BIOLOGY
12th Edition
ISBN: 9781260169614
Author: Raven
Publisher: RENT MCG
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Textbook Question
Chapter 13, Problem 2A
As real genetic distance increases, the distance calculated by recombination frequency becomes an
a. overestimate due to multiple crossovers that cannot be scored.
b. underestimate due to multiple crossovers that cannot be scored.
c. underestimate due to multiple crossovers adding to recombination frequency.
d. overestimate due to multiple crossovers adding to recombination frequency.
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For single crossovers, the frequency of recombinant gametes is half the frequency of crossing over because
a. a testcross between a homozygote and a heterozygote produces ½ heterozygous and ½ homozygous progeny.b. the frequency of recombination is always 50%.c. each crossover takes place between only two of the four chromatids of a homologous pair.d. crossovers take place in about 50% of meioses.
What is/are the phenotypes of the recombinant offspring of the F2generation?a. red eyes, long wingsb. white eyes, miniature wingsc. red eyes, long wings and white eyes, miniature wingsd. red eyes, miniature wings and white eyes, long wings
[Answer the multiple-choice questions based on the following experiment:P generation: True-breeding flies with red eyes and long wings werecrossed to flies with white eyes and miniature wings. All F1 offspringhad red eyes and long wings.The F1 female flies were then crossed to males with white eyes and miniaturewings. The following results were obtained for the F2 generation:129 red eyes, long wings133 white eyes, miniature wings71 red eyes, miniature wings67 white eyes, long wings]
. A three-point testcross was made in corn. The results and a recombination analysis are shown in the display below, which is typical of three-point testcrosses (p = purple leaves, + = green; v = virus-resistant seedlings, + = sensitive; b = brown midriff to seed, + = plain). Study the display, and answer parts a through c.
Chapter 13 Solutions
BIOLOGY
Ch. 13.1 - Describe sex-linked inheritance in fruit flies.Ch. 13.1 - Prob. 2LOCh. 13.2 - Prob. 1LOCh. 13.2 - Explain the genetic consequences of dosage...Ch. 13.3 - Prob. 1LOCh. 13.4 - Prob. 1LOCh. 13.4 - Explain the relationship between frequency of...Ch. 13.4 - Prob. 3LOCh. 13.5 - Prob. 1LOCh. 13.5 - Prob. 2LO
Ch. 13.5 - Prob. 3LOCh. 13 - Inquiry question Mendel did not examine plant...Ch. 13 - What would Mendel have observed in a dihybrid...Ch. 13 - Prob. 2DACh. 13 - Why is the white-eye phenotype always observed in...Ch. 13 - In an organisms genome, autosomes are a. the...Ch. 13 - What cellular process is responsible for genetic...Ch. 13 - The map distance between two genes is determined...Ch. 13 - How many map units separate two alleles if the...Ch. 13 - How does maternal inheritance of mitochondrial...Ch. 13 - Which of the following genotypes due to...Ch. 13 - Prob. 1ACh. 13 - As real genetic distance increases, the distance...Ch. 13 - Down syndrome is the result of trisomy for...Ch. 13 - Genes that are on the same chromosome can show...Ch. 13 - The A and B genes are 10 cM apart on a chromosome....Ch. 13 - Prob. 6ACh. 13 - Color blindness is caused by a sex-linked,...Ch. 13 - Assume that the genes for seed color and seed...Ch. 13 - A low frequency of calico cats are male (about...
