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. 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?X-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 maternal-effect mutation bicoid (bcd) is recessive. Inthe absence of the bicoid protein product, embryogenesis isnot completed. Consider a cross between a female heterozygousfor the bicoid mutation (bcd+/ bcd-) and a homozygousmale(bcd-/ bcd-). Predict the outcome (normal vs. failed embryogenesis) inthe F1 and F2 generations of the cross described.X-linked Recessive Inheritance A gene is described as X-linked when it occurs on the X chromosome and not the Y. Our convention is to indicate X-linkage by attaching the appropriate gene symbol as a superscript on the letter X. Commonly, the wild-type (+) allele is indicated with only a "+" to avoid having to type a superscript on a superscript. For example, a female that is heterozygous and carrying a recessive mutant allele is indicated as X+Xm. Note the convenience of the shorthand + for m+ in this situation. A mutant male has the genotype XmY. When working with X-linked inheritance, always include the X and Y chromosomes in the descriptions of genotypes, and include the sex (male or female) in the descriptions of the phenotypes (e.g., mutant male, wild-type female, etc.). Here are the genotypes and associated phenotypes for X-linked recessive inheritance: X+X+ Wild-type female X+Xm Wild-type female xmxm Mutant female X+Y xmy Wild-type male Mutant maleIn 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+vg
- The maternal-effect mutation bicoid (bcd) is recessive. Inthe absence of the bicoid protein product, embryogenesis isnot completed. Consider a cross between a female heterozygousfor the bicoid mutation (bcd+/ bcd-) and a homozygousmale(bcd-/ bcd-). How is it possible for a male homozygous for the mutationto exist?Whyte type canaries are yelllow. A dominant mutant allel of the color gene, designated W causes white feathers. Inheriting two dominant alleles is lethal to the embryo. If a yellow canary is crossed to a white canary, what is the probability that an offspring will be yellow? What is the probability that it will be white? [ autosomal special condition]Consider this pedigree showing an autosomal dominant rare disorder. What is the degree of penetrance? Show your work. na оп 16 19 fa 16 R 9X
- Tightly curled hair is caused by a dominant autosomal gene in humans. If a heterozygous curly-haired person marries a person with straight hair, what phenotypes (and in what proportions) are expected in the offspring? O 1 curly: 1 straight 1 curly: 2 straight O 1 curly: 3 straight O 2 curly: 1 straight O 3 curly: 1 straight Submit Previous Answers Request Answer X Incorrect; Try Again; One attempt remaining "In humans, the ABO blood type is under the control of autosomal multiple alleles. Color blindness is a recessive X-linked trait. If two parents who are both type A and have normal vision produce a son who is color-blind and is type O, what is the probability that their next child will be a female who has normal vision and is type O?Cross C: Homozygote wild type flies (+/+) were crossed with white eye flies. The table shows the results for the F1 and F2 generations. White eye females Wild type females White eye males Wild type males 50 F1 phenotypes (observed) F2 phenotypes (observed) F2 - expected phenotypes 50 60 60 60 60 What does the F1 tell you about the trait? Is the gene on an autosome or sex chromosome? What was the genotypes of each parent, male and female? What are the genotypes of the F1 flies? The F1 were crossed with each other and the phenotypes found in F2 are shown in the table above. Are these results consistent with the F1 and your theory of inheritance of this gene? Do a X2 test to see if the observed numbers are consistent with your theory.