Polydactly is an autosomal dominant phenotype. In this pedigree, I-2 has one copy of the dominant allele and has polydactyly. Individuals I-2 and Ill-2 have extra thumbnails or their hands. Individuals Il-3, III-1 and III-4 have extra digits on both hands and feet, Il-4 has an extra digit on one hand. 1-2 Il-3 Il-4 III-1 III-2 III-3 II-4 Is this an example of variable expressivity? Explain why or why not.
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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. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?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?This pedigree shows the inheritance of an autosomal dominant disorder. Il-1 and II-2 plan to have four children. What is the probability that 2 children will be affected and 2 will not? Hint: First find the probability that a child has the disease, then use the formula for binomial distribution. Success not affected. -p*q"-x P= I II n! (n-x)!x! 6/8 3/8 14/32 5/16 1/16 1 ? 2
- Cystic 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 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. %DThe following pedigree represents the inheritance of an autosomal recessive disease in a certain family. 2 5 2 4 5 8 3 4. 6. 1 1- How many males are affected by the disease? 3- How many children did the couple Il-4 and Il-5 have? 4- How many sisters did III-8 have? 5- What is the genotype of III-8? 40
- Adenike and her partner, Debare, are expecting their first child. Adenike is healthy as are her two sisters but her brother has PKU, a recessive disorder of the PAH gene. Genotypically he must be pah-/pah-. Neither of Adenike's parents has PKU but both of her grandmothers did. Debare is a carrier of Sickle Cell trait, which means he has one allele for wildtype hemoglobin (HbA) and one allele for Sickle Cell (HbS), making him HgA/HbS. More importantly, he has two alleles for Marfan Syndrome, a pleiotropic dominant disorder that can affect up to 30 different traits (M+/M+). No one in Debare's family has ever had PKU and no one in Adenike’s family has ever had Sickle Cell or Marfan Syndrome. They consult a genetics counselor over concern of the health of their baby. The genetics counselor collects the medical histories of three generations; their grandparents, their parents, and them. In her calculations, she produces a phenotypic key that indicates M = Marfan, Hb = sickle cell, and pah-…Adenike and her partner, Debare, are expecting their first child. Adenike is healthy as are her two sisters but her brother has PKU, a recessive disorder of the PAH gene. Genotypically he must be pah-/pah-. Neither of Adenike's parents has PKU but both of her grandmothers did. Debare is a carrier of Sickle Cell trait, which means he has one allele for wildtype hemoglobin (HbA) and one allele for Sickle Cell (HbS), making him HgA/HbS. More importantly, he has two alleles for Marfan Syndrome, a pleiotropic dominant disorder that can affect up to 30 different traits (M+/M+). No one in Debare's family has ever had PKU and no one in Adenike’s family has ever had Sickle Cell or Marfan Syndrome. They consult a genetics counselor over concern of the health of their baby. The genetics counselor collects the medical histories of three generations; their grandparents, their parents, and them. In her calculations, she produces a phenotypic key that indicates M = Marfan, Hb = sickle cell, and pah-…Adenike and her partner, Debare, are expecting their first child. Adenike is healthy as are her two sisters but her brother has PKU, a recessive disorder of the PAH gene. Genotypically he must be pah-/pah-. Neither of Adenike's parents has PKU but both of her grandmothers did. Debare is a carrier of Sickle Cell trait, which means he has one allele for wildtype hemoglobin (HbA) and one allele for Sickle Cell (HbS), making him HgA/HbS. More importantly, he has two alleles for Marfan Syndrome, a pleiotropic dominant disorder that can affect up to 30 different traits (M+/M+). No one in Debare's family has ever had PKU and no one in Adenike’s family has ever had Sickle Cell or Marfan Syndrome. They consult a genetics counselor over concern of the health of their baby. The genetics counselor collects the medical histories of three generations; their grandparents, their parents, and them. In her calculations, she produces a phenotypic key that indicates M = Marfan, Hb = sickle cell, and pah-…
- 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)1/12Please consider the pedigree below. There are no cases of false paternity. I II III IV в 1 a. Which individual/s definitely has/have Bombay phenotype in the descendants of I-1 and I-2? b. What are the genotypes of individuals II-2 and III-2 at the AB0 and H loci? Please label your answers a and b, Il-2: and Ill-2:.Two autosomal mutations include albinism and dwarfism. Albinism (a) is recessive, and dwarfism (D) is dominant. Complete a dihybrid cross of two people that are heterozygous for normal pigmented skin. One person is normal height and has no genetic trace of dwarfism in the family. The other person has dwarfism but has a mother who is normal height. 21. Complete a full dihybrid Punnett square (6pts).