genetics practice exam

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genetics practice exam
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  1. 1. Being the brilliant geneticist that you are, you have just completed a mutagenesis screen on a red flowering plant. 
    You have isolated 2 mutant alleles that produce white flowering plants.
    When you cross each of the mutant plants (white) to wild-type (red), in both
    cases, the F1 progeny are all red flowering plants and the F2 progeny are 75%
    red and 25% white flowering plants.

    a. You don’t want to spend months mapping the alleles, so you decide to use a
    relatively quick test to determine if the mutant alleles you isolated are in
    the same or different genes? Name and describe the test.

    b.Outline the crosses you will set up and show the results you expect if the alleles are
    in different genes as well as the results that you would expect if the alleles
    are in the same gene.
    • a. Complementation test: used to
    • determine if 2 alleles are in the same gene. Involves crossing two homozygous
    • mutant lines to produce a heterozygote. 

    •  If the heterozygote is wild-type
    • then the mutations must be in different genes. Complementation occurred.

    • If the heterozygote has a mutant
    • phenotype, mutations must be in the same gene.
    • b.
    •  
  2. 2. In corn, the alleles c+ and c result in colored versus colorless seeds, wx+ and wx
    in nonwaxy versus waxy endosperm, and sh+ and sh in plump versus shrunken
    endosperm.  When plants grown from seeds
    heterozygous for each of these pairs of alleles (from crosses of pure breeding
    colorless waxy plump parents with pure breeding colored, nonwaxy, shrunken
    parents) were testcrossed with plants from colorless, waxy, shrunken seeds, the
    progeny seeds were as follows:

    colorless, nonwaxy, shrunken            55       

    colorless, nonwaxy, plump                105

    colorless, waxy, shrunken                    5

    colorless, waxy, plump                       334

    colored, waxy, shrunken                     98

    colored, waxy, plump                          61

    colored, nonwaxy, shrunken              339

    colored, nonwaxy, plump                      3

                                              Total:1000

     
    a. What is the order of the 3 genes?
    b. What is the distance in map units between genes c and sh?
    c. What is the distance in map units between genes wx and c? 
    d.What is the distance in map units between genes wx and sh? 
    e. Draw a map to illustrate the distances between the three loci.
    f. Calculate interference.
    • a. c, sh, wx
    • b. 55+61+5+3 = 124/1000 x 100
    •   = 12.4 mu
    • c. 55+61+10+6+105+98 = 335/1000 x 100
    • = 33.5 mu
    • d. 105+98+5+3=211/1000 x 100 = 21.1 mu
    • e. c ---12.4---sh-----21.1---wx (total: 33.5)
    • f.  E= (0.124 x 0.211) x 1000 = 26
    •   C = O/E = 8/26 = 0.308
    •   I = 1-C = 1 – 0.308 = 0.692 or 69.2%   positive interference
  3. 2.For each of the following mutations, is it a transition, transversion, addition or deletion?  
     Original DNA Sequence:                          5’-GACTACTGTTAA-3’

    a. 5’-GACGACTGTTAA-3’ 
    b. 5’-GACTACTATTAA-3’
    • a. T->G transversion
    •   pyr -> pur
    • b. G->A transition
    •   pu -> pu
  4. 4.For each of the following mutations, is it a silent, missense, nonsense or frameshift
    mutation?

    Original DNA strand (coding strand) is:       5’-ATGTGGTCTAGAACC-3’
    (First amino acid is Met).
    Met, Trp, Ser, Arg, Thr
    a. 5’-ATGTGGTCCAGAACC-3’
    b. 5’-ATGTGGTCTAGAGCC-3’
    c. 5’-ATGTGATCTAGAACC-3’
    • a. Silent mutation – UCU & UCC both code for     Ser
    • b. Missense mutation Thr -> Ala
    • c. UGA = Stop codon, nonsense mutation
  5. 5. During mismatch repair, why is it necessary to distinguish between the template strand
    and the newly made daughter strand?  How
    is this accomplished?
    • DNA in E.coli is methylated. To distinguish the old template from the newly synthesized strand the mismatch repair mechanism takes advantage of a delay in methylation of the new strand.This makes sense as replication errors produce mismatches only on the
    • newly synthesized strand, so the mismatch repair system replaced the wrong base
    • on that strand.
  6. 6. As you remember, the occurrence of calico cats is due to X-inactivation.
    a.With this in mind, why is the occurrence of male calico cats unusual? 

