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 38.22P
Summary Introduction
To determine: The reason why problem state "centromere or centromeres" and not just "centromere".
Introduction: The centromere is the specialized DNA sequence of a chromosome that links a pair of sister chromatids. During mitosis, spindle fibers attach to the centromere via the kinetochore.
Summary Introduction
To determine: The general method for mapping centromeres in tetrad analysis.
Introduction: In some algae and
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Recombination frequencies between four genetically-linked loci in corn are shown in the following table:
Loci Recombination Frequency (%)
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R and L 30
Q and W 13
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What is the order of the genes on the chromosome?
(note: The same answer can be represented forward or backwards. e.g. A B C D = D C B A)
LQWR
RQWL
LRQW
QRLW
RLWQ
With regard to the segregation of centromeres, why is adjacent-2segregation less frequent than alternate or adjacent-1 segregation?
Map distances were determined for four differentgenes (MAT, HIS4, THR4, and LEU2) on chromosome III of the yeast Saccharomyces cerevisiae:HIS4 ↔ MAT 37 cMTHR4 ↔ LEU2 35 cMLEU2 ↔ HIS4 23 cMMAT ↔ LEU2 16 cMMAT ↔ THR4 20 cMWhat is the order of genes on the chromosome?
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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- In Brinjal eggplants, purple fruit is incompletely dominant to white fruit, with the heterozygote being light violet. What is/are the genotype(s) of: the megaspore mother cell of a plant with purple flowers? the polar nucleus of a plant with light violet flowers? the megasporocyte of a plant with white flowers? the companion cell of a plant with light violet flowers?arrow_forwardOn Neurospora chromosome 4, the leu3 gene is just to theleft of the centromere and always segregates at the firstdivision, whereas the cys2 gene is to the right of the centromere and shows a second-division segregation frequency of 16 percent. In a cross between a leu3 strain anda cys2 strain, calculate the predicted frequencies of thefollowing seven classes of linear tetrads where l = leu3 andc = cys2. (Ignore double and other multiple crossovers.)arrow_forwardInversions are known to affect crossing-over. The following homologs have the indicated order (the filled and open circles indicate centromeres): • (A B C D E) o (A D C B E) a. Considering the position of the centromere, what is this sort of inversion called?arrow_forward
- This problem leads you through the derivation of acorrected equation for RF in yeast tetrad analysis thattakes into account double crossover (DCO) meioses. A yeast strain that cannot grow in the absence ofthe amino acid histidine (his−) is mated with a yeaststrain that cannot grow in the absence of the aminoacid lysine (lys−). Among the 400 unordered tetrads resulting from this mating, 233 were PD, 11 wereNPD, and 156 were T.a. What types of spores are in the PD, NPD, andT tetrads?b. Are the his and lys genes linked? How do you know? c. Using the simple equation RF = 100 × [NPD +(1/2)T]/total tetrads, calculate the distance in mapunits between the his and lys genes.d. If you think about all the kinds of meiotic eventsthat could occur (refer to Fig. 5.24), you can seethat the calculation you did in part (c) may substantially underestimate RF. What kinds of meioses(NCO, SCO, or DCO) generated each of the tetradtypes in this cross? e. What incorrect assumptions does the simple RFequation…arrow_forwardExplain how ordered and unordered tetrad analysis canmap the positions of genes and (for ordered tetrads)centromeres.arrow_forwardThe a, b, and c loci are all on different chromosomesin yeast. When a b+ yeast were crossed to a+ b yeastand the resultant tetrads analyzed, it was found thatthe number of nonparental ditype tetrads was equal tothe number of parental ditypes, but there were no tetratype asci at all. On the other hand, many tetratypeasci were seen in the tetrads formed after a c+ wascrossed with a+ c, and after b c+ was crossed withb+ c. Explain these results.arrow_forward
- Give typed full explanation You are studying three linked genes in snapdragons. The flower color locus is in the center. There are 13.8 cM between the flower color locus and the plant height locus. There are 14.5 cM between the flower color locus and the leaf type locus. The coefficient of coincidence is 0.8. Pure-breeding tall, red-flowered plants with fuzzy leaves were crossed to pure-breeding dwarf, blue-flowered plants with smooth leaves. The F1 were testcrossed. Calculate the proportion of the testcross progeny that are expected to be dwarf with red flowers. Round properly to 4 decimal digits.arrow_forwardIn a haploid fungus, the genes al-2 and arg-6 are 30 m.u.apart on chromosome 1, and the genes lys-5 and met-1are 20 m.u. apart on chromosome 6. In a crossal-2 + ; + met-1 × + arg-6 ; lys-5 +what proportion of progeny would be prototrophic + + ;+ +?arrow_forwardmapping three linked genes, the heterozygous AaBbCc was crossed to aabbcc individual the following phenotypes were obtained: ABC 4 abc 2 ABc 150 abC 141 aBc 18 AbC 20 Abc 30 aBC 35 A) What are the alleles found on each parental chromosome(genotypes of the parents)? B) How many double recombinants(cross over) were there? c) what is the linear arrangement of loci on the chromosome?arrow_forward
- When stained with Giemsa (G-Banding staining), the centromere will be stained: Question 28 options: more robustly (dark band) less robustly (light band)arrow_forward6). To investigate the genetic control of locomotive behavior in the diploid nematode, Caenorhabditis elegans, eight "wiggly" mutants were isolated, which wiggle in an uncoordinated way rather than smoothly gliding. These eight mutants were crossed together in all possible pairwise combinations, yielding the results summarized in the table above (+ means that all the progeny were wild type, and m means that all the progeny showed the wiggly mutant phenotype). Organize the eight mutations into con lementation groups, each consisting of a group of mutations affecting the same gene. 1 2 3 4 5 6 7 8 1 E N+E 2 3+1 m + m 4 5 + m m E+ + + EE 6 ++E E+ +EE m m + + + + + + + + + m 7 8 ++4 ∞+E +E m m ++ Earrow_forwardFor this particular family, what is the recombination rate between the D17S74 marker and the breast cancer gene? Lets say that at age 45 the third granddaughter III-3, is diagnosed with breast cancer. Now recalculate the recombination rate between the marker locus D17S74 and the breast cancer gene? Use only the 5 women in your analysis:the two daugthers in generation II and the three granddaughters in generation IIarrow_forward
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