A researcher is interested in whether a disease gene is linked to an RFLP marker. A cross is made betweenpure breeding blind cavefish and pure breeding sighted cavefish. The lines are pure breeding. The resultingRFLP markers of the F1 offspring and the F2 (from an F1XF1 cross) offspring are shown.F1F2A. If the disease gene is 10 m.u. from the RFLP marker, what proportion of the F2 will have the gelpattern shown above AND be blind? Show your workB. If the disease gene is 10 mu. from the RFLP marker, what proportion of the F2 will have the gelpattern above AND NOT be blind? Show your workBlind cavefishSighted cavefishSighted cavefish1Sighted cavefish

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Description: A researcher is interested in whether a disease gene is linked to an RFLP marker. A cross is made betweenpure breeding blind cavefish and pure breeding sighted cavefish. The lines are pure breeding. The resultingRFLP markers of the F1 offspring and

If the genes are present in close proximity to each other they have chances to be inherited together…

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A researcher is interested in whether a disease gene is linked to an RFLP marker. A cross is made between pure breeding blind cavefish and pure breeding sighted cavefish. The lines are pure breeding. The resulting RFLP markers of the F1 offspring and the F2 (from an F1XF1 cross) offspring are shown. Blind cavefish X P Sighted cavefish - F1 Sighted cavefish F2 Sighted cavefish

A researcher is interested in whether a disease gene is linked to an RFLP marker. A cross is made between pure breeding blind cavefish and pure breeding sighted cavefish. The lines are pure breeding. The resulting RFLP markers of the F1 offspring and the F2 (from an F1XF1 cross) offspring are shown. Blind cavefish X P Sighted cavefish - F1 Sighted cavefish F2 Sighted cavefish

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 3TYU: The type of cross described in question 2 is (a) an F2 cross (b) a dihybrid cross (c) a test cross...

