3. Based on the following data which of the following would be considered: a. Recessive? b. Dominant? c. A weak loss-of-function mutation? с. d. A null mutation? e. A haploinsufficiency ? Data - Gene "B" is required for formation of body segments. A wild type fly had 3 body segments. bb (Homozygous recessive) Lacks all body segments 1 body segment Bb (Heterozygous) BB (Homozygous dominant) Wild Type Mutant 1 Wild Type Mutant 2 Wild Type 2 body segments Mutant 3 Wild Type Wild Type 2 body segments

3. Based on the following data which of the following would be considered: a. Recessive? b. Dominant? c. A weak loss-of-function mutation? с. d. A null mutation? e. A haploinsufficiency ? Data - Gene "B" is required for formation of body segments. A wild type fly had 3 body segments. bb (Homozygous recessive) Lacks all body segments 1 body segment Bb (Heterozygous) BB (Homozygous dominant) Wild Type Mutant 1 Wild Type Mutant 2 Wild Type 2 body segments Mutant 3 Wild Type Wild Type 2 body segments

Biology 2e

2nd Edition

ISBN:9781947172517

Author:Matthew Douglas, Jung Choi, Mary Ann Clark

Publisher:Matthew Douglas, Jung Choi, Mary Ann Clark

Chapter12: Mendel's Experiments And Heredity

Section: Chapter Questions

Problem 21RQ: Which of the following situations does not follow the Law of Independent Assortment? A blond man and...

