Consider the results of a three point linkage mapping testcross between a female (AaBbCc) and a male (aabbcc). Phenotype | no. Of progeny ABc 849 abC 820 Abc 58 aBC 60 AbC 102 aBc 98 ABC 6 abc 7 a) what is the correct order of the gene loci? ( which gene is in the middle) B) what was the configuration of recessive alleles in the triple heterozygous mother? (Use a cis-trans statement) C) calculate the map distances for each segment of the map.

Consider the results of a three point linkage mapping testcross between a female (AaBbCc) and a male (aabbcc). Phenotype | no. Of progeny ABc 849 abC 820 Abc 58 aBC 60 AbC 102 aBc 98 ABC 6 abc 7 a) what is the correct order of the gene loci? ( which gene is in the middle) B) what was the configuration of recessive alleles in the triple heterozygous mother? (Use a cis-trans statement) C) calculate the map distances for each segment of the map.

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Description: Consider the results of a three point linkage mapping testcross between a female (AaBbCc) and a male (aabbcc). Phenotype | no. Of progenyABc 849abC 820 Abc 58aBC

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter15: Genomes And Genomics

Section: Chapter Questions

Problem 2QP: Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To...

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Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?
A purple snow pea plant known to be hybrid for three linked genes (a, b, and d) is testcrossed. The progeny phenotypes and frequencies are: +a+ 86 b+d 56 +ad 78 b++ 49 +++ 12 bad 6 ba+ 515 ++d 460 a) Draw the chromosomes of the trihybrid and tester lined mentioned above. b) Give a map of the gene order with distance.
A trihybrid individual with the genotype AaDdEe is testcrossed with an aaddee individual. The resulting offspring are as follows: ADe 7 aDE 157 AdE 688 ade 153 ADE 125 aDe 680 Ade 182 adE 8 Total = 2000 a) b) Determine the allele arrangement and gene order in the trihybrid parent. Give your answer using the PQr/pqR notation. Draw a chromosome map for these genes. Give your map distances to two decimal places. (Use underline or underscore (_) to draw the line for the map)
What would justify the following ratio appearing after phenotyping the outcome of a crossing trial: 8.9: 2.9: 3.2:1? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a b C d Obviously this represents independent assortment based on crossing dihybrid heterozygotes. Obviously this represents gene linkage based on test crossing a dihybrid heterozygote. Obviously this represents the results of a trihybrid test cross. Obviously this represents independent assortment based on crossing monohybrid heterozygotes.
A trihybrid individual with the genotype QqRrTt is testcrossed with a qgrtt individual. The resulting offspring are as follows: QrT 11 50 343 76 grT qRT grt QRt 76 357 78 9 QRt Qrt QRT Total - 1000 a) Determine the allele arrangement and gene order in the trihybrid parent. Give your answer using the AbC/aBc notation. b) Draw a chromosome map for these genes. Give your map distances to one decimal place. (Use underline or underscore () to draw the line for the map)
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?
Consider the following dihybrid testcross: B/b • E/e × b/b • e/e For the progeny from this testcross, determine the relative proportions (from 0% to 100%) of each genotype if the two genes: a) are linked (dominant alleles in cis conformation) with no crossing over: Be/be: be/be: BE/be: bE/be: b) assort independently. B/b; E/e: B/b; e/e: b/b; E/e: b/b; e/e: c) are linked (dominant alleles in cis conformation) and 20 map units apart. Be/be: be/be: BE/be: bE/be:
A cross is made between a heterozygote, +++/abc, and a recessive homozygote, abc/abc. Analysis of the progeny gave the following results: + + + 450 + + C 10 a + c 70 + b c 210 + b + 65 a + + 200 a bc 460 ab + 15 What is the map distance between the a and c genes? O 25.12 map units O 27.71 map units O 29.39 map units O 32.53 map units O None of the above Next» • Previous PgUp End F10 PrtScn Home 23 F2 DII F3 F4 F5 FB F9 F1 @ %23 %24 1. 3. 4. 8. To W R. Y U S D G H. C M B
Three autosomal recessive mutations in yeast, all producing the same phenotype (m1, m2, and m3), are subjected to complementation analysis. Of the results shown below, which, if any, are alleles of one another? Predict the results of the cross that is not shown—that is, m2 * m3. Cross 1: m1 * m24 F1: all wild-type progeny Cross 2: m1 * m34 F1: all mutant progeny
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