In tomato the mutant genes o (oblate=flattened fruit), p (peach=hairy fruit)and s (compound inflorescence) were found to be in chromosome 2. Thetest cross results are:73110+34822+306p9663a) Construct the linkage mapb) What are the genotypes of the homozygous parents used in making the F1heterozygote?c) Compute for the coefficient of coincidence.O o O O+

Parents- Parent 1= + + s= 348 (the highest number) Parent 2= o p + =306 (complementary of parent 1)…

In tomato the mutant genes o (oblate=flattened fruit), p (peach=hairy fruit) and s (compound inflorescence) were found to be in chromosome 2. The test cross results are: 73 110 348 2 p + 2 306 p 96 63 a) Construct the linkage map b) What are the genotypes of the homozygous parents used in making the F1 heterozygote? c) Compute for the coefficient of coincidence. O o o o

In tomato the mutant genes o (oblate=flattened fruit), p (peach=hairy fruit) and s (compound inflorescence) were found to be in chromosome 2. The test cross results are: 73 110 348 2 p + 2 306 p 96 63 a) Construct the linkage map b) What are the genotypes of the homozygous parents used in making the F1 heterozygote? c) Compute for the coefficient of coincidence. O o o o

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter6: Cytogenetics: Karyotypes And Chromosome Aberrations

Section: Chapter Questions

Problem 24QP: A geneticist discovers that a girl with Down syndrome has a Robertsonian translocation involving...

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From a cross between e+ f+ g+ and e− f − g− strains ofNeurospora, recombination between these linkedgenes resulted in a few octads containing the followingordered set of spores:e+ f+ g+e+ f+ g+e+ f − g+e+ f − g+e− f − g−e− f − g−e− f − g−e− f − g−a. Where was recombination initiated?b. Did crossing-over occur between genes e and g?Explain.c. Why do you end up with 2 f+ : 6 f − but 4 e+: 4 e−and 4g+: 4g−?d. Could you characterize these unusual octads as MIor MII for any of the three genes involved?Explain.
In a diploid organism of 2n=10, assume that you can label all the centromeres derived from its female parent and all the centromeres derived from its male parent. When this organism produces gamestes, how many male and female-labeled centromere combinations are possible in the gametes?
In a series of two-point mapping crosses involving five genes located on chromosome II in Drosophila, the following recombinant (single-crossover) frequencies were observed: pr–adp 29% pr–vg 13 pr–c 21 pr–b 6 adp–b 35 adp–c 8 adp–vg 16 vg–b 19 vg–c 8 c–b 27 (a) Given that the adp gene is near the end of chromosome II (locus 83), construct a map of these genes. (b) In another set of experiments, a sixth gene, d, was tested against b and pr: d–b 17% d–pr 23% Predict the results of two-point mapping between d and c, d and vg, and d and adp.
Consider the first category of test-cross offspring shown in figure 8.2 (+b, LS). Consider also that the parents of the heterozygous female flies in the test cross had the following genotypes: bb, SS, and +, LL. A. What would be the physical phenotype of these flies? B. If PC was conducted with the DNA of one of these flies using the primers for the molecular marker, what would be the appearance of the bands on an electrophoresis gel with the PC products? C. If the gene for black body and the locus for the molecular marker (L long or S short) were unlinked, what proportion of the test-cross progeny would be black flies that are heterozygous for the molecular marker? What proportion would be flies with normal body color, which are homozygous for one form of the molecular marker? D. If the gene for black body and the locus for the molecular marker were linked, how would the proportion of flies be different?
From a series of two-point crosses, the following mapdistances were obtained for the syntenic genes A, B,C, and D in peas:B ↔ C 23 m.u.A ↔ C 15 m.u.C ↔ D 14 m.u.A ↔ B 12 m.u.B ↔ D 11 m.u.A ↔ D 1 m.u.Chi-square analysis cannot reject the null hypothesis of no linkage for gene E with any of theother four genes.a. Draw a cross scheme that would allow you todetermine the B ↔ C map distance.b. Diagram the best genetic map that can be assembled from this data set.c. Explain any inconsistencies or unknown features inyour map.d. What additional experiments would allow you toresolve these inconsistencies or ambiguities?
In a series of two-point map crosses involving five genes locatedon chromosome II in Drosophila, the following recombinant (single-crossover) frequencies were observed:pr–adp 29pr–vg 13pr–c 21pr–b 6adp–b 35adp–c 8adp–vg 16vg–b 19vg–c 8c–b 27 If the adp gene is present near the end of chromosome II(locus 83), construct a map of these genes.
A geneticist discovers that a girl with Down syndrome has a Robertsonian translocation involving chromosomes 14 and 21. If she has an older brother who is phenotypically normal, what are the chances that he is a translocation carrier?
Females of wild-type Strain A and males of mutant Strain B, as well as females of mutant Strain B and males of wild-type Strain A, make reciprocal crosses. Explain why reciprocal crosses are needed in genetics experiments involving Drosophila fruit flies.
An individual is heterozygous for a reciprocal translocation, with the following chromosomes: A • B C D E F A • B C V W X R ST • U D E F R ST • U V W X Q. Explain why the fertility of this individual is likely to be less than the fertility of an individual without a translocation.
In a series of two-point map crosses involving five genes locatedon chromosome II in Drosophila, the following recombinant (single- crossover) frequencies were observed:pr–adp 29pr–vg 13pr–c 21pr–b 6adp–b 35adp–c 8adp–vg 16vg–b 19vg–c 8c–b 27 In another set of experiments, a sixth gene (d) was testedagainst b and pr, and the results were d - b = 17% andd - pr = 23%. Predict the results of two-point mapsbetween d and c, d and vg, and d and adp.
The genotype of EB27 and EB67 are unknown. Based on pedigree, what are the most likely genotype of each individual?
In the example shown , what is the underlying cause of nonallelichomologous recombination?