Help me create a pedigree of this information: Pedigree analysis: Generation 1: Normal parents (AA x AA) Generation 2: Carrier parents (AA x AS) Generation 3: Affected child (AS x AS) Generation 4: Affected grandchild (SS) This pedigree has two normal parents in the first generation. Second generation carriers carry the sickle cell trait from one parent. The disease is 25% more likely to be inherited in the third generation if both parents have the 'S' allele. If both parents have the 'S' allele, their children will have sickle cell anemia in the fourth generation
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Help me create a pedigree of this information:
Pedigree analysis:
Generation 1: Normal parents (AA x AA)
Generation 2: Carrier parents (AA x AS)
Generation 3: Affected child (AS x AS)
Generation 4: Affected grandchild (SS)
This pedigree has two normal parents in the first generation. Second generation carriers carry the sickle cell trait from one parent. The disease is 25% more likely to be inherited in the third generation if both parents have the 'S' allele. If both parents have the 'S' allele, their children will have sickle cell anemia in the fourth generation
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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?A proband female with an unidentified disease seeks the advice of a genetic counselor before starting a family. Based on the following data, the counselor constructs a pedigree encompassing three generations: (1) The maternal grandfather of the proband has the disease. (2) The mother of the proband is unaffected and is the youngest of five children, the three oldest being male. (3) The proband has an affected older sister, but the youngest siblings are unaffected twins (boy and girl). (4) All the individuals who have the disease have been revealed. Duplicate the counselors feat
- PLEASE tell me what each pedigree diagram is. so which one is most likely to show a family with Haemophilia A? most likely to show a family with Gaucher Disease? most likely to show a family with Sickle Cell Anaemia? most likely to show a family with Achondroplasia? most likely to show a family with Goltz Syndrome? What is the most likely inheritance pattern shown in image B, below?Cystic Fibrosis (CF) is an autosomal recessive condition. Therefore, heterozygous (Cc) carriers do not display symptoms. Two parents who are carriers plan to start a family and you are a genetic counselor helping to advise them about their chances of having children affected by CF. a) Suppose the couple has 4 children, each one year apart. What is the probability that all 4 children will inherit CF? b) What is the probability that any 3 of their 4 children will not inherit CF, but 1 will be affected? c) What is the probability that their first child will not inherit CF, but the younger 3 children will inherit CF?Niemann Pick Type C disease is a recessive disorder that causes the accumulation of cholesterol and other lipids in lysosomes, ultimately affecting both the liver and the nervous system. Below are the genotypes and phenotypes of offspring of a family with a history of Niemann Pick. 7 NN ( all normal phenotype) 3 Nn (all normal phenotype) 4 nn (1 early onset dementia, 1 mid-life onset dementia, 2 late-onset dementia). From this information, Niemann-Pick disease is an example of: A) variable expressivity B) incomplete dominance C) incomplete penetrance D) variable expressivity and incomplete penetrance E) multiple alleles
- 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)1/12My 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.Please help me with the following question: Von Willebrand disease is an inherited bleeding disorder. People with von Willebrand disease take a much longer period for blood to clot/stop than others. von Willebrand disease is either inherited in an autosomal dominant pattern or in an autosomal recessive pattern. Question: what is the genotype of the disorder? what are the phenotypic effects of the disorder? What is happening with the DNA to cause the phenotypic effects?
- While sitting at home during Movement Control Order (MCO) because of pandemic covid19, observe two different traits of a couple in your family (eg. your mom & dad or your sister & her husband or your brother & his wife, etc). Draw a genetic cross that involves cross of the parents with the chosen 2 pairs of their contracting traits. Imagine that the cross obeys the Mendelian Laws, show the cross and gametes production for each generation (P, F1 and F2). By Using a Punnet square as symbolic representation of the results for the cross, determine the phenotypes, genotypes, phenotypic ratio and genotypic ratio of F2 generation in the family.A research team genotyped hundreds of families affected by haemophilia. Among those families, they found many that had a pattern of RFLP like that of daughter1, her husband and her parents. When they genotyped the children in those families, they found the following association between the RFLP and hemophilia: The numbers at the bottom indicate the number of progeny with that combination of hemophilia phenotype and RFLP polymorphism found among these families. What is the map distance between the RFLP and the hemophilia gene? Show your work.Please consider the following pedigree. Assume that people who marry in to the family do not carry the allele unless otherwise indicated. Assume complete penetrance. I II 5 6 III 6 IV 1 2 a. Is it possible for the inheritance pattern for the trait illustrated in this pedigree to be as a result of each of the following? Answer yes or no. (i) an autosomal recessive allele (AR) (ii) an autosomal dominant allele (AD) (iii) a X-linked recessive allele (XR) (iv) a X-linked dominant allele (XD) b. Provide a genotype for individual III-6 for the most likely mode of inheritance as determined in (a).