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Below is the mapping data from three different crosses of Hfr strains with separate samples of an F- strain. Appearance of Genes in F- cells Hfrl: Genes d+ c+ ft et g+ Time (min) 4 15 26 44 58 Hfr2: Genes b+ d+ c+ f+ g+ 59 Time (min) 3 5 16 27 Hfr3: Genes et f+ c+ d+ b+ Time (min) 6 24 35 46 48 Use the mapping data from these interrupted conjugation crosses to construct a genetic map for these genes, indicating their order on the bacterial chromosome and the distances between them.
Wild-type strains of the haploid fungus Neurospora canmake their own tryptophan. An abnormal allele td renders the fungus incapable of making its own tryptophan.An individual of genotype td grows only when its medium supplies tryptophan. The allele su assorts independently of td; its only known effect is to suppress the tdphenotype. Therefore, strains carrying both td and su donot require tryptophan for growth.a. If a td ; su strain is crossed with a genotypically wildtype strain, what genotypes are expected in the progenyand in what proportions?b. What will be the ratio of tryptophan-dependent totryptophan-independent progeny in the cross of part a?
Based on the table below, what is the order of the genes examined? a b с d A B C/C B A Selected answer will be automatically saved. For keyboard navigation, press up/down arrow keys to select an answer. A C B/B C A CAB/BAC Phenotype of testcross ССА/АСС A,C C A, B, C A B, C A, B B Genotype of hybrid gamete Abc ab C A B C Abc a B C abc е Not enough information A B C a B c # of progeny 388 5 50 64 51 47 5 390 1000
A series of crosses were performed between pure breeding lines of mice with different coat colours. white agouti white agouti ВВсcDD bbCCdd bbccDD bbCCdd black brown BbCcDd bbCcDd white agouti white agouti ВВсcdd bbCCdd bbccdd bbCCdd gray agouti BbCcdd bbCcdd A) Based on the results of the crosses, describe the type of gene interaction that gives the white phenotype. Explain your reasoning. B) Determine the order of the function.
The fungus Melampsora lini causes a disease known as flax rust.Different strains of M. lini cause varying degrees of the rust disease.Conversely, different strains of flax are resistant or sensitive to thevarious varieties of rust. The Bombay variety of flax is resistant toM. lini-strain 22 but sensitive to M. lini-strain 24. A strain of flaxcalled 770B is just the opposite; it is resistant to strain 24 but sensitive to strain 22. When 770B was crossed to Bombay, all F1 individuals were resistant to both strain 22 and strain 24. When F1individuals were self-fertilized, the following data were obtained:43 resistant to strain 22 but sensitive to strain 24 Explain the inheritance pattern for flax resistance and sensitivityto M. lini strains.9 sensitive to strain 22 and strain 2432 sensitive to strain 22 but resistant to strain 24110 resistant to strain 22 and strain 24
A RFLP is discovered that is linked to the gene for Duchenne’s muscular dystrophy (DMD). DMD is an X-linked, recessive trait. The RFLP is 2 map units from the gene for DMD. Consider the following pedigree and Southern blot using a probe that hybridizes to the RFLP. Which band/s is/are associated with DMD? What is the genotype for individuals 3 and 4? (Remember, this is an X linked disease, so use X’s and Y’s to denote). Individual 9 married a man who does NOT have muscular dystrophy, and she is pregnant. DMD is an X-linked trait. What is the probability for their child to have DMD? An amniocentesis is performed and it is determined that 9’s child in utero has only a 10 kb band that hybridizes to the same probe used above. What can you say about the child now?
Mapping three genes in a single cross: generic a, b, c mutant loci (a+, b+, c+ = wild type). a, b, c x a, b, c a+, b+, c+ a, b, c Offspring: a, b, c = 870 a+, b+, c+ = 855 a+, b, c = 65 a, b+, c+ = 63 a+, b+, c = 67 a, b, c+ = 65 a+, b, c+ = 8 a, b+, c = 7 Determine the map distance between each gene and draw a map showing the map distance and the order of the three genes.
Three linked loci are evaluated to determine the recombination frequency of the loci. To set up the test cross, an individual that is true breeding dominant for all three phenotypes is crossed with an individual that is true breeding recessive for all three phenotypes. F1 offspring are crossed with individuals that are recessive for all phenotypes. The phenotypic results are shown in the table below. Phenotype Number of Offspring Dominant 1, Dominant 2, Dominant 3 322 Dominant 1, Dominant 2, Recessive 3 82 Recessive 1, Dominant 2, Recessive 3 5 Dominant 1, Recessive 2, Dominant 3 2 Recessive 1, Dominant 2, Dominant 3 28 Recessive 1, Recessive 2, Dominant 3 89 Recessive 1, Recessive 2, Recessive 3 312 Dominant 1, Recessive 2, Recessive 3 25 1. Which two phenotypes are parental? 2. What is the map distance between locus 1 and locus 2? 3. What is the map distance between locus 1 and locus 3? 4. What is the map distance between…
Suppose that you have obtained the following data from your complementation crosses with the M.5.2/CyO mutant stock. Chromosomal Complement? Yes/No Genotype of deficiency stock Df(2R)1/CyO Df(2R)2/CyO Df(2R)3/CyO Df(2R)4/CyO Deficiency stock Chromosomal breakpoint 2 120,400 193,070 161,250 143,175 number breakpoint 1 1 Yes 100,000 Yes 158,630 No 128,100 4 No 116,300 1. What phenotypic ratio is seen in the crosses that complement? Explain this ratio.
A three point cross is carried out and the following 1000 offspring are recovered. PHENOTYPES NUMBER OBSERVED CP+ 5 CPR 330 C++ 115 C+R 70 +++ 385 ++R 10 +PR 70 +P+ 15 - A. B. C. D. E. What is the correct gene map for these genes? (A-E) F. G. H. I. J. What is the interference value? (F-J) A. P----20----C----10----R B. P----20----R----10----C C. C----20----R----10----P D. R----20----C----10----P E. C----20----P----10----R…
Use 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: 100
Tick all the essential steps to demonstrate a genetic linkage between a disease and a molecular marker in humans. identify the alleles of the genetic marker only for diseased individuals in the pedigree enumerate parental type individuals sequence the wild-type and mutant alleles to find the mutation no correct answer calculate a Lod score calculate the recombination frequency between the mutation and the molecular marker identify the alleles of the genetic marker for each individual in the pedigree pedigree analysis cloning the defective gene enumerate recombinant individuals