The pedigree below shows the inheritance pattern of an autosomal dominant trait in a family. Given this nformation, which individuals are heterozygous for the gene encoding this trait? I II III по O 1-2, 11-3, 11-7 3 O More information needed 4 6 3 O 1-1, 11-4, 11-6 All of the individuals expressing this trait are heterozygous O All of the individuals not expressing this trait are heterozygous 7 4 8 5 9 10 11 12 13
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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?Examine the following pedigrees. Which is the most likely mode of inheritance of each disorder? (a) autosomal recessive (b) autosomal dominant (c) X-linked recessive (d) a, b, or c (e) a or c 10.
- What is the most likely inheritance pattern shown in image B, below? B A E KEY Homozygous Homozygous Heterozygous Heterozygous Wild Type Male Female Male Female Male Note: Completely red symbol denotes an individual exhibiting the phenotype of interest CI 11 III IV V 1/4 A Autosomal Dominant Autosomal Recessive Sex-linked Dominant Sex-Linked Recessive Mitochondrial 1/2 1/2 1/2 1/2 Wild Type Female 1/4 1/2 B Affected Known carrier Affected female Normal female Affected male Normal maleWhich of the Pedigree Diagrams below is most likely to show a family with Hereditary haemorrhagic telangiectasia? A Affected female Normal female Atfected mate ormal male KEY Homozygous Homazygous Heterozygous Heterazygous Wild Type Male Wild Type Female Male Female Female Mala Note: Completely red symbol denotes an individual exhibiting the phenotype of interest C 3 12 1/2 IV 14 14 Affected 12 12 Known camerIn this case a family history revealed a genetic basis for the disorder. The pedigree is shown in Fig. 1 Below. Key Ø Female: affected Female: unaffected || IV V 5600 orize 077808 15 10 9 10 CHO વ Male: affected Male: unaffected Deceased Disease status not given Dizygotic twins Monozygotic twins Fig. 1 Disease pedigree. Five generations I, II, III, IV, V are shown. Females are represented by circles, males by squares, dizygotic (non-identical) twins by diagonal lines originating from the same point, Monozygotic (identical) twins by diagonal lines originating from the same point and joined symbols and deceased by a diagonal line through the symbol. Filled symbols indicate that the individual displays the disease phenotype. Unfilled symbols indicate that the individual does not display the disease phenotype. Carriers of the disease are not indicated. Information on disease status is not known for generation I and is omitted for the individuals represented by a symbol with an asterisk.…
- The pedigree shows inheritance of an autosomal recessive disease in an extended family. Assume unrelated individuals marrying into the family do not carry the disease, unless there is reason to believe otherwise. What is the chance that IV-3 and IV-4 will have a child with the disease? Individuals I-1, Il-5, III-5 and III-16 have the disease. 2 1 7. III-18 2 3 5 6 17 8. 9 10 11 12 13 14 15 16 17 III-19 IV IV-3 IV-4 IV-5 IV-6 IV-1 IV-2 O a. 1/8 b. 1/12 C. 1/16 d. 3/16 e. 1/24 f. 1/32 g. 3/32 h. 1/64 FEB 17 MacBook Air 6, ... 5. %DWhich of the Pedigree Diagrams above is most likely to show a family with Achondroplasia? A KEY Homozygous Homozygous Heterozygous Heterozygous Wild Type Male Female Male Female Male с. Note: Completely red symbol denotes an individual exhibiting the phenotype of interest 11 IV V 1/4 1/2 1/2 ₂₂( 1/2 1/2 5 1/4 Wild Type Female 1/2 B Affected Known carrier Affected female Normal female Affected male Normal male DI II III II-1 11-2 Aa 1-1 11-3 Aa 1-2 II-4 III-1 Example: Given the pedigree above, what is the probability that the granddaughter (III-1) will be heterozygous? In order to determine the probability that III-1 is heterozygous, we need to determine the probability of the possible genotypes for her mother (II-2). Because the grandparents are both heterozygous, we would expect the following genotypic ratios in their offspring: ¼ AA, ½ Aa, and 14 aa. However, we have to use all possible information, and the circle representing III-1 is not shaded in, so she cannot be aa. Therefore, we eliminate the probability that II-2 is aa, and the final probability for II-2 is 1/3 AA and 2/3 Aa. We assume II-1 is AA because he is marrying into the family and we assume everyone marrying into a family is homozygous wild-type, unless proven otherwise. If II-2 is AA, and we assume II-1 is AA, then there is no chance that their daughter (III-1) is Aa If II-2 is Aa, and we assume II-1 is AA, then there is a…
- In the following pedigree of an autosomal recessive disorder, what is the probability that IV-1 will be affected? I II III IV 1/2 1/12 O 3/4 2/3 O 1/4 Rr 1 R 2 Rr 2 R 3 RR 3 R 1 5 Rr 4 2Cystic fibrosis (CF) is an autosomal recessive trait. A three-generation pedigree is shown below for a family that carries the mutant allele for cystic fibrosis. Note that carriers are not colored in to allow you to figure out their genotypes. Normal allele = F CF mutant allele = f What is the genotype of individual #8?The following pedigree shows a family in which an inherited condition is apparent. The muscle biopsy from the one of the affected persons shows ragged red fibers and parking lot inclusions on microscopy. What is the most likely mode of inheritance for this condition? Answers A - E A Autosomal Dominant B Autosomal Recessive C Mitochondrial D X-linked Dominant E X-linked Recessive O O TO 0 ☐ Q