10 11 12 13 14 15 Given that one gene pair is involved use the pedigree ibove to answer the following questions. Is the inheritance pattern X-linked or autosomal, recessive or dominant? Give the genotype of each of the following individuals in the pedigree. If more than one genotypic possibility exists, present all possible alternatives. b. Individual #1 - f. Individual # 2 – d. Individual #5 – e. Individual # 8 -

10 11 12 13 14 15 Given that one gene pair is involved use the pedigree ibove to answer the following questions. Is the inheritance pattern X-linked or autosomal, recessive or dominant? Give the genotype of each of the following individuals in the pedigree. If more than one genotypic possibility exists, present all possible alternatives. b. Individual #1 - f. Individual # 2 – d. Individual #5 – e. Individual # 8 -

Name: 10111213 14 15Given that one gene pair is involved use the pedigree i
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Description: 10111213 14 15Given that one gene pair is involved use the pedigree ibove to answer the following questions.Is the inheritance pattern X-linked or autosomal, recessive or dominant?Give the genotype of each of the following individuals in the pedigre

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter4: Pedigree Analysis In Human Genetics

Section: Chapter Questions

Problem 3QP: Pedigree Analysis Is a Basic Method in Human Genetics Using the pedigree provided, answer the...

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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?
Pedigree Analysis Is a Basic Method in Human Genetics Using the pedigree provided, answer the following questions. a. Is the proband male or female? b. Is the grandfather of the proband affected? c. How many siblings does the proband have, and where is he or she in the birth order?
Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?
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?
Help me create a pedigree of this information: Pedigree analysis: Generation 1: Normal parents (AA x AA) Generation 2: Carrier parents (AA x AS) Generation 3: Affected child (AS x AS) Generation 4: Affected grandchild (SS) This pedigree has two normal parents in the first generation. Second generation carriers carry the sickle cell trait from one parent. The disease is 25% more likely to be inherited in the third generation if both parents have the 'S' allele. If both parents have the 'S' allele, their children will have sickle cell anemia in the fourth generation
A. What is the inheritance pattern of the pedigree illustrated above? Explain in 1 sentence minimum citing two pieces of evidence from the pedigree to support your answer. B. Assuming that the disorder portrayed in this pedigree is very rare, based on your answer in a., what are the most likely genotypes of I-2, Ill-7 and I1-3. C. Based on your answer in a, what are the odds that IV-1 and IV-2 would have an affected male child?
Look at the pedigree below and answer the following questions related to the human genetic trait depicted in this pedigree. 1. Indicate whether the pattern of inheritance associated with this human trait is most likely to be (i) rare X-linked recessive, (ii) sex-influenced, or (iii) sex-limited. You may assume that the gene is fully penetrant. Then, provide a specific reason that justifies your selection of this pattern of inheritance as the correct one, and also explain why each of the other two alternatives are not correct. As part of your answer, include the phenotypic ratio, including the sex of the offspring, that you would expect to find in each of the three possible scenarios.
Consider an example of two autosomal recessive gene a and b where a heterozygote (AaBb) is testcrossed to a double homozygous mutant (AaBb) predict the phenotype ratio under the following conditions 1.a and b are located on separate autosomes 2.a and b are linked on the same chromosome but are far apart that a crossover always occurs 3.a and b are linked on the same chromosome but very close together so that a crossover never occur
The following pedigree shows the inheritance of deafmutism.a. Provide an explanation for the inheritance of thisrare condition in the two families in generations I andII, showing the genotypes of as many persons aspossible; use symbols of your own choosing.b. Provide an explanation for the production of onlynormal persons in generation III, making sure that yourexplanation is compatible with the answer to part a
Figure 5.1 presented abbreviated pedigrees looking at the inheritance of X-linked human conditions. El Figure 5.1a initiated with a male who was color blind and had hemophilia A. The grandsons and great-grandsons of this male were either color blind and had hemophilia A, or they had neither condition. a. Which if any individuals in the G Fig. 5.1a pedigree is a double heterozygote? b. Are any of the grandsons or great-grandsons shown in the pedigree the product of recombinant type gametes? c. Do the genes responsible for color blindness and hemophilia A segregate independently? d. If you examined many pedigrees, each starting with a man who was simultaneously color blind and had hemophilia A, do you think any grandsons or great-grandsons would ever have one condition but not the other? L Figure 5.1b initiated with a male who was color blind and had hemophilia B. The grandsons of this male were either only color blind, only had hemophilia B, had both conditions, or they had neither…