Introduction to Genetic Analysis
11th Edition
ISBN: 9781464109485
Author: Anthony J.F. Griffiths, Susan R. Wessler, Sean B. Carroll, John Doebley
Publisher: W. H. Freeman
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Chapter 4, Problem 44P
a.
Summary Introduction
To determine: The linkage relations of these two genes to their centromere.
Introduction: The association in inheritance of two or more non-allelic GENES due to their being located more or less closely on the same chromosome is called a chromosome.
b.
Summary Introduction
To determine: A diagram to show the origin of the ascus type with only one single representative.
Introduction: Linked genes are genes that are so physically close to each other that it is rare for a homologous recombination event to separate them. Thus their alleles tend to be co-inherited.
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A yeast geneticist irradiates haploid cells of a strain thatis an adenine-requiring auxotrophic mutant, caused bymutation of the gene ade1. Millions of the irradiatedcells 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 ofresults.b. Write the genotypes of the prototrophs in each case.c. What progeny phenotypes and ratios do you predictfrom crossing a prototroph of type 2 by the original ade1auxotroph?
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?
The 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.
Chapter 4 Solutions
Introduction to Genetic Analysis
Ch. 4 - Prob. 1PCh. 4 - Prob. 5PCh. 4 - Prob. 12PCh. 4 - Prob. 13PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - Prob. 16PCh. 4 - Prob. 17PCh. 4 - Prob. 18PCh. 4 - Prob. 19P
Ch. 4 - Prob. 20PCh. 4 - Prob. 21PCh. 4 - Prob. 21.1PCh. 4 - Prob. 21.2PCh. 4 - Prob. 21.3PCh. 4 - Prob. 21.4PCh. 4 - Prob. 21.5PCh. 4 - Prob. 21.6PCh. 4 - Prob. 21.7PCh. 4 - Prob. 21.8PCh. 4 - Prob. 21.9PCh. 4 - Prob. 21.10PCh. 4 - Prob. 21.11PCh. 4 - Prob. 21.12PCh. 4 - Prob. 21.13PCh. 4 - Prob. 21.14PCh. 4 - Prob. 21.15PCh. 4 - Prob. 21.16PCh. 4 - Prob. 21.17PCh. 4 - Prob. 21.18PCh. 4 - Prob. 21.19PCh. 4 - Prob. 21.20PCh. 4 - Prob. 21.21PCh. 4 - Prob. 21.22PCh. 4 - Prob. 21.23PCh. 4 - Prob. 21.24PCh. 4 - Prob. 21.25PCh. 4 - Prob. 21.26PCh. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - Prob. 29PCh. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - Prob. 32PCh. 4 - Prob. 33PCh. 4 - Prob. 34PCh. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 38.1PCh. 4 - Prob. 38.2PCh. 4 - Prob. 38.3PCh. 4 - Prob. 38.4PCh. 4 - Prob. 38.5PCh. 4 - Prob. 38.6PCh. 4 - Prob. 38.7PCh. 4 - Prob. 38.8PCh. 4 - Prob. 38.9PCh. 4 - Prob. 38.10PCh. 4 - Prob. 38.11PCh. 4 - Prob. 38.12PCh. 4 - Prob. 38.13PCh. 4 - Prob. 38.14PCh. 4 - Prob. 38.15PCh. 4 - Prob. 38.16PCh. 4 - Prob. 38.17PCh. 4 - Prob. 38.18PCh. 4 - Prob. 38.19PCh. 4 - Prob. 38.20PCh. 4 - Prob. 38.21PCh. 4 - Prob. 38.22PCh. 4 - Prob. 38.23PCh. 4 - Prob. 38.24PCh. 4 - Prob. 39PCh. 4 - Prob. 40PCh. 4 - Prob. 41PCh. 4 - Prob. 42PCh. 4 - Prob. 43PCh. 4 - Prob. 44PCh. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - Prob. 48PCh. 4 - Prob. 49PCh. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52PCh. 4 - Prob. 53PCh. 4 - Prob. 54PCh. 4 - Prob. 55PCh. 4 - Prob. 56PCh. 4 - Prob. 57PCh. 4 - Prob. 58PCh. 4 - Prob. 59PCh. 4 - Prob. 60PCh. 4 - Prob. 62PCh. 4 - Prob. 63PCh. 4 - Prob. 64PCh. 4 - Prob. 65PCh. 4 - Prob. 66PCh. 4 - Prob. 67PCh. 4 - Prob. 68PCh. 4 - Prob. 69P
