Y-: yellow yy: black S-: star ss: starless All the sneetches want their children to have stars on their bellies. An embryo with either YYss or yySS genotype is lethal. Y and S are two independently assorting autosomal genes. A true breeding yellow star bellied sneetch mated with a true breeding black starless sneetch and produced 20 F₁ sneetches. The F₁ then mated with each other and produced 420 F2 progeny.
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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 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-linked
- In humans, ABO blood types refer to glycoproteins in the membranes of red blood cells. There are three alleles for this autosomal gene: IA, IB, and i. The IA allele codes for the A sugar, The IB allele codes for the B sugar, and the i allele doesn't code for any sugar. IA and IB are codominant, and i is recessive to both IA and IB. If an individual with type AB blood has a child with an individual with type O blood, what blood types could their children possibly have?In Drosophila, the white gene located on the X chromosome affects eye color; an autosomal gene, wingless, is on an autosomal chromosome. Use the following allele symbols: Xw+ _ , Xw+Y = wild type red eyes; X-linked dominant allele Xw Xw , XwY = white eyes; X-linked recessive allele Y = Y sex chromosome vg+ = wild type wings; autosomal dominant vg = wingless; autosomal recessive Predict ratios/proportions of genotypes and phenotypes of offspring from the following cross, of a white-eyed male with wild type wings and a wild type red eyed female with wild type wings: indicate sex of offspring along with phenotypes. XwY vg+ vg x Xw+Xw vg+vgX-linked ichthyosis is an X-linked recessive trait that manifests in part as dry, scaly skin (“ichthy-” = fish or fish like). Suppose a couple are considering having a child together. Parent A is heterozygous for the ichthyosis allele while Parent B is hemizygous negative for the ichthyosis allele. What is the probability their child would be unafflicted with ichthyosis but be a carrier of the ichthyosis-causing allele? a.0% b.25% c.50% d.75% e.100%
- The most common form of colorblindness is a recessive, sex-linked hereditary con dition caused by a defect on the X chromosome. Females are XX, while males are XY. Individuals inherit one chromosome from each parent, with equal probability; for example, an individual has a 50% chance of inheriting their father's X chromosome, and a 50% chance of inheriting their father's Y chromosome. If a male has an X chromosome with the defect, he is colorblind. However, a female with only one defective X chromo some will not be colorblind. Thus, colorblindness is more common in males than females; 7% of males are colorblind but only 0.5% of females are colorblind. (a) Assume that the X chromosome with the wild-type allele is X+ and the one with the disease allele is X. What is the expected frequency of each possible female genotype: X+X+, X+X¯, and X-X-? What is the expected frequency of each possible male genotype: X+ Y and X-Y? (b) Suppose that two parents are not colorblind. What is the…Hemophilia is another example of a X-linked disease caused when a recessive allele (Xh) is expressed. If a normal male reproduces with a heterozygous normal female, what are the expected genotypes and phenotypes? Will any of their daughters develop hemophilia? you must also give the gender of the child in your genotype and phenotype descriptions here.A type of red-green colorblindness is inherited recessively on the X chromosome. A woman who is a carrier had children with a male that is not affected by the disease. Show your answers with the use of a Punnett square • What is the probability (% or likelihood) that a child is affected by the disease? • What is the probability (% or likelihood) that a son is a carrier? (Careful!) • Is there a chance that a girl of the couple is affected by the disease?
- Hillary is 8 years old and has neuronal ceroid lipofuscinosis, also known as Batten disease. It is autosomal recessive and always affects children before the age of 2. Her parents are not affected. Her younger brother Jaden, age six, is unaffected. What is the probability that Jaden is homozygous dominant for this trait? Be sure to construct a pedigree and punnet squate to defend your answer.(Please label your pedigree to show the name of each individual, or write in mom and dad )Red-green color blindness is inherited as an X-linked recessive (Xc). If a color-blind man marries a woman who is heterozygous for normal vision, what would be the expected phenotypes of their children with reference to this character? In your answer, specify in your phenotype descriptions the gender of the children. (For example, don’t just say 75% of the children would be colorblind – you would instead say 100 % of the daughters would be colorblind and 50% of the sons would be colorblind. Note that this is not a correct answer; it is just to give you an idea of how to explain the correct phenotypes of the cross.)___In mice the autosomal genotype yy is for gray fur, YY is for brown fur and Yy is for yellow fur. What is the ratio of offspring from the cross between a brown mouse and a yellow mouse? ABO blood groups, an autosomal trait, in Humans are genetically determined. A woman with type O blood has a son with type O. What are the possible genotype(s) of the son, his mother and possible father? Mother: Father: Son: Polydactylism (extra digits) is due to a dominant autosomal allele, what is the likelihood that a couple with the normal number of digits have a son with polydactylism? Flower color in snapdragons is due to a gene with incomplete dominance: CRCR plants have red flowers, CRCW have pink flowers, and CWCW plants have white flowers. Which cross is expected to yield progeny that all have pink flowers? A couple is expressing a child, the man has hemophilia (a sex-linked blood disorder due to a recessive allele) and the woman is a carrier…