Please answer all parts along with the reason. I'll definitely give a like. Thank you in advance!1A) From the cross Ab/aB x ab/ab, what is the recombination frequency if the progeny numbers are 17 AB/ab, 72 Ab/ab, 68 aB/ab, and 21 ab/ab?1B)In human gene mapping, a LOD score is calculated to see if a gene causing a rare disease is linked to a known SNP. The LOD score is -4. This means that1C) A three-point testcross is used to determine the order of three linked genes. The following crossover progeny result: single crossovers, double crossovers, and no crossovers. To determine the order, the no-crossover progeny must be compared to what other class of progeny? A)From the cross Ab/aB x ab/ab, what is the recombination frequency if the progeny numbersare 17 AB/ab, 72 Ab/ab, 68 aB/ab, and 21 ab/ab?O 0.213O 0.271O 0.500O 0.787O Cannot be determinedIB)In human gene mapping, a LOD score is calculated to see if a gene causing a rare disease islinked to a known SNP. The LOD score is -4. This means thatO The disease gene is unlinked to the SNP.The disease gene is linked to the SNP.O The disease is caused by the SNP,O It is not clear yet whether the gene is linked or unlinked to the SNP.1C)A three-point testcross is used to determine the orderof three linked genes. The following crossover progenyresult: single crossovers, double crossovers, and nocrossovers. To determine the order, the no-crossoverprogeny must be compared to what other class ofprogeny?The double-crossover classO The single-crossover classO The no-crossover classO All classes need to be considered simultaneously todetermine gene order

If the genes are located on the same chromosome then they are considered as linked genes. The…

Please answer all parts along with the reason. I'll definitely give a like. Thank you in advance! 1A) From the cross Ab/aB x ab/ab, what is the recombination frequency if the progeny numbers are 17 AB/ab, 72 Ab/ab, 68 aB/ab, and 21 ab/ab? 1B)In human gene mapping, a LOD score is calculated to see if a gene causing a rare disease is linked to a known SNP. The LOD score is -4. This means that 1C) A three-point testcross is used to determine the order of three linked genes. The following crossover progeny result: single crossovers, double crossovers, and no crossovers. To determine the order, the no-crossover progeny must be compared to what other class of progeny?

Please answer all parts along with the reason. I'll definitely give a like. Thank you in advance! 1A) From the cross Ab/aB x ab/ab, what is the recombination frequency if the progeny numbers are 17 AB/ab, 72 Ab/ab, 68 aB/ab, and 21 ab/ab? 1B)In human gene mapping, a LOD score is calculated to see if a gene causing a rare disease is linked to a known SNP. The LOD score is -4. This means that 1C) A three-point testcross is used to determine the order of three linked genes. The following crossover progeny result: single crossovers, double crossovers, and no crossovers. To determine the order, the no-crossover progeny must be compared to what other class of progeny?

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter5: The Inheritance Of Complex Traits

Section: Chapter Questions

Problem 5QP: As it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women....

