Aristapedia is a lethal allele that is also dominant. Individuals with this trait must be heterozygous (Aa) because the homozygous condition (AA) is lethal. This is not a sex-linked trait. Wild-type flies do not carry the allele for aristopedia (aa). 4. Predict what the outcome of a cross between a wild-type fly and one with aristopedia. 5. Perform the cross and determine if your prediction is correct using statistical analysis. Summarize your results and indicate whether your prediction is confirmed. 6. Suggest a reason why a typical punnett square prediction for a heterozygous cross might be accurate in this case.

Aristapedia is a lethal allele that is also dominant. Individuals with this trait must be heterozygous (Aa) because the homozygous condition (AA) is lethal. This is not a sex-linked trait. Wild-type flies do not carry the allele for aristopedia (aa). 4. Predict what the outcome of a cross between a wild-type fly and one with aristopedia. 5. Perform the cross and determine if your prediction is correct using statistical analysis. Summarize your results and indicate whether your prediction is confirmed. 6. Suggest a reason why a typical punnett square prediction for a heterozygous cross might be accurate in this case.

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter10: From Proteins To Phenotypes

Section: Chapter Questions

Problem 3CS: A couple was referred for genetic counseling because they wanted to know the chances of having a...

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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. What if the couple wanted prenatal testing so that a normal fetus could be aborted?
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?
Neurospora of genotype a + c are crossed withNeurospora of genotype + b +. (Here, + is shorthandfor the wild-type allele.) The following tetrads areobtained (note that the genotype of the four sporepairs in an ascus are listed, rather than listing alleight spores):a + c a b c + + c + b c a b + a + ca + c a b c a + c a b c a b + a b c+ b + + + + + b + + + + + + c + + ++ b + + + + a b + a + + + + c + b +137 141 26 25 2 3a. In how many cells has meiosis occurred to yieldthese data?b. Give the best genetic map to explain these results.Indicate all relevant genetic distances, both betweengenes and between each gene and the centromere.c. Diagram a meiosis that could give rise to oneof the three tetrads in the class at the far right inthe list
Dihybrid Cross gray hair dominant In rabbits, is dominant to white hair. Also in rabbits, black red eyes. These letters the eyes are to represent genotypes of the rabbits: GG = gray hair Gg = gray hair gg = white hair BB = black eyes Bb = black eyes bb = red eyes 1. What the phenotypes (descriptions) of rabbits that have the are following genotypes? Ggbb ggBB ggbb GgBb
y 301 Amelogenesis imperfecta is X-linked dominant. Affected XY individuals have extremely thin enamel on the teeth while XX carriers have grooved teeth from uneven deposition of enamel. If an unaffected XY individual were to produce children with a XX carrier partner, a. what would be the expected chance of a XY child being affected with the disease? b. what would be the expected chance of a XY child being affected with the disease?
Polygenic trait Pleiotropic condition 1:1:1:1 9:3:3:1 Expressivity III III III III Digenic Mendelian ratio in the offspring when a heterozygote is test crossed. A single trait that is coded for by multiple genes. The degree to which a condition shows up in a population as a result of P = E + G Digenic Mendelian ratio in the offspring when two heterozygotes are mated. A single gene that can affect multiple traits.
7-8. Assume that white color is dominant over yellow color in squash. If pollen from the anthers of a heterozygous white-fruited plant is placed on the pistil of a yellow- fruited plant, show using ratios the genotypes and phenotypes you would expect the seeds from this cross to produce. Genotypes = 1/2 Ww 1/2 Ww = 1:1 ratio | Phenotypes = All white = 1:0 ratio Genotypes = 1/2 WW 1/2 ww = 1:1 ratio | Phenotypes = 1/2 white 1/2 yellow = 1:1 ratio Genotypes = 3/4 Ww 1/4 ww = 1:1 ratio | Phenotypes = 3/4 white 1/4 yellow = 1:1 ratio Genotypes = 1/2 Ww 1/2 ww = 1:1 ratio | Phenotypes = 1/2 white 1/2 yellow = 1:1 ratio
An individual has the following genotype. Gene loci (A) and (B) are 15 m.u. apart. What are the correct frequencies of some of the gametes that car be made by this individual? Bl a O A. Ab = 7.5%; AB = 42.5% B. ab = 25%; aB = 50% O C. AB = 7.5%; aB = 42.5% O D. aB = 15%; Ab = F0% E. aB = 70%; Ab = 15% Reset Selection OMark for Review What's This?
2/7 - <. Hair texture is an incompletely dominant trait in humans. The three phenotypes are curly, wavy and straight. Straight only occurs when both parents have straight hair. Curly hair also tends to follow this pattern. Only two wavy haired parents produce all three phenotypes. Please explain this observation using punnett squares. Curly x Curly straight x straight Wavy x wavy
Miniature wings, X in Drosophila melanogaster result from an X-linked allele that is recessive to the allele for long wings, 9 + X. Match the genotypes for each parent in the crosses. Male parent phenotye long miniature miniature 111 long long Female parent phenotype long long long miniature long m m X X Male offspring phenotypes 231 long, 250 miniature 610 long 410 long, 417 miniature 753 miniature 625 long m X Y Female offspring phenotypes 560 long 632 long 412 long, 415 miniature 761 long 630 long Answer Bank ++ X X Male parent genotype + X Y Female parent genotype + X X m 00
BB AB AB A. None B.3 с.1 D.2 ВВ OE. The father AB In the pedigree presented above, an autosomal dominant disease which causes significant visual loss and eventual blindness, is segregating in the family. The disease gene causing this sight loss looks as if it may be linked to a marker locus. The alleles of this marker locus that are present in this family are allele A and allele B. Are there any recombinant individuals in this pedigree? AB ВВ AB ВВ AB ВВ AB AB ВВ