Sickle-cell anemia is a human disorder caused by a recessive allele (HbS) for the hemoglobin gene found on chromosome #11. The normal allele (HbA) is dominant. Mary has sickle-cell anemia, and she marries Steve, who does not have the disease. Mary and Steve have one kid Rosa, who has sickle-cell anemia. What is Steve's genotype? a. HbS Ob. HbSHbS c. HbAHbS Od. HbAHbA Oe. Cannot be determined
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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?Mike was referred for genetic counseling because he was concerned about his extensive family history of colon cancer. That family history was highly suggestive of hereditary nonpolyposis colon cancer (HNPCC). This predisposition is inherited as an autosomal dominant trait, and those who carry the mutant allele have a 75% chance of developing colon cancer by age 65. Mike was counseled about the inheritance of this condition, the associated cancers, and the possibility of genetic testing (on an affected family member). Mikes aunt elected to be tested for one of the genes that may be altered in this condition and discovered that she did have an altered MSH2 gene. Other family members are in the process of being tested for this mutation. Once a family member is tested for the mutant allele, is it hard for other family members to remain unaware of their own fate, even if they did not want this information? How could family dynamics help or hurt this situation?
- Mike was referred for genetic counseling because he was concerned about his extensive family history of colon cancer. That family history was highly suggestive of hereditary nonpolyposis colon cancer (HNPCC). This predisposition is inherited as an autosomal dominant trait, and those who carry the mutant allele have a 75% chance of developing colon cancer by age 65. Mike was counseled about the inheritance of this condition, the associated cancers, and the possibility of genetic testing (on an affected family member). Mikes aunt elected to be tested for one of the genes that may be altered in this condition and discovered that she did have an altered MSH2 gene. Other family members are in the process of being tested for this mutation. Is colon cancer treatable? What are the common treatments, and how effective are they?Mike was referred for genetic counseling because he was concerned about his extensive family history of colon cancer. That family history was highly suggestive of hereditary nonpolyposis colon cancer (HNPCC). This predisposition is inherited as an autosomal dominant trait, and those who carry the mutant allele have a 75% chance of developing colon cancer by age 65. Mike was counseled about the inheritance of this condition, the associated cancers, and the possibility of genetic testing (on an affected family member). Mikes aunt elected to be tested for one of the genes that may be altered in this condition and discovered that she did have an altered MSH2 gene. Other family members are in the process of being tested for this mutation. Seventy-five percent of people who carry the mutant allele will get colon cancer by age 65. This is an example of incomplete penetrance. What could cause this?State whether each of the following genetic defects is inherited as an autosomal recessive, autosomal dominant, or X-linked recessive trait: phenylketonuria (PKU), sickle cell anemia, cystic fibrosis, Tay-Sachs disease, Huntingtons disease, and hemophilia A.
- Aav AaBbCc Normal No Spacing Heading 1 Paragraph Styles In man, two abnormal conditions, cataracts (C) in the eyes and excessive fragility (F) in the bones, seem to depend on separate dominant genes located on different chromosomes. Normal vision and normal bones are recessive traits. A man with cataracts and nomal bones, whose father had normal eyes, married a woman free from cataracts but with fragile bones. Her father had normal bones. 11. What is the genotype of the man with cataracts and nomal bones? What is the genotype of the woman with normal vision and fragile bones? What type of offspring might this couple expect? Genotypes Phenotypes What is the probability that their first child will, (a) be free from both abnormalities (b) have cataracts but not fragile bones (c) have fragile bones but not cataracts (d) have both cataracts and fragile bones? liliRecessive hemophilia, x-linked. The man has it, and his wife is a carrier. Draw a Punnett square to show this. What is the probability that their daughter has hemophilia? What is the probability that their son does?i). pronde. clear- cut. explenations.in differences/Similarīties, if ony, between." Extensnons of mendelren. mhentonce Vs. Non.- mendellen. nhentnce.
- 2) Indicate the pattern of inheritance for the human genetic disorders. Use letter symbols for your answers (AR, AD, XR,XD, M) where appropriate. Table 2. HUMAN GENETIC DISORDER PATTERN OF INHERITANCE Marfan Syndrome Sickle Cell Anemia Classical Hemophilia Hypophosphatemia Cystic Fibrosis Phenylketonuria Huntington’s Disease Tay Sachs Disease Neurofibromatosis Alkaptonuria Xeroderma pigmentosum Kearns-Sayre Syndrome Achondroplasia Beta thalassemia Duchene Muscular DystrophyConsider this pedigree showing an autosomal dominant rare disorder. What is the degree of penetrance? Show your work. na оп 16 19 fa 16 R 9Xtion 8: below is the pedigree of inheritance of phenylketonuria (PKU). We will designate the letter Caven for the dominant allele and "p" for the recessive allele. 4 The pedigree shows that the pattern of inheritance for the allele for phenylk ylketonuria is: I. II. 1 III. IV. Autosomal dominant Autosomal recessive X-linked dominant X-linked recessive b. The parents in generation I have how many children: I. 3 Boys II. 3 Girls III. IV. 3 Boys and 1 Girl 3 Girls and 1 Boy c. What is the genotype of individual 1 in generation III: I. PP II. pp III. Pp " O 1 III. 50% E III 1 ▬ 2 2 IV. 25% 1 3 IV. Can be PP or Pp ii. Suppose that a man having type AB blood marries a woman having type O blood. What is the probability that their child will have type A blood? I. 100% II. 75% 2 4 3