Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN: 9780134580999
Author: Elaine N. Marieb, Katja N. Hoehn
Publisher: PEARSON
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In a mutant screen in Drosophila, you identified a gene related to memory, as evidenced by the inability of recessive homozygotes to learn to associate a particular scent with the availability of food. Given another line of flies with an autosomal mutation that produces orange eyes, design a series of crosses to determine the map distance between these two loci. You do not need to calculate recombination frequency.
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- First used in Drosophila, recombination frequencies can help map relative positions of genes along chromosomes. The greater the distance between two loci on a single chromosome, the greater the chance that they will be separated by a crossing-over event during meiosis I. If two genes are recombined in x% of gametes, they are separated by a genetic map distance of xmap units. In humans, such mapping located disease-causing mutations near single-nucleotide polymorphisms (SNPS). Recombination frequencies among six Drosophila chromosome genes (yellow-body (yb), white-eye (we), singed-bristle (sb), miniature wings (mw), round eyes (re), and hairy body (hb) are given in the table below. Using this information, place the genes in the correct relative order along the chromosome. Genes we and yb have been filled in to get you started. we yb sb mw re hb we 1.4 19.6 34.7 6.1 12.3 yb 1.4 21 36.1 7.5 13.7 sb 19.6 21 15.1 13.5 7.3 mw 34.7 36.1 15.1 28.6 22.4 re 6.1 7.5 13.5 28.6 6.2 hb 12.3 13.7 7.3…arrow_forwardYou are working with a hypothetical fly and have found color and wing mutants. Preliminary work indicates that the mutant traits are recessive and the associated genes are not sex-linked, but beyond that, you have no information. You first look at 2 genes, each with 2 alleles. "B" or “b" for body color and "W" or "w" for wing surface. The red-body phenotype is dominant to the yellow-body phenotype and smooth wings are dominant to crinkled wings.arrow_forwardIn the fruit fly, dumpy wings (d) and purple eyes (p) are encoded by mutant alleles that are recessive to those that produce wild type traits; long wings (d+) and red eyes (p+). These two genes are on the same chromosome. In a particular lab, two researchers Walt and Jesse crossed a fly homozygous for dumpy wings and purple eyes with a fly homozygous for the wild type traits. The F1 progeny, which had long wings and red eyes, was then crossed with flies that had dumpy wings and purple eyes. Unfortunately, the progeny of this cross somehow escaped. To prevent their other projects from contamination, they decided to spend an exceptionally boring hour in the lab catching and counting the progeny and found the following: long wings, red eyes – 482 dumpy wings, purple eyes – 473 long wings, purple eyes – 23 dumpy wings, red eyes - 22 What is the genetic distance between these two loci? a. 4.5 cM b. 55 cM c. 45 cM d. 49.5 cM e. 4.7 cMarrow_forward
- In a series of mapping experiments, the recombination frequencies for four different linked genes of the Drosophila fly were determined, as shown in the figure. Based on this information, what is the most probable order of these genes on the chromosome? b 0 cn 90 rb 3.5 6.5 0 vg 19 9.0 16 0 b cn rb vg The numbers in the boxes are the recombination frequencies in between the genes (in percent). O b-rb-cn-vg O vg-cn-b-rb Orb-cn-vg-b b = black body cn = cinnabar eyes rb = reduced bristles vg = vestigial wings O vg-b-rb-cnarrow_forwardConsider a maize plant: Genotype C/cm ; Ac/Ac+ where cm is an unstable colorless allele caused by Ds insertion. What phenotypic ratios would be produced and in what proportions when this plant is crossed with a mutant c/c Ac+/Ac+? Assume that the Ac and c loci are unlinked, that the chromosome-breakage frequency is negligible, and the C allele encodes pigment production.arrow_forwardIn c. elegans, genetics model organism, movement problems (unc) and small body size (sma) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (unc+ and sma+). A worm homozygous for movement problems and small body is crossed with a worm homozygous for the wild-type traits. The F1 have normal movement and normal body size. The F1 are then crossed with worms that have movement problems and small body size in a testcross. The progeny of this testcross is: Normal movement, normal body size 210 Movement problems, normal body size 9 Normal movement, small body size 11 Movement problems, small body size 193 a)From the test cross results, can you tell if the two genes are on the same chromosome or not? Explain your reasoning. b)What phenotypic proportions would be expected if the genes for round eyes and white body were located on different chromosomes? (please explain hot to get to these conclusions)arrow_forward
- In Drosophila, the X-linked recessive mutation vermilion (v) causes bright red eyes, in contrast to the brick-red eyes of wild type. A separate autosomal recessive mutation, suppressor of vermilion (su-v), causes flies homozygous or hemizygous for v to have wildtype eyes. In the absence of vermilion alleles, su-v has no effect on eye color. Determine the F1 and F2 phenotypic ratios from a cross between a female with wild-type alleles at the vermilion locus, but who is homozygous for su-v, with a vermilion male who has wildtype alleles at the su-v locusarrow_forwardPart A You start your experiments with the eyeless mutation on chromosome IV. You cross the reciprocal translocation strain to the eyeless pure line to generate F, flies that are both translocation heterozygotes and Ee heterozygotes. You decide to testcross F, males and females in two separate experiments to take advantage of the fact that crossing over does not occur in male Drosophila. F, males x pure eyeless females (Note: There is no crossing over F, females x pure eyeless males (Note: Crossing over in the F, males.) Cross may occur in the F, females.) 4 eyeless, fully fertile 4 wild-type, semi-sterile 4 eyeless, semi-sterile 4 wild-type, fully fertile 4 eyeless, fully fertile 4 wild-type, semi-sterile 4 eyeless, semi-sterile 4 wild-type, fully fertile F, Progeny What can you conclude from these results? Select the two correct statements. > View Available Hint(s) O The F2 progeny in both experiments contain recombinants. O The F2 progeny contain recombinants only when F, females…arrow_forwardIn Figure 3-11, if the input genotypes were a • B and A • b,what would be the genotypes colored blue?arrow_forward
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