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- Known deletions (1–6) were crossed to each other. The following chart shows which crosses produced wild-type offspring (+) and which crosses produced no wild-type offspring (−). Point mutants (a–f) were then crossed to these deletions. The results of these crosses are shown in the second chart. Which two deletions have the greatest distance between them? a. 3, 6 b. 2, 6 c. 2, 3 d. 4, 5 e. 1, 5arrow_forwardWhen you test cross DdSsNn, you obtain the following numbers of offspring with the following phenotypes out of a total of 1449 offspring: DSN 427 dsn 447 dSN 218 Dsn 210 DSn 68 dsN 71 dSn 6 DsN 2 A. Calculate the distances between D and S. Show your work. B. Calculate the distances between S and N. Show your work. C. How many double crossovers would you expect if there were no interference? Show your work. D. Calculate the coefficient of coincidence and interference based upon your answer in part C. Show your work.arrow_forwardCalculate the recombination frequency from the following data collected from a breeding experiment with butterflies. Parent 1: dihybrid, long legs and blue wings. Parent 2: double recessive, short legs and white wings Offspring: 50 long legs and white wings 200 long legs and blue wings 50 short legs and blue wings 200 short legs and white wings Recombination frequency is: Select one: a. 0.25 (25%) b. 0.10 (10%) c. 0.20 (20%) d. 0.33 (33%) e. 0.50 (50%)arrow_forward
- You cross two plants and get the following results: Recombinant progeny type A: 43 Recombinant progeny type B: 51 Nonrecombinant progeny type C: 129 Nonrecombinant progeny type D: 135 What is the recombination frequency? (Enter your answer as a % without the sign. For example, if your answer is 0.534, you would enter 53.4)arrow_forwardThe following recombination frequencies are calculated for four linked genes in the image attached: What is the frequency of crossing-over between A and D? a. 8% b. 6% c. 15% d. 11%arrow_forwardShort hair (S) in rabbits is dominant over long hair (s). The following crosses are carried out, producing the progeny shown. Give all possible genotypes of the parents in each cross. The Remaining subparts to be solve is Letter D and E.arrow_forward
- I am studying a pair of genes, A and B, that I think are both located on the same chromosome. After counting offspring from a test cross of an AaBb individual, I notice that recombination frequency for genes A and B is very low (~4%). From this, I can conclude that Select one: a. A and B are sex-linked b. A and B are actually on different chromosomes c. A and B are very close together on the same chromosome d. A and B on the same chromosome, but far apartarrow_forwardWhat is the answer for number 4 and how do you calculate this?arrow_forwardUse the data provided in the figure to calculate the recombination frequency between the two genes. Parents Wild Type (gray with normal wings) bb vg* vg Black with vestigial wings bb vg vg recombination frequency: % + TOOLS x10 Wild Type Black nogmal Black vestigial Gray vestigial b'bvg vg bb vg vg bb vg vg bb vg vg The number of each type of offspring is • wild type: 865 • black vestigial: 875 • gray vestigial: 160 • black normal: 100arrow_forward
- E. W. Lindstrom crossed two corn plants with green seedlings and obtained the following progeny: 3583 green seedlings, 853 virescentwhite seedlings, and 260 yellow seedlings (E. W. Lindstrom. 1921. Genetics 6:91–110). a. Give the genotypes for the green, virescent-white, and yellow progeny. b. Explain how color is determined in these seedlings. c. Is there epistasis among the genes that determine color in the corn seedlings? If so, which gene is epistatic and which is hypostatic?arrow_forwardWhich of the following statements regarding interference is false? Group of answer choices 1)Interference = 1 - (observed number of double recombinants/expected 2)number of double recombinants) 3)If interference is 1, then the observed and expected number of double crossovers is the same 4)Interference describes how likely it is for one crossing over event to influence the likelihood of another 5)If interference is 1, then none of the expected double crossovers were observedarrow_forwardYou breed two monster parents (green, hairless) GGhh and ggHH (white, hairy) to produce F1 offspring (green, hairy). Genes G and H are on the same chromosome. a. What are the dominant phenotypes? b. What are ALL of the potential gamete genotypes that could be created by an F1? Clearly indicate the recombinant genotypes. c. Explain how recombinants are created. d. What are the potential offspring genotypes from the cross of a F1 monster with a homozygous recessive monster. e. From this test cross you get 152 green, hairless; 23 green, hairy; 148 white, hairy, and 27 white, hairless. Calculate the chi squared to test the independent assortment hypothesis and determine if these genes assort independently.arrow_forward
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genetic recombination strategies of bacteria CONJUGATION, TRANSDUCTION AND TRANSFORMATION; Author: Scientist Cindy;https://www.youtube.com/watch?v=_Va8FZJEl9A;License: Standard youtube license