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- Please label the tetrad type in the table as PD (parental ditype), NPD (non parental ditype) or T (tetratype) and answer the following questions a) Are the genes linked? Please explain SPECIFICALLY how you can distinguish between linked and unlinked genes in this instance. b) If the two genes are linked, calculate the % recombination between ser and thr. Show the formula used, as well as all of your calculations. c) Draw a single map illustrating the arrangement of the two genes on the chromosome with respect to each other and to the centromere of the chromosome. Make sure to map ALL three distancesarrow_forwardConsider a maize plant: Genotype C/cm ; Ac/Ac+ where cm is an unstable colorless allele caused by Ds insertion. What phenotypic ratios would be produced and in what proportions when this plant is crossed with a mutant c/c Ac+/Ac+? Assume that the Ac and c loci are unlinked, that the chromosome-breakage frequency is negligible, and the C allele encodes pigment production.arrow_forwardAlleles of genes A and B were analyzed in Neurospora according to the cross shown below. Ordered tetrads are summarized in each horizontal row with the number of tetrads in each category listed alongside. A) Analyze the data to determine the recombination frequency (RF) between A and B, along with any additional information that is available from this data. B) Draw a map of the chromosome or chromosomes with appropriate map distances. C) Use the Perkins formula to reanalyze any relationship between A and B.arrow_forward
- Time mapping is performed in a cross involving the genes his,leu, mal, and xyl. The recipient cells are auxotrophic for all fourgenes. After 25 minutes, mating is interrupted, with the resultsin recipient cells shown below. Diagram the positions of thesegenes relative to the origin (O) of the F factor and to one another.(a) 90% are xyl+(b) 80% are mal+(c) 20% are his+(d) None are leu+arrow_forwardAn Hfrstrain that is a *b*c*d* e*f* g *h* is mated with an F strain that is a b e d e f gh. The mating is interrupted at 5 minutes interval, and the genotypes of the F recombinants are determined. The results obtained are tabulated in Table 2. Draw the map of the Hfrchromosome and indicate the position of the origin of transfer, the direction of the transfer and the minutes between genes. Table 2:Entry time of Hfr chromosome into recipient cell. Time a d e f h 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +arrow_forwardTwo morphotypes of the newly discovered plant in Mt. Banahaw de Lucban were tested for linkage of three genes such as the presence of tendril (t+), dense trichomes (d+), and the presence of secretory cells (sc+). The loci for the mutant genes have been mapped and are separated by the following map distances: t and d = 20CM; d and sc = 12CM. The interference between these genes is 0.4. The first morphotype is located at the lower elevation and is characterized by the absence of tendril, sparse trichome, and absence of secretory cells. On the other hand, the second morphotype is located at the higher elevation and is characterized by the presence of tendril, dense trichome, and presence of secretory cells. Further genetic analysis showed that the morphotypes were true breeding. The two morphotypes were intercrossed and the resulting F1 is testcrossed with the first morphotype. The cross resulted in 1800 progeny. Give the genotypes, phenotypes, and the expected numbers of phenotypes in…arrow_forward
- Time mapping is performed in a cross involving the genes his, leu, mal, and xyl. The recipient cells were auxotrophic for all four genes. After 25 minutes, mating was interrupted with the following results in recipient cells. Diagram the positions of these genes relative to the origin (O) of the F factor and to one another. (a) 90% were xyl+ (b) 80% were mal+ (c) 20% were his+ (d) none were leu+arrow_forwardThe synthesis of flower pigments is known to be dependent on enzymatically controlled biosynthetic pathways. For the crosses shown here, postulate the role of mutant genes and their products in producing the observed phenotypes: (a) P1: white strain A * white strain B F1: all purple F2: 9/16 purple: 7/16 white (b) P1: white * pink F1: all purple F2: 9/16 purple: 3/16 pink: 4/16 whitearrow_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_forward
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