e. If couple I-1 and I-2 will have a son, what is the of having the disorder? probability f. If couple III-8 and III-9 will have another child, what is the probability of having the disorder? g. Theoretically, if individual IV-3 and individual IV-5 will marry and will have a child, what is the probability of having a child without the X-linked disorder?
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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?Analysis of X-Linked Dominant and Recessive Traits Suppose a couple, both phenotypically normal, have two children: one unaffected daughter and one son affected with a genetic disorder. The phenotype ratio is 1:1, making it difficult to determine whether the trait is autosomal or X-linked. With your knowledge of genetics, what are the genotypes of the parents and children in the autosomal case? In the X-linked case?
- Pedigree Analysis Is a Basic Method in Human Genetics Using the pedigree provided, answer the following questions. a. Is the proband male or female? b. Is the grandfather of the proband affected? c. How many siblings does the proband have, and where is he or she in the birth order?Analysis of X-Linked Dominant and Recessive Traits As a genetic counselor investigating a genetic disorder in a family, you are able to collect a four-generation pedigree that details the inheritance of the disorder in question. Analyze the information in the pedigree to determine whether the trait is inherited as: a. autosomal dominant b. autosomal recessive c. X-linked dominant d. X-linked recessive e. Y-linkedAchondroplasia is a rare dominant autosomal defect resulting in dwarfism. The unaffected brother of an individual with achondroplasia is seeking counsel on the likelihood of his being a carrier of the mutant allele. What is the probability that the unaffected client is carrying the achondroplasia allele?
- Could someone please help me with this grade 11 bio dihybrid cross problem in detail and how to solve this question, using a strategy? Assume that curly hair (C) is dominant to straight hair. Albinism (P ) is recessive to normal skin pigmentation. A woman who is heterozygous for curly hair and albinism has a child. The father is homozygous dominant for curly hair and has albinism. (a) Determine the possible phenotypes for their child. (double-crossing, Puneet square)(b) Calculate the four different probabilities of a child beingboth a male and of each phenotype.(c) What is the probability that the child will expressalbinism and have curly hair like his father?Please answer all of them, they are all connected. PEDIGREE ANALYSIS and SYMBOLOGY: Examine the pedigree which has X linked Dominant inheritance of disorder. Use letter X* (asterisk denotes disorder) as genotype of the individuals which can be XX, XY, X*X*, X*X and X*Y. a. What is the genotype of IV-6? b. What is the genotype of III-6? c. What is the genotype of II-3? d. What is the genotype of III-8? e. If couple I-1 and I-2 will have a son, what is the probability of having the disorder? f. If couple III-8 and III-9 will have another child, what is the probability of having the disorder? g. Theoretically, if individual IV-3 and individual IV-5 will marry and will have a child, what is the probability of having a child without the X-linked disorder?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)1/12
- Could someone please help me with this grade 11 bio dihybrid cross problem on how to solve this question, using a strategy in detail! Assume that curly hair (C) is dominant to straight hair. Albinism (P ) is recessive to normal skin pigmentation. A woman who is heterozygous for curly hair and albinism has a child. The father is homozygous dominant for curly hair and has albinism. (a) Determine the possible phenotypes for their child. (double-crossing, Punnet square) (how to display the traits) (b) Calculate the four different probabilities of a child beingboth a male and of each phenotype. (steps in solving this) (c) What is the probability that the child will expressalbinism and have curly hair like his father? (steps in solving this)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?PEDIGREE: Shaded individuals in the pedigree have a genetic disease. Individuals marrying into the family, that is individuals II–1, II–4 and II–6, have no history of the disease in their families.1. Determine the mode of inheritance ______________________________2. Give the genotypes of the following individuals