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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?Alberta is phenotypically normal, but her brother (Rodrigo) has albinism, which is caused by an autosomal recessive mutation. The probability that Alberta is a carrier (i.e., heterozygous for albinism) is [express your answer as a fraction]
- 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)Pedigree attached shows an autosomal recessive genetic disease. G is the normal allele and g is the disease-causing allele. Individual 1’s father is heterozygous (*) and his mother is homozygous dominant. Other individuals in the pedigree may be carriers, but are not marked. The question mark (?) indicates that you do not yet know anything about this individual’s phenotype with regard to the disease. part a) What is the probability that individuals 1 and 2 will have a child (5) who is a boy with the disease (the child is unborn and the sex is not yet known)? a)1/8 b)1/4 c)0 d)1/16 part b) What is the probability that the daughter (6) that individual 3 and 4 just had will have the disease? a)1/8 b)1/6 c)1/4 d)1/12The following pedigree illustrates the inheritance of Nance–Horan syndrome, a rare genetic condition in which affected people have cataracts and abnormally shaped teeth.
- A woman who sought genetic counseling is found to be heterozygousfor a chromosomal rearrangement between the second andthird chromosomes. Her chromosomes, compared to those in anormal karyotype, are diagrammed on the next page:(a) What kind of chromosomal aberration is shown?Neurofibromatosis-1 (NF1) is an autosomal dominant disorder where tumours form in the base layer of the skin or in nerve tissues. What is the probability that individuals II-1 and II-2 will have a genetic son with NF1? Find the image attached.Shown above is a family pedigree tree in which family members afflictedwith the disease Haemophilia are shown with filled-in squares (male) or circles (females). A couple is trying to determine the likelihood of passingon the disease to their future children (represented by the ? symbolabove) because the hemophilia runs in the woman’s family. Turner syndrome is a disease in which an individual is bornwith only a single X chromosome. Suppose the woman in thecouple is a carrier for hemophilia and has a child with Turnersyndrome. Would this child have the disease?
- From the powerpoint presentation, choose five (5) terms or concepts pertaining to chromosomal abnormality in the number. Describe each in a sentence. https://www.slideshare.net/farhanali911/chromosomal-abnormalities-33461290My Question is what is the probability their first child will have hemophilia and drawn pedigrees for family members with genotypes. My explantion so far: A man has both X and Y chromosomes as sex chromosomes in his body. Here, though the brother of the man is hemophiliac, a man can’t be a carrier of hemophilia. So, it can be said that his chromosome is “XnY”.Here, the “n” stands for “normal”.Though the paternal uncle is hemophiliac, a man cannot be a carrier of hemophilia, his niece will not be a career. So it can be said that the woman is also not a carrier and has the “XnX” chromosome.So, as the mother is not a carrier, their first child does not have a chance of having hemophilia. This can be determined as it is known that there is no hidden carrier of hemophilia in the family.The mother of a family with 10 children has blood typeRh+. She also has a very rare condition (elliptocytosis,phenotype E) that causes red blood cells to be oval rather than round in shape but that produces no adverseclinical effects. The father is Rh− (lacks the Rh+ antigen)and has normal red blood cells (phenotype e). The children are 1 Rh+ e, 4 Rh+ E, and 5 Rh− e. Information isavailable on the mother’s parents, who are Rh+ E andRh− e. One of the 10 children (who is Rh+ E) marriessomeone who is Rh+ e, and they have an Rh+ E child.a. Draw the pedigree of this whole family.b. Is the pedigree in agreement with the hypothesisthat the Rh+ allele is dominant and Rh− is recessive?c. What is the mechanism of transmission ofelliptocytosis?d. Could the genes governing the E and Rh phenotypesbe on the same chromosome? If so, estimate the mapdistance between them, and comment on your result