Answered: Pedigree #2 shows the inheritance of a…

Name: 1. Pedigree #2 shows the inheritance of a X-linked trait. What is the
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Description: 1. Pedigree #2 shows the inheritance of a X-linked trait. What is the genotype of individual 11 from this pedigree?Group of answer choicesA. aaB. AaC. AAD. XaXaE. XAXaF. XAXA 2.Refer to Pedigree #2 determine the mode of inheritance for the indicated

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Pedigree #2 shows the inheritance of a X-linked trait. What is the genotype of individual 11 from this pedigree? Group of answer choices

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter10: From Proteins To Phenotypes

Section: Chapter Questions

Problem 3CS: A couple was referred for genetic counseling because they wanted to know the chances of having a...

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The Meeting Sarah stared blankly at the blue paisley wallpaper. Her husband Mike sat by her side, bending and unbending a small paper clip. “Sarah and Michael, it’s good to meet you,” welcomed the genetic counselor, as she entered the room. “I apologize for being late, but I was just meeting with another couple. Let’s see, you’d like to have a child, but you’re concerned because of your family history of cystic fibrosis.” “Yes,” Sarah replied softly. “Mike and I met at a CF support group meeting a few years ago. He had a younger brother who died of cystic fibrosis, and I had a younger sister. We saw the painful lives they had—difficulty breathing, the constant respiratory infections. Although the treatments for CF are better now, we just don’t know if we can…” she trailed off. “I can certainly understand your concern,” the genetic counselor responded sympathetically. “That’s where I hope to help, by providing as much information and advice as I can. I’m glad that you came to see me…
Consider this pedigree showing an autosomal dominant rare disorder. What is the degree of penetrance? Show your work. na оп 16 19 fa 16 R 9X
Name Sofia Falcione P Pedigree Analysis Practice - for each pedigree, write the genotypes of the individuals The disorder shown on the pedigree is Maple Syrup Urine Disease (MSUD) which is a metabolic disorder that affects the body's ability to process certain proteins. It was named after a distinctive odor of a baby's urine. 1. What is the inheritance pattern of this gene? a) autosomal dominant b) autosomal recessive c) X-linked recessive 2. Provide at least one piece of evidence for your claim. This pedigree shows the inheritance Leber congenital amaurosis (LCA) which is a type of hereditary blindness. Individuals with this disease lose their vision during childhood. 3. What is the inheritance pattern shown? 4. Highlight one individual whose genotype is unknown. What additional information would you need to determine his/her genotype? Marfan syndrome affects the connective tissue and causes individuals to have long, thin, arms, legs, fingers and toes. 5. What is the inheritance…
d/1n5NtidRwTwUzcDkDPi5Z9P SHPZ9IA-XH-pfftLbhNc/edit 1) @ Is Add-ons Help Last edit was 15 minutes ago | Calibri в I UA 12 + 3I | II 6 1 I 7. Construct a Punnett square for a cross between two heterozygous pea plants with violet flower color. a. What genotypes would you expect in the offspring? b. What percentage or ratio of each genotype would you expect in the offspring? !!!
The pedigree below represents a family in which some members (shaded) have a monogenic genetic disease. Assume it is fully penetrant. 4999 a. If the disease were X-linked recessive, what would be the probability that a male child born to II- 3 and II-4 will be affected? Why? (Note: we already know the child is male.) b. If the disease were autosomal dominant what would be the probability that a male child born to II-3 and II-4 will be affected? Why? (Note: we already know the child is male.)
As a genetic counselor, you inform Susan and John that a blood test for cystic fibrosis is available. would you recommend generic testing for Susan and John?
5 & :56M ******* 24 DIHYBRID CROSSES DRV 0 Stv T alı A @ zladenA 9160p2-id2 bns obidalbaneoviene da II\ MOD YR 21 $59A ... Create a dihybrid cross and determine the expected phenotypic percentages of the offspring of two corn plants both of which are heterozygous for colour and texture (RrTt X RrTt). Don't forget to include clear let statements, and follow the all six steps taught on solving genetics problems. insig moni nellog: bna. zoom
A genetic counsellor is working with a couple who have just had a child who has a debilitating autosomal recessive form of a disease termed epidemolysis bullosa (which causes severe external and internal skin fragility resulting in blistering etc). Neither parent has epidemolysis bullosa, nor does anyone in their families. What should the counsellor say to this couple? OA. "Because no one in either of your families has epidemolysis bullosa, you are not likely to have another baby with epidemolysis bullosa. You can safely have another child." OB. "Because you have had one child with epidemolysis bullosa, you must each carry the allele. Any child you have has a 50% chance of having the disease." OC. "Because you have had one child with epidemolysis bullosa, you must each carry the allele. Any child you have has a 25% chance of having the disease." OD. "Because you have had one child with epidemolysis bullosa, you must both carry the allele. However, since the chance of having an affected…
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?
Using the pedigree below for an autosomal dominant disorder to determine the phase (which ones are recombinants and which are non-recombinants? Identify them by pedigree position (11, 112, etc.) (SLO4) 1 11 III 2/2 1/2 1/1 -O 2/12 2/2 2/2 2/2 2/2 1/2 2/2 1/2 2/2 1/2 2/2 a. 0.01 b. 0.1 c. 0.2 Calculate the LOD score for each theta below for the pedigree above. Note: keep lots of places behind the decimal until the very end for accuracy. (SLO4)