Pedigree 1: What is the most likely mode of inheritance of this disease given itspattern on the pedigree?Autosomal DominantAutosomal RecessiveO X-Linked DominantO X-Linked Recessive Complete the following questions BEFORE ATTEMPTING THE HW 11 BLACKBOARD ASSIGNMENT.In the following human pedigrees, the filled symbc epresent the affected individuals who suffer fromthe disease. Use A/a to represent alleles for autosomal traits and XA/Xa/Y to represent alleles forX-linked traits se the uppercase letter to represent the dominant allele and the lowercase lettertorepresent the recessive allele.atlected female O Unaffected femaleaffected maleUnaffected malePedigree 1:A. What is the most likely mode ofinheritance of this disease?Choose from: autosomaldominant, autosomal recessive,X-linked dominant, X-linkedrecessive.B State the genotypes of individuals#1 - #3.C. What is the probability thatindividual #4 is a carrier of thisdisease if his mother ishomozygous dominant and hisfather is heterozygous?Pedigree 2What is the most liely mode of

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Description: Pedigree 1: What is the most likely mode of inheritance of this disease given itspattern on the pedigree?Autosomal DominantAutosomal RecessiveO X-Linked DominantO X-Linked Recessive Complete the following questions BEFORE ATTEMPTING THE HW 11 BLACKB

Pedigree chart shows the inheritance of traits from parents to their offspring.

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Pedigree 1: What is the most likely mode of inheritance of this disease given its pattern on the pedigree? Autosomal Dominant Autosomal Recessive O X-Linked Dominant O X-Linked Recessive

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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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 3: What is the most likely mode of inheritance of this disease given its pattern on the pedigree? Autosomal Dominant O Autosomal Recessive O X-Linked Dominant O X-Linked Recessive
Pedigree 2: A. What is the most likely mode of inheritance of this disease? Choose from: autosomal dominant, autosomal recessive, X-linked dominant, X-linked recessive. B State the genotypes of individuals # 1 #4. C If individual #3 has another daughter with the same partner, what is the probability that this daughter will be affected (show the disease)?
Pedigree 2: A. What is the most likely mode of inheritance of this disease? Choose from: autosomal dominant, autosomal recessive, X-linked dominant, X-linked recessive. B State the genotypes of individuals # 1 #4. C If individual #3 has another daughter with the same partner, what is the probability that this daughter will be affected (show the disease)?
Pedigree 4: A. What mode of inheritance supports the pattern of this disease in this family? Choose from: autosomal dominant or autosomal recessive. B. State the genotypes of individuals #1 - #4 for this mode of inheritance. C. Ifindividual #3 has another child with the same partner, what is the probability that this child will have the disease? hp fg f10 トト %24 4. & 7 8 T
autosomal recessive allele (not sex-linked). Omplete the following monohybrid crosses for different types neritance pattefh autosomal dominant, sex linked recessive, and dominant inheritance. Inheritance of autosomal recessive traits Female parent phenolype: Example: Albinism Albinism (lack of pigment in hair, eyes and skin) is inherited as an Male parent phenatype: Using the codes: PP Pp (normal) (albino) la) Enter the parent phenotypes and complete the Punnett square for a cross between two carrier genotypes. A Give the ratios for the phenotypes from this cross. Pp (carrier) eggs sperm Phenotype ratios: Inheritance of autosomal dominant traits Example: Woolly hair Woolly hair is inherited as an autosomal dominant allele. Each affected individual will have at least one affected parent. Using the codes: WW (woolly hair) Female parent phenotype: Male parent phenotype: Ww (woolly hair, heterozygous) W w (normal hair) (a) Enter the parent phenotypes and complete the Punnett square for a…
2/7 - <. Hair texture is an incompletely dominant trait in humans. The three phenotypes are curly, wavy and straight. Straight only occurs when both parents have straight hair. Curly hair also tends to follow this pattern. Only two wavy haired parents produce all three phenotypes. Please explain this observation using punnett squares. Curly x Curly straight x straight Wavy x wavy
Complete the following queatione BEFORE ATTEMPTING THE HW 1 BLACKBOARD ASSIGNMENT, In the following human pedigrees, the filled symbols represnent the affected individuals who suffer from the disease. Use A/a to represent alleles for autosomal traits and XIX/Ytorepresent alleles for X-linked traits. Use the uppercase letter to represent the dominant allele and the lowercase letterto represent the receasive allele. afected fomale Unafected female affected male Unaffected male Pedigree 1: A. What is the most likely mode of inheritance of this disease? Choose from: autosomal dominant, autosomal recessive, X-linked dominant, X-linked recessive. B. State the genotypes of individuals #1-#3. C. What is the probability that individual #4 is a carrier of this disease if his mother is homozygous dominant and his father is heterozygous? Pedigree 2: What is the most likely mode of inhentance of this disease? Choose from: autosomal dominant, autosomal recessive X-linked dominant X-linked recessive.…
Consider the following pedigree. 하 3 10 (5 3 2 (a) What pattern of transmission is most consistent with this pedigree? (1) autosomal recessive, (2) autosomal dominant, (3) X-linked recessive, (4) X-linked dominant. (b) If individual V-2 marries a normal individual, and if the condition has a pene-trance of 85 percent, what is the probability that their second child will express the trait? (c) On the third line, what does the diamond with a 10 in the middle mean?
X-linked Recessive Inheritance A gene is described as X-linked when it occurs on the X chromosome and not the Y. Our convention is to indicate X-linkage by attaching the appropriate gene symbol as a superscript on the letter X. Commonly, the wild-type (+) allele is indicated with only a "+" to avoid having to type a superscript on a superscript. For example, a female that is heterozygous and carrying a recessive mutant allele is indicated as X+Xm. Note the convenience of the shorthand + for m+ in this situation. A mutant male has the genotype XmY. When working with X-linked inheritance, always include the X and Y chromosomes in the descriptions of genotypes, and include the sex (male or female) in the descriptions of the phenotypes (e.g., mutant male, wild-type female, etc.). Here are the genotypes and associated phenotypes for X-linked recessive inheritance: X+X+ Wild-type female X+Xm Wild-type female xmxm Mutant female X+Y xmy Wild-type male Mutant male
Match the disease with the pattern of inheritance Phenylketonuria Achondroplasia Autosomal Autosomal X-linked X-linked dominant recessive recessive dominant O Y-linked