     

    b. What kind of aberrant events could give rise to a male calico cat?  Describe/draw how such an aberrant event could occur.
    • a. X-inactivation occurs when more
    • than 1 X chromosome is present. Only 1 X chromosome is left active in each
    • cell.  Since males are normally XY, there
    • is no X inactivation and thus no calico pattern.

    • b. Nondisjunction at MI in female
    • **picture**
  7. 2. Five different Hfr strains were allowed to conjugate with F-minus cells.  After appropriate screening, it was
    determined that genes transferred in the following order for each Hfr strain:

    Strain 1: L N O H I X
    Strain 2: X I H O
    Strain 3: X J K A P
    Strain 4: A P L N O
    Strain 5: J X I H O N 
    a. What is the order of the genes on the circular bacterial chromosome?

    b. For each Hfr strain, give the location of the F factor in the chromosome and its polarity.
    • a. L N O H I X J K A P - circular
    • **picture**
    • b.
  8. 8. A cross is made between 2 E.coli
    strains:
    Hfr: met+ bio+ pro+    x    F-: met- bio- pro-

    Interrupted mating studies show that met enters last, so met+ recombinants are selected on medium containing only bio and pro.These recombinants are tested for the
    presence of bio+ pro+.The following numbers of individuals are found for each genotype:

    met+ bio+ pro+           600

    met+ bio+ pro-                2

    met+ bio- pro+              32

    met+ bio- pro-               66

    a. What is the order of these genes? 
    b. What is the map distance between pro       and bio? 
    c. What is the map distance between met       and pro?
    • a. bio, pro, met
    • b. (32 + 2)/700 x 100 = 4.86 mu
    • c. (66 + 2)/700 x 100 = 9.71 mu
  9. 9.
    a.Describe and/or illustrate the epigenetic regulation of the Igf2 and H19 locus in a
    wild-type mouse.  Indicate paternal and
    maternal chromosomes in your discussion/drawing. 

    b. Predict the phenotype of a mouse embryo that is homozygous for a deletion in the DMR
    (differentially methylated region) of the Igf2
    locus.  Discuss the affects this deletion
    would have on the expression of Igf2 and
    H19. You may assume the deletion of the DMR does not disrupt the promoter of H19.
    • a. Maternal chromosome: Insulator able
    • to bind to DMR region, Igf2 gene expression is turned off.  H19 gene expression is on.

    Paternal chromosome: Insulator protein unable to bind DMR due to methylation of the region.The H19 promoter is also disrupted by methylation. Igf2 gene is on, H19 gene is off.

    • b. Maternal chromosome: without DMR
    • region insulator cannot bind, both Igf2 and H19 are on.

    •  Paternal chromosome: No DMR
    • methylation, but no insulator binding either. 
    • Igf2 and H19 are both on.

     Phenotype of mouse: larger than normal
  10. 10. Consider a heterozygous female fly that contains an inversion spanning three X-linked genes: forked (f), vermillion (v) and yellow (y). The normal X chromosome is f, v, y and the inversion chromsome is y+, v+, f+. The centromere is not located within the inversion. 
    Note: the lack of + means that each allele is a recessive null mutation and + indicates a normal, wild-type allele. This female is mated to a male that contains the normal X chromosome (no inversion) with all three recessive mutations (f, v, y).  Almost all the F1 male progeny are either f, v, y or y+, v+, f+.  There are very few recombinants (f, v+, y+ or f, v, y+, etc). provide an explanation for this suppression of recombination.  A diagram might help.
    • Only viable male progeny are of the
    • parental genotype because crossover products contain dicentric and acentric products which are not viable.This is a result of heterozygote paracentric inversion.
  11. 11. A chromosome has the following segments, where * represents the centromere:

    A B C D E F * G H I

    What types of chromosome mutations are required to change this chromosome into each of the following chromosomes? In some cases more than one chromosome mutation
    may be required.
    a. A E D C B F * G H I
    b. A B C D E F * I

    c. A B C D E F D E F * G H I
    d. A B C D E I H G * F        
    e. A B C C B A D E F * G H I
    • a. Paracentric inversion of BCDE
    • b. Deletion of GH
    • c. Tandem duplication DEF
    • d. Pericentric inversion F*GHI
    • e. Duplication & paracentric inversion ABC
  12. 12.The following diagrams represent two non-homologous chromosomes:

     A B C * D E F G H 
    TU * V W X Y Z

    What type of chromosome mutation would produce the following chromosomes? 

    a. A B C * D E F
    T U * V W X Y Z G H I

    b. A B C * D E F X Y Z

     T U * V W G H I
    • a. Simple translocation
    • b. Reciprocal translocation

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