Question is in the image. Thank you so much! 3.A human pedigree for inheritance of a disease determined by a mutation in a mitochondrial geneis shown. To be affected, an individual must be homoplasmic for mutant mtDNA. Using the figureas a reference, classify each statement below as true or false.All the children of affected mothers will have this disease.If an affected female has a child with an unrelated, normal male, the chance that thechild will have the trait is 50%.ㅇㅇㅇㅇㅇㅇㅇ65SNone of the children of affected fathers can have the disease.If an affected male has a child with an unrelated, normal female, the chance that thechild will have the disease is 50%.Males can be unaffected even if they are homoplasmic for mutant mtDNA.If an unaffected female is heteroplasmic, some of her children could develop the trait.All affected fathers have unaffected offspring if the mother is normal.The phenotype of the father is unrelated to the phenotype of the offspring.

Name: Question is in the image. Thank you so much! 3.A human pedigree for in
Uploaded: 2024-03-20T06:43:04.000Z
Channel: Bartleby
Description: Question is in the image. Thank you so much! 3.A human pedigree for inheritance of a disease determined by a mutation in a mitochondrial geneis shown. To be affected, an individual must be homoplasmic for mutant mtDNA. Using the figureas a reference,

Mitochondria in a cell have their own chromosome. Though the size is small and it codes for proteins…

Science

Biology

3.A human pedigree for inheritance of a disease determined by a mutation in a mitochondrial gene is shown. To be affected, an individual must be homoplasmic for mutant mtDNA. Using the figure as a reference, classify each statement below as true or false. 0 TO что ㅇㅇㅇㅇㅇㅇㅇㅇㅇㅇㅇ 658 58 All the children of affected mothers will have this disease. If an affected female has a child with an unrelated, normal male, the chance that the child will have the trait is 50%. None of the children of affected fathers can have the disease. If an affected male has a child with an unrelated, normal female, the chance that the child will have the disease is 50%. Males can be unaffected even if they are homoplasmic for mutant mtDNA. If an unaffected female is heteroplasmic, some of her children could develop the trait. All affected fathers have unaffected offspring if the mother is normal. The phenotype of the father is unrelated to the phenotype of the offspring.

3.A human pedigree for inheritance of a disease determined by a mutation in a mitochondrial gene is shown. To be affected, an individual must be homoplasmic for mutant mtDNA. Using the figure as a reference, classify each statement below as true or false. 0 TO что ㅇㅇㅇㅇㅇㅇㅇㅇㅇㅇㅇ 658 58 All the children of affected mothers will have this disease. If an affected female has a child with an unrelated, normal male, the chance that the child will have the trait is 50%. None of the children of affected fathers can have the disease. If an affected male has a child with an unrelated, normal female, the chance that the child will have the disease is 50%. Males can be unaffected even if they are homoplasmic for mutant mtDNA. If an unaffected female is heteroplasmic, some of her children could develop the trait. All affected fathers have unaffected offspring if the mother is normal. The phenotype of the father is unrelated to the phenotype of the offspring.

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter11: Genome Alterations: Mutation And Epigenetics

Section: Chapter Questions

Problem 16QP: Familial retinoblastoma, a rare autosomal dominant defect, arose in a large family that had no prior...