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Homozygous wild-type male mice (AA BB CC) were crossed with triplemutant female mice (aa bb cc), forming an F1 generation with the followinggenotype (Aa Bb Cc). The F1 males were crossed with triple mutantfemales, forming the following F2 phenotypes”“a B c” 3“A b C” 3“a b c” 8“A B c” 5“a b C” 5“A B C” 8“a B C” 6“A b c” 6 44 Determine the sequence of the genes
In a diploid plant species, an F1 with the genotype Mm Rr Ss is test crossed to a pure breeding recessive plant with the genotype mm rr ss. The offspring genotypes are as follows: Genotype Number Mm Rr Ss 687 Mm Rr ss 5 Mm rr Ss 68 Mm rr ss 196 mm Rr Ss 185 mm Rr ss 72 mm rr Ss 8 mm rr ss 679 Total 1900 1. Calculate the recombination frequency between each pair of genes
An individual heterozygous for four genes, A/a • B/b •C/c • D/d, is testcrossed with a/a • b/b • c/c • d/d, and 1000progeny are classified by the gametic contribution ofthe heterozygous parent as follows:a • B • C • D 42A • b • c • d 43A • B • C • d 140a • b • c • D 145a • B • c • D 6A • b • C • d 9A • B • c • d 305a • b • C • D 310a. Which genes are linked?b. If two pure-breeding lines had been crossed toproduce the heterozygous individual, what would theirgenotypes have been?c. Draw a linkage map of the linked genes, showing theorder and the distances in map units.d. Calculate an interference value, if appropriate
In a cross between mice the genotypes AB/ab  ab/ab, what is the recombination frequency if the progeny numbers are 72 AB/ab, 68 ab/ab, 17 Ab/ab, and 21 aB/ab? The alleles are shown for each chromosome, separated by a slash (/).
In a diploid plant species, an F1 with the genotype Mm Rr Ss is test crossed to a pure breeding recessive plant with the genotype mm rr ss. The offspring genotypes are as follows: Genotype Number Mm Rr Ss 687 Mm Rr ss 5 Mm rr Ss 68 Mm rr ss 196 mm Rr Ss 185 mm Rr ss 72 mm rr Ss 8 mm rr ss 679 Total 1900 1. Why is the recombination frequency for the outside pair of genes not equal to the sum of the recombination between the adjacent gene pairs?
An organism of the genotype AaBbCc was testcrossed to a triplyrecessive organism (aabbcc). The genotypes of the progeny are inthe following table.AaBbCc 20 AaBbcc 20aabbCc 20 aabbcc 20AabbCc 5 Aabbcc 5aaBbCc 5 aaBbcc 5 a.) Assuming simple dominance and recessiveness in each genepair, if these three genes were all assorting independently,how many genotypic and phenotypic classes would result inthe offspring, and in what proportion? b.) Answer part (a) again, assuming the three genes are sotightly linked on a single chromosome that no crossovergametes were recovered in the sample of offspring. c.) What can you conclude from the actual data about thelocation of the three genes in relation to one another?
In letter B: If the map distance equals the number of recombinant/total of offspring, wouldn't it be 24/806 x 100? Wouldn't we add both recombinants? Can you explain letter C? I don't grasp that concept well. And since I'm using my question already, would you be able to answer D. Thank you!
In a diploid plant species, an F1 with the genotype Mm Rr Ss is test crossed to a pure breeding recessive plant with the genotype mm rr ss. The offspring genotypes are as follows: Genotype Number Mm Rr Ss 687 Mm Rr ss 5 Mm rr Ss 68 Mm rr ss 196 mm Rr Ss 185 mm Rr ss 72 mm rr Ss 8 mm rr ss 679 Total 1900 1. What is the gene order of these linked genes?
Cross Cross A Cross A Cross B Cross B Phenotype F1 generation F2 generation F1 generation F2 generation Male red eyes 132 150 0 99 Female red eyes 135 295 110 101 Male white eyes 0 147 105 93 Female white eyes 0 0 0 95 Using “+” to indicate the wildtype red-eyed allele and “w” to indicate the mutant white-eyed allele, state the genotypes of the following: Wildtype red-eyed and white-eyed parental flies from cross A and cross B. Males and females from the F1 generation flies from cross A and cross B Males and females, F2 generation flies from cross A and cross B.
In a diploid plant species, an F1 with the genotype Mm Rr Ss is test crossed to a pure breeding recessive plant with the genotype mm rr ss. The offspring genotypes are as follows: Genotype Number Mm Rr Ss 687 Mm Rr ss 5 Mm rr Ss 68 Mm rr ss 196 mm Rr Ss 185 mm Rr ss 72 mm rr Ss 8 mm rr ss 679 Total 1900 1. What is the interference value for this data set?
. A geneticist mapping the genes A, B, C, D, and E makestwo 3-point testcrosses. The first cross of pure lines isA/A ⋅ B/B ⋅ C/C ⋅ D/D ⋅ E/E × a/a ⋅ b/b ⋅ C/C ⋅ d/d ⋅ E/EThe geneticist crosses the F1 with a recessive tester andclassifies the progeny by the gametic contribution ofthe F1:A ⋅ B ⋅ C ⋅ D ⋅ E 316a ⋅ b ⋅ C ⋅ d ⋅ E 314A ⋅ B ⋅ C ⋅ d ⋅ E 31a ⋅ b ⋅ C ⋅ D ⋅ E 39A ⋅ b ⋅ C ⋅ d ⋅ E 130a ⋅ B ⋅ C ⋅ D ⋅ E 140A ⋅ b ⋅ C ⋅ D ⋅ E 17a ⋅ B ⋅ C ⋅ d ⋅ E 131000The second cross of pure lines is A/A • B/B • C/C • D/D •E/E × a/a • B/B • c/c • D/D • e/e.The geneticist crosses the F1 from this cross with arecessive tester and obtainsA ⋅ B ⋅ C ⋅ D ⋅ E 243a ⋅ B ⋅ c ⋅ D ⋅ e 237A ⋅ B ⋅ c ⋅ D ⋅ e 62a ⋅ B ⋅ C ⋅ D ⋅ E 58A ⋅ B ⋅ C ⋅ D ⋅ e 155a ⋅ B ⋅ c ⋅ D ⋅ E 165a ⋅ B ⋅ C ⋅ D ⋅ e 46A ⋅ B ⋅ c ⋅ D ⋅ E 341000The geneticist also knows that genes D and E assortindependently.a. Draw a map of these genes, showing distances inmap units wherever possible.b. Is there any evidence of interference?
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?