Question 3 Listen You want to determine the distance between 2 mice genes F and G. You take mice which are FG/fg and cross them to mice which are fg/fg. You get 80 mice which are FG/fg 75 mice which are fg/fg 6 mice which are Fg/fg 4 mice which are fG/fg What is the distance between the genes? (rounded off to the nearest whole number) a) 80 map units b) 4 map units O c) 6 map units d) 94 map units

Question 3 Listen You want to determine the distance between 2 mice genes F and G. You take mice which are FG/fg and cross them to mice which are fg/fg. You get 80 mice which are FG/fg 75 mice which are fg/fg 6 mice which are Fg/fg 4 mice which are fG/fg What is the distance between the genes? (rounded off to the nearest whole number) a) 80 map units b) 4 map units O c) 6 map units d) 94 map units

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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QUESTION 6 Genes a, b, and c are linked, and pairwise crosses gave the following distances: a-b 20 cM b-c 2 cM a-c 22 cM Which gene is in the middle? O a O It's not possible to tell without additional information. QUESTION 7 A man and woman both have tpe AB blood and are carriers for the Bombay mutation. What is the probability their first child will be phenotypically O (Bombay)? O 1/4 O 1/16 O 1/2 O 3/16 QUESTION 8 The Bombay phenotype occurs because: O Rh antibodies can travel across the placenta and cause an incompatibility reaction. O an individual is homozygous recessive at the Rh locus. O the H substance is not present and so no A or Bantigens can be produced. O an individual does not haveor alleles, and so cannot produce the corresponding antigens.
Question:- 3. Wild type Drosophila has red eyes and gray body*. Autosomal recessive mutations sepia (se) and ebony (e) modify the eye and body color, respectively. What kind of a cross (or crosses) can you set up to determine whether the two genes are linked? You have several lines to choose from: a pure-breeding wild type line (where alleles for both, sepia and ebony are wild type), a double homozygous recessive line, a pure breeding sepia mutant line and a pure breeding ebony mutant line. You don’t have to use all of them. (a) Write down the cross (or crosses, if necessary) which will allow you to determine linkage; include genotypes of both crossed individuals (b) What kind of results (offspring) do you expect if the two genes are linked? (c) What kind of results (offspring) do you expect if the two genes are not linked? For (b) and (c), state the phenotypes, genotypes and numbers.
Question 5 [HINT: This problem wants you to calculate gene number from the proportion of individuals in the extreme classes, not from the number of phenotypes.] A farmer has two pure-breeding varieties of tomatoes: Little Pete (average fruit diameter 2cm) and Big Boy (average fruit diameter 14 cm). The farmer crosses Little Pete and Big Boy and obtains F1 plants with fruit averaging 8cm in diameter. When he intercrosses the F1s to obtain F2s, he examines 2000 F2 plants and finds two plants that produce fruit even smaller than Little Pete and three plants that produce fruit even larger than Big Boy. From these data, how many genes can we infer are involved in controlling fruit size in these plants? What is the genotype of Little Pete? What is the genotype of Big Boy? What is the genotype of the F1s? What are the genotypes that are bigger than Big Boy? What are the genotypes that are smaller than Little Pete?
Question 9 Consider that a certain gene is a maternal effect gene and that the allele for dark brown pigment is incompletely dominant to the allele for no pigment (white). The incomplete dominant phenotype is light tan. If a heterozygous female is crossed with heterozygous male, what will be the phenotypic ratio of the progeny? A) 1 dark brown: 2 light brown: 1 white. B) 3 dark brown: 1 white C). all dark brown D) all light tan E) cannot be determined CS Scanned with CamScanner
Question #1 Since experimental crosses are not performed in humans, how do we know how traits are inherited? State the name of the analysis/ chart and explain in at least one sentence.
5 & :56M ******* 24 DIHYBRID CROSSES DRV 0 Stv T alı A @ zladenA 9160p2-id2 bns obidalbaneoviene da II\ MOD YR 21 $59A ... Create a dihybrid cross and determine the expected phenotypic percentages of the offspring of two corn plants both of which are heterozygous for colour and texture (RrTt X RrTt). Don't forget to include clear let statements, and follow the all six steps taught on solving genetics problems. insig moni nellog: bna. zoom
Question 7 Which of the following is inconsistent (not consistent) with inheritance of a maternal effect trait? A) Progeny from the same parents have the same maternal effect phenotype as each other but not their mother. B) Progeny can show a recessive phenotype even though the progeny have dominant genotypes. C) Both parents contribute recessive alleles to an offspring yet it has a dominant phenotype for the maternal effect trait. D) Progeny from the same parents have different traits related to the maternal effect gene in question. CS Scanned with CamScanner
You are given a Drosophila female that looks wild-type but is heterozygous for mutations intan body (t), miniature wings (m), and white eyes (w). You test cross this female with a tanbodied, miniature winged, and white-eyed homozygous mutant male, and you obtain thefollowing 1400 progeny: Phenotype : number+ + + : 608t m w : 516+ m w : 2t + + : 6+ m + : 39t + w : 46+ + w : 81t m + : 102 Calculate the distance between each pair t-m, m-w, and t-w only using the number ofrecombinants between them (i.e. ignoring the gene in the middle). Draw a linear map with thedistances between genes.
Question 4. Recalling the eye pigment pathway in Drosophila, match the following fly genotypes with the appropriate phenotypes (w=white, bw-brown, cn=cinnabar red, st=scarlet red, se=sepia brown) a. b. d. W W W W ; ; ; ; bw + bw + + cn bw + Genotype A + cn bw cn bw cn Genotype B Genotype C + cn Genotype D ; ; ; ; st+ st se st se + se st se + se st se + se Scarlet Sepia Sepia White < < <
QUESTION 6 In Drosophila, sepia eyes (se) and stubble bristles (sb) are recessive to the wildtype eyes and bristles (se* and sb+). A female heterozygous for both genes was test crossed to a male homozygous recessive for both genes. The progeny of this cross are given below: Phenotypes wildtype eyes and wildtype bristles (set, sbt) # of Progeny 334 widltype eyes and stubble bristles (set, sb) 162 sepia eyes and wildtype bristles (se, sb+) 158 sepia eyes and stubble bristles (se, sb) 346 What is the genetic distance (in map units) between these two genes? Are the alleles in the heterozygous female in the cis (coupling) or trans (repulsion) conformation? Are the two genes in this problem unlinked, completely linked or incompletely linked?
Question 4: Imagine that you isolated X-receptor proteins from mutant cells with the nonsense mutation in exon 3. You also isolate X-receptor proteins from wild-type (normal) cells. If you ran these protein molecules on a gel which are your predicted results? A В C Ladder Wild-type Mutant Ladder Wild-type Mutant Ladder Wild-type Mutant
QUESTION 6 Assume that a researcher is studying coat color in voles. Three strains of white vole have been isolated: milky, blanc, and weiss. White is a recessive trait in each strain. Homozygous white voles are obtained for each strain. Consider the following crosses: milky X weiss = all white progeny (mutant) milky X blanc = all brown (wild-type vole color) blanc X weiss = all brown (wild-type vole color) The conclusion most consistent with these results is O A all three strains have mutations in the same gene. O B. all three strains have mutations in different genes. OC. milky and blanc have mutations on the same gene, weiss has a mutation in a different gene. D. milky and weiss have mutations on the same gene, blanc has a mutation in a different gene. O E. weiss and blanc have mutations on the same gene, milky has a mutation in a different gene.