See similar textbooks

Similar questions

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?
6. A person is simultaneously heterozygous for two autosomal genetic traits. One is a recessive condition foralbinism (alleles A and a); this albinism gene is foundnear the centromere on the long arm of an acrocentricautosome. The other trait is the dominantly inheritedHuntington disease (alleles HD and HD+). TheHuntington gene is located near the telomere of oneof the arms of a metacentric autosome. Draw all copies of the two relevant chromosomes in this person asthey would appear during metaphase of (a) mitosis,(b) meiosis I, and (c) meiosis II. In each figure, labelthe location on every chromatid of the alleles forthese two genes, assuming that no recombinationtakes place.
1. The pedigree below shows the incidence of rare, autosomal dominant disorder called Ehlers-Danlos disease. The pedigree covers three generations of a particular family and also shows individual genotypes at a potential marker locus (M). a) Indicate the phase of all gen II and III individuals. DdM1M3 ddM2M6 II DDM3M6 ddM4M5 III DdMзM4 DdMЗМ5 DDM3M4 ddM3M5 DDM3M4 ddM5M6 DDM3M4 ddM4M6 ddM5M6 ddM5M6 b) Which, if any, of the gen III individuals are recombinants? c) Calculate the LOD score as a test of physical linkage between the marker (M) and the disease locus. d) What do you conclude about linkage between D and M?
6. Identify the mode of inheritance for the following pedigree. Provide the genotypes of indicated individuals. ? Genotypes: Il-1 IV-1 V-6 Il-2 IV-2 V-12 Il-3 IV-3 V- 14 III-1 IV-4 V-16 III-13 V- 1 V-19
In individuals affected by cystic fibrosis, salt crystals may appear afterperspiration dries up. In addition, the disease causes respiratory disorderswhich can be both debilitating and lethal. It occurs in individuals homozygousfor the recessive gene. Two normal parents had a daughter with thesymptoms of this disease, and a normal son who marries a normal womanwith an afflicted A test (salt concentration in perspiration of heterozygotes ishigher than normal) disclosed that both are indeed carriers of the gene. If thefirst child born to the mating in (b) was defective, what is the probability thatthe 2nd child would also be defective?Express answer in fraction form
9. Make a pedigree for each of the following situations. For each individual, write the individual's genotype (when possible) next to the individual's symbol (e.g. O xty, I Gg): a. Two parents do not have cystic fibrosis and they have a daughter with cystic fibrosis and a son who does not have cystic fibrosis. The daughter grows up and she mates with a male who does not have cystic fibrosis. Their only child is a boy and he has cystic fibrosis. b. A man with hemophilia mates with a female without hemophilia. They have one son and one daughter. The daughter has hemophilia and the son does not have hemophilia. The son grows up, and he marries and mates with a female. Their only child is a boy, and he has hemophilia.
The XG locus on the human X chromosome has twoalleles, XG+ and XG. The XG+ allele causes the presence of the Xg surface antigen on red blood cells,while the recessive XG allele does not allow antigento appear. The XG locus is 10 m.u. from the STSlocus. The STS+ allele produces normal activity ofthe enzyme steroid sulfatase, while the recessive STSallele results in the lack of steroid sulfatase activityand the disease ichthyosis (scaly skin). A man withichthyosis and no Xg antigen has a normal daughterwith Xg antigen. This daughter is expecting a child.a. If the child is a son, what is the probability he willlack Xg antigen and have ichthyosis?b. What is the probability that a son would have boththe antigen and ichthyosis?c. If the child is a son with ichthyosis, what is theprobability he will have Xg antigen?
age 105 of 114 3 5 € BUD > D Q Autosomal Genes 1. Parental Cross: It is possible that your mutation is autosomal (dominant or recessive). In this case, you only need one Punnett square to make predictions for the F1 generation that will result from the parental cross, as it does not matter if the female or male in the parental generation is the true breeding mutant. a. Provide a Punnett square to predict the genotype of the F1 generation. b. Q Search If the mutant allele is dominant, predict the phenotype of the F1 generation. Make sure to also summarize the prediction as a ratio, comparing Wild-type : Mutant. c. If your mutant allele is recessive, predict the phenotype of the F1 generation. Make sure to also summarize the prediction as a ratio, comparing Wild-type : Mutant. 2. F1 Self-Cross: Take the F1 results from question #1 to create a Punnett square predicting the F2 generation genotypes. 104 a. What are the predicted F2 phenotypes if the mutant allele is dominant? Make sure to…
6. Below is a pedigree of a rare human genetic disease. The filled in symbols indicate affected individuals. Assume that the disease is caused by a mutant allele of gene A. a. Based on this pedigree, what is the most likely mode of inheritance? b. What is/are the possible genotype(s) of person 1, 2, 3 and 4? c. If individuals 1 and 4 have a child together, what is the probability that the child will exhibit the disease? 50% d. If individuals 2 and 3 have a child together, what is the probability that the child will exhibit the disease? 2 3 4
1. Construct pedigree charts using the inheritance of hemophilia in figure 92. This is X-linked inheritance so you are required to label XX for females and XY for males. The gene responsible for the trait is represented by the superscript which should be specified in the legend.