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Concept explainers
(a)
To determine:
The mutation that was induced by X-rays.
Introduction:
A haploid strain with wild-type
(b)
To determine:
The reason that two spores in each ascus die.
Introduction:
The ascomycetes
(c)
To determine:
Whether the genes w, x, y, or z are located on the same chromosome.
Introduction:
Chromosomes are present in the eukaryotic organisms. They are “thread-like” in appearance. A chromosome is made up of two units: DNA and proteins. The genes can either be present on the same chromosome or a different chromosome.
(d)
To determine:
The order of genes that are present on the same chromosomes.
Introduction:
The genes that are present on the same chromosome have a similar type of function. The genes w, x, y, or z are located on the same chromosome. They are the pseudo-dominant genes. They are all recessive genes, but they dominate each other.
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Chapter 12 Solutions
Genetics: From Genes to Genomes, 5th edition
- A yeast geneticist irradiates haploid cells of a strain that is an adenine-requiring auxotrophic mutant, caused by mutation of the gene ade1. Millions of the irradiated cells are plated on minimal medium, and a small number of cells divide and produce prototrophic colonies. These colonies are crossed individually with a wildtype strain. Two types of results are obtained:(1) prototroph × wild type : progeny all prototrophic(2) prototroph × wild type : progeny 75% prototrophic, 25% adenine-requiring auxotrophsa. Explain the difference between these two types of results.b. Write the genotypes of the prototrophs in each case.c. What progeny phenotypes and ratios do you predict from crossing a prototroph of type 2 by the original ade1auxotroph?arrow_forwardFor each genotype in the table below, determine whether or not functional B-gal will be produced in the presence or absence of the inducer. Write a plus (+) if B-gal is produced or a minus (-) if it is not. ma Chromosome F' lac (plasmid) - Inducer + Inducer ---- --- --- I*O*Z I*O°Z+ I*O*Z* I* = wildtype repressor |- - no functional repressor produced O* - wildtype operator OC = operator mutation prevents repressor from binding %3Darrow_forwardA yeast strain with a mutant spo11- allele has been isolated. The mutant allele is nonfunctional; it makes no spo11 protein. What do you suppose is the phenotype of this mutant strain?arrow_forward
- In yeast, LYS5, ADE1, and URA2 are genes required to synthesize lysine, adenine and uracil, respectively. A MATa lys5 ADE1 URA2 haploid was mated to a MATA LYS5 ade1 ura2 haploid to make a triply heterozygous diploid. This diploid was put through meiosis, 100 tetrads were dissected and the spore colonies transferred to medium lacking either lysine, adenine or uracil to determine which spores were prototrophic or auxotrophic for those nutrients. The following results were obtained: LYS5-ADE1 ADE1-URA2 LYS5-URA2 PD NPD Ι 20 22 58 18 20 62 50 8 42 Based on this information, select the statements below that are TRUE. Select 4 correct answer(s) A) The LYS5 and URA2 genes are linked. B) The genotypes of the four spores in the ADE1-URA2 tetratypes is ade1 ura2 ade1 URA2 ADE1 ura2 ADE1 URA2 C) The map distance between ADE1 and URA2 is 51 CM. D) ADE1 and URA2 are likely located on the same chromosome. E) The map distance between LYS5 and URA2 is 29 CM. ☐ F) LYS5 and ADE1 independently assort…arrow_forwardAn Hfr strain that is hisE + and pheA + was conjugated to a strain that is hisE − and pheA −. The conjugation was interrupted at different times, and the percentage of recombinants for each gene was determined by streaking on media that lacked either histidine or phenylalanine. The following results were obtained: A. Determine the map distance (in minutes) between these twogenes.B. In a previous experiment, it was found that hisE is 4 minutesaway from pabB and that PheA is 17 minutes from pabB. Drawa genetic map showing the locations of all three genes.arrow_forwardThe DNA of every individual in the pedigree shown in image B (below) has been sequenced at the causative locus, all the non-shaded individuals are wild type apart from III.1 and III.6. III.1 and III.6 have both been proven to have the causative allele for the condition but they do not exhibit any of the phenotypic signs or symptoms. Based on this pedigree, what is the level of penetrance for the condition? Please give your answer as a percentage to one decimal place, give the number only, no percentage symbol. Given the information above I calculate the level of penetrance seen in image B to be "Blank" 1 percent.arrow_forward