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2. Null mutations are valuable genetic resources becausethey allow a researcher to determine what happens to anorganism in the complete absence of a particular protein. However, it is often not a trivial matter to determinewhether a mutation represents the null state of the gene.a. Geneticists sometimes use the following test forthe nullness of an allele in a diploid organism: If theabnormal phenotype seen in a homozygote for theallele is identical to that seen in a heterozygote(where one chromosome carries the allele in question and the homologous chromosome is known tobe completely deleted for the gene) then the alleleis null. What is the underlying rationale for thistest? What limitations might there be in interpreting such a result?
1. In humans, widow’s peak (W) is dominant over having a straight hairline (w). If a homozygous man with a widow’s peak has a child with a woman with a straight hairline, what are the odds the child will have a widow’s peak? Show work. 2. Ellis-van Creveld syndrome is a recessive inherited condition that results in 6-fingered dwarfism. Two parents without this condition have with Ellis-van Creveld syndrome. What are the genotypes of the parents and offspring? What are the odds that the son will have a 6-fingered child if his partner is a woman who is heterozygous for the condition? Show work. 3. The gene for dangling earlobes is dominant over the gene for attached earlobes. A woman with attached earlobes has a child with a man with dangling earlobes whose mother had attached earlobes. What is the probability that this child will have attached lobes? Show work. 4. In humans, normal pigmentation is due to a dominant gene, while albinism is recessive. A normally pigmented man has a…
6. A diploid strain of yeast was made by mating a haploidstrain with a genotype w−, x−, y−, and z− with a haploidstrain of opposite mating type that is wild type for thesefour genes. The diploid strain was phenotypically wildtype. Four different X-ray-induced diploid mutantswith the following phenotypes were produced fromthis diploid yeast strain. Assume a single new mutation is present in each strain.Strain 1 w− x+ y− z+Strain 2 w+ x− y− z−Strain 3 w− x+ y− z−Strain 4 w− x+ y+ z+When these mutant diploid strains of yeast go throughmeiosis, each ascus is found to contain only two viablehaploid spores.a. What kind of mutations were induced by X-rays tomake the listed diploid strains?b. Why did two spores in each ascus die?
3. Albinism in humans is controlled by a recessive gene (c). From marriages between two normally pigmented carriers (Cc), what is the probability of having a normal child? An albino? Assume that from such a marriage (Cc x Cc) four children will be produced. (@ What is the probability that all four will be normal? (b) Three will be normal and one albino? (c) Two normal and two albino?
1. The enzymes required to breathe fire (BFi2) is expressed in throat tissues of dragons from the Known World (comprising Westeros, Essos, Sothyros, and Ulthos) at various times during their life cycles. A test very similar to DNase I hypersensitivity is conducted on embryonic (E) and adult (A) throat tissue samples of Valeriyan, Ice and Sea dragons. In the analysis, a radioactive label is attached to one end of each chromosome fragment, and the samples from each tissue are exposed to DNase I to determine if regions of BFi2 are DNase I hypersensitive. The content from each sample is then separated by gel electrophoresis and the results are shown below.
The ability of humans to taste the bitter chemical phenylthiocarbamide (PTC) is a genetic trait. People with at least one copy of the normal, dominant allele of the PTC gene can taste PTC; those who are homozygous for a mutant, recessive allele cannot taste it. Could two parents able to taste PTC have a nontaster child? Could nontaster parents have a child able to taste PTC? A pair of taster parents, both of whom had one parent able to taste PTC and one nontaster parent, are expecting their first child. What is the chance that the child will be able to taste PTC? Unable to taste PTC? Suppose the first child is a nontaster. What is the chance that their second child will also be unable to taste PTC?
Which of the following situations does not follow the Law of Independent Assortment? A blond man and a brunette woman produce three offspring over time, all of who have blond hair A white cow crossed with a brown bull produces roan cattle Mating a hog with a sow produces six female piglets Men are more likely to experience hemophilia than women.
Achondroplasia is an autosomal dominant form of dwarfism caused by a single gene mutation. Calculate the mutation rate of this gene given the following data: 10 achondroplastic births to unaffected parents in 245,000 births.
When two protein-coding genes have very similar nucleotide sequences and are located right next to each other on a chromosome, we can hypothesize that: a. one of them is a duplicate of the other, copied by aretrotransposon. b. they are nonhomologous. c. one of them is a pseudogene. d. they were produced by unequal crossing-over. e. they are transcribed in the same cell types.
A major difference between hereditary information in eukaryotes and prokaryotes is: a. in prokaryotes, the hereditary information is distributed among individual, linear DNA molecules in the nucleus. b. in eukaryotes, the hereditary information is encoded in a single, circular DNA molecule. c. in prokaryotes, the hereditary information is usually distributed among multiple circular DNA molecules in the cytoplasm. d. in eukaryotes, the hereditary information is distributed among individual, linear DNA molecules in the cytoplasm. e. in eukaryotes, the hereditary information is distributed amoung individual, linear DNA molecules in the nucleus.
3.Tay Sachs is a lysosomal storage disease that is inherited in an autosomal recessive manner (T= dominant allele; t = recessive allele). Individuals with Tay Sachs have a defect in the gene encoding hexosaminidase A, a lysosomal enzyme that breaks down cell membrane components called gangliosides. The inability to break down gangliosides causes harmful quantities of this lipid to accumulate in nerve cells of the brain, eventually leading to premature death of affected cells. There is no known cure for the disease, and death usually occurs by age 4. Jack and Jill have a child who has Tay-Sachs. What do you know about the genotypes of Jack and Jill? A.has the genotype DD B.has the genotype Dd C.is still of unknown genotype since there were offspring of both deaf and hearing phenotypes. D.has the genotype dd
2 Color Blindness with a Twist: As a genetic counselor, you may face some ethical dilemmas. Imagine that a couple came to your office to discuss potential color blindness in their children. The man is not color blind, but his wife is. They already have one daughter that is color blind and want to know the probability of future children having the disorder. First, you will need to explain to the couple the difference between sex chromosomes and autosomal chromosomes and that color blindness is caused by a recessive mutation in a gene found on the X chromosome. Once they understand this, you can then explain to them the probability of other offspring having the disorder. What is the dilemma in this case and why might you need to suggest to the couple that they see a marriage counselor?