An individual heterozygous for d e f (D/d E/e F/f) was crossed to a homozygous recessive individual (d/d e/e f/f). The following offspring were recovered. d e + 39 + e f 372 d + f 7 + + f 47 + e + 5 d + + 412 What is the correct order of these genes? EDF DFE DEF

An individual heterozygous for d e f (D/d E/e F/f) was crossed to a homozygous recessive individual (d/d e/e f/f). The following offspring were recovered. d e + 39 + e f 372 d + f 7 + + f 47 + e + 5 d + + 412 What is the correct order of these genes? EDF DFE DEF

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Description: An individual heterozygous for d e f (D/d E/e F/f) was crossed to a homozygous recessive individual (d/d e/e f/f). The following offspring were recovered.d e + 39+ e f 372d + f 7+ + f 47+ e + 5d + +

Biology (MindTap Course List)

11th Edition

ISBN:9781337392938

Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Publisher:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Chapter11: The Basic Principles Of Heredity

Section: Chapter Questions

Problem 11TYU: Individuals of genotype AaBb were mated to individuals of genotype aabb. One thousand offspring were...

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Individuals of genotype AaBb were mated to individuals of genotype aabb. One thousand offspring were counted, with the following results: 474 Aabb, 480 aaBb, 20 AaBb, and 26 aabb. What type of cross is it? Are these loci linked? What are the two parental classes and the two recombinant classes of offspring? What is the percentage of recombination between these two loci? How many map units apart are they?
As it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women. During his tenure as guard, he had numerous clandestine affairs. In each case, children resulted. Subsequently, some of the childrenwho had no way of knowing that they were relatedmarried and had children of their own. Assume that two pairs of genes determine height. The genotype of the 7-foot-tall Potsdam Guard was A9A9B9B9, and the genotype of all of his 5-foot clandestine lovers was AABB. An A9 or B9 allele in the offspring each adds 6 inches to the base height of 5 feet conferred by the AABB genotype. a. What were the genotypes and phenotypes of all the F1 children? b. Diagram the cross between the F1 offspring, and give all possible genotypes and phenotypes of the F2 progeny
The F1 from a cross of AABB x aabb is testcrossed, resulting in the following phenotypic ratios: AB 308 Ab 190 ab 292 aB 210 (1) what are the genotypes for the listed phenotypes? (2) why aren't the phenotypes in a 1:1:1:1 ratio? (3) what is the frequency of recombination between genes A and 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 gene order of these linked 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
The parental genotypes for a series of crosses are wild-type male fruit flies mated to females with white eyes (wh) and miniature (min) wings. The phenotypes of the F1 generation were wild-type females, and males with white eyes, and miniature wings. These flies were allowed to mate with each other and produced the following offspring: Red eyes, long wings White eyes, miniature wings Red eyes, miniature wings White eyes, long wings 770 716 401 318 Total 2205 A. Are these genes linked? Why or why not?
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?
. 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?
In corn, we cross the pure line s t to the pure line u. It is known that the genes are linked in the order t s u. If the F1 is testcrossed, of the options given, which phenotypic class will have the lowest percentage? A.s u B.s C.tu D.s u t E. WT
In corn, the genes for tassel length (alleles T and t) andrust resistance (alleles R and r) are known to be on separate chromosomes. In the course of making routinecrosses, a breeder noticed that one T/t ; R/r plant gaveunusual results in a testcross with the double-recessivepollen parent t/t ; r/r. The results wereProgeny: T/t ; R/r 98t/ t ; r/r 104T/t ; r/r 3t/t ; R/r 5Corncobs: Only about half as many seeds as usuala. What key features of the data are different from theexpected results?b. State a concise hypothesis that explains the results.c. Show genotypes of parents and progeny.d. Draw a diagram showing the arrangement of alleleson the chromosomes.e. Explain the origin of the two classes of progenyhaving three and five members.
The Labrador Retriever’s coat colours are black, yellow, and chocolate.Yellow is produced by the presence of a recessive epistatic gene, which has the effect of masking the black and chocolate colour genes. Black (B) is dominant to chocolate (b).Therefore, the possible genotypes and corresponding phenotypes are: Phenotype Possible Genotypes Black G__B__ Brown G__bb Yellow gg__ Predict the percentage of offspring that would be black in colour for each of the following crosses. Record your answer as a whole number. Cross: GgBbGgBB GGBbGGBb Percentage ofOffspring: Answer% Answer%
In corn the genes for tassel length (alleles T and t) and rust resistance (alleles R and r) are known to be on separate chromosomes. In the course of a routine mating check experiment, a breeder got the following results upon crossing T/t:R/r with a double recessive strain (t/t;r/r): genotype Number of progeny T/t; R/r 98 t/t; r/r 104 T/t; r/r 3 t/t; R/r 5 In addition to these data she noted that the cobs contain only about 50% of the expected kernels. Note that in corn, each kernel is the result of one mating or can be considered one progeny. Hence 50% of the matings are not producing viable offspring. Draw a diagram showing the arrangement of the alleles on the parental chromosomes. Based on your hypothesis from part b, reassign genotypes to the progeny classes. Enter your new genotypes into the table above as an additional column. If the karyotype was done on the progeny what unusual feature would be observed?