be affected if she inherits an AB marker genotype equals 0.96. OE. The probable pattern of inheritance shown by the disease in this family is X-linked recessive.
Please consider the pedigree below. There are no cases of false paternity. AB 2 I II 2 3 A (AB A)B 1 III 2 3 4 IV B 1 2 a. Which individual/s definitely has/have Bombay phenotype in this extended family? b. What are the genotypes of individuals II-2 and II-3 at the ABO and H loci?
The individuals with a certain disease are shown in this pedigree. The disease is caused by an autosomal recessive allele, q. Give the genotype of the following individuals: (a) III-1 (the girl at lower left corner) (b) II-1 (the girl’s mother) (c) II-2 (the girl’s father)