- The DNA of every individual in the pedigree shown in image B (below) has been sequenced at the causative locus, all the non- shaded individuals are wild type apart from III.1 and III.6. III.1 and III.6 have both been proven to have the causative allele for the condition but they do not exhibit any of the phenotypic signs or symptoms. Based on this pedigree, what is the level of penetrance for the condition? Please give your answer as a percentage to one decimal place, give the number only, no percentage symbol. ANSWER: Given the information above I calculate the level of penetrance seen in image B to be Blank 1 percent. A KEY Homozygous Homozygous Heterozygous Heterozygous Wild Type Male Female Male Female Male Note: Completely red symbol denotes an individual exhibiting the phenotype of interest CI || III IV V 1/4 1/2 1/2 1/2 1/2 Wild Type Female 1/4 1/2 Affected Known carrier Affected female Normal female Affected male Normal male D ●●●arrow_forwardThe DNA of every individual in the pedigree shown below has been sequenced at the causative locus. All the non-shaded individuals are wild type apart from III.1. III.1 has been proven to have the causative mutation for this autosomal dominant condition, but they exhibit no symptoms. Based on this small pedigree, what is the level of penetrance for the condition? Please give your answer as a WHOLE percentage, give the number only, no percentage symbol. Answer: The level of penetrance for the condition shown in the pedigree below is Blank 1 percent. 1:1 1:2 Il:1 I1:2 I1:3 Il:4 I1:5 I1:6 II:1 I:2 III:3 III:4 III:3 III:6 III:7 III:8 III:9 III:10 III:11 III12 II:13 III:14 IV:1 | IV:2 IV:3 IV:4 IV:5 IV:6 IV:7 IV:8 IV:9 IV:10 IV:11 IV:12 IV:13 IV:14 IV:15 IV:16 IV:17 IV:18 IV:19 V:1 V:2 V:3 V:4 V:5 V:6 V:7 V:8 V:9 V:10 V:11 V:12arrow_forwardWhen a female melanotic fly is crossed with a normal male, the progeny are produced: 123 normal females, 125 melanotic females, and 124 normal males. In subsequent crosses between melanotic females and normal males, melanotic females are frequently obtained, but never any melanotic males. Provide a possible explanation for the inhertiacne of the melanotic mutation (Hint: The cross produces twice as many female progeny as male progeny)arrow_forward
- For a haploid fungus, the starting point in the biosynthesis of the amino acid arginine is Compound X, which is always present in and absorbed from the environment. The arginine biosynthetic pathway is: Enzyme A Enzyme B Enzyme Ç Compound X It is know that genes encoding enzymes A and C are on two different chromosomes. Compound Y Compound Z- Arginine A mutant strain of genotype a (lacking only enzyme A) is crossed to a mutant strain of genotype c (lacking only enzyme C) to generate a diploid strain. Sporulation (i.e. meiosis) is subsequently induced in the resulting diploid strain. What proportion of the spores (haploids formed by sporulation) is expected to grow on medium without arginine but supplemented with Compound Y? O 100% 50% 0% 25%arrow_forwardThe following results are derived from crosses between Neurospora strain xy and strain ++: Tetrad Class 3 4 ху x+ x+ xy ++ ++ ++ +y +y ху +y 25 ++ 3 124 4 (i) Name the ascus type of each class from 1 to 4 as P, NP or T. (ii) Are genes x and y linked? Explain your answer. (iii) If they are linked, determine the map distance between the two genes. If they are unlinked, provide all the information you can about why you draw this conclusion.arrow_forwardThe wild-type (normal) fruit fly, Drosophila melanogaster, has straight wings and long bristles. Mutant strains have been isolated that have either curled wings or short bristles. The genes representing these two mutant traits are located on separate chromosomes. Carefully examine the data from the following five crosses shown below (running across both columns). (a) Identify each mutation as either dominant or recessive. In each case, indicate which crosses support your answer. (b) Assign gene symbols and, for each cross, determine the genotypes of the parents.arrow_forward
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