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1. Chromosomes, colored bodies, are composed of proteins and DNA. In human somat

ID: 10319 • Letter: 1

Question

1.       Chromosomes, colored bodies, are composed of proteins and DNA. In human somatic cells there are 23 chromosome pairs. In human somatic cells there are ___ nuclear DNA       molecules.       a. 23   b.1023  c. 46  d.1046   e. 92

 

2.       Mitotic cell division produces identical daughter cells by ___ DNA replication and _____ cell division(s).  a. 1 & 1    b. 1 & 2   c. 2 & 1  d. 2 & 1  e. none of these is correct  

 

3.       If we follow meiosis in a diploid heterozygote, the two alleles are, following the first cell division,

 

             a. no longer together in one cell    b. together on a chromatid     c. present only as

 

                 mutants           d. blended into a new mix                  e. chromatids

 

4.       Mendel's Law of Independent Assortment basically states that the alleles of two or more genes segregate independently within and  between loci; the chromosomal explanation of this observation is

 

            a. alignment of nonhomologous chromosomes at metaphase I

 

            b. alignment of nonhomologous chromosomes at metaphase II

 

            c. alignment of mitochondria at anaphase

 

            d. misalignment of tetrads during mitosis

 

            e. very much like transformation

 

   5.    Knowing that the segregation ratios for a set of loci are all  1/2  :  1/2, we can

 

          predict the frequencies of all possible gametic types by multiplying, if the loci are

 

           a. on different chromosomes and independent   b. linked  c. on the same            

 

           chromosome and dependent   d. mitochondrial       e. both c & d must hold

 

  6.   Watson and Crick used model building and the data of other workers to propose a

 

        double helical model for DNA. The data of Maurice Wilkins and Rosalind Franklin

 

        was critical and  showed that

 

a.       phosphodiester linkages are covalent

 

b. purines = pyrimidines  c. the two chains are held by  hydrogen bonds  

 

 d. the DNA molecule is a double helix           e. bases face the axis

 

7.     The DNA structure proposed by Watson and Crick suggested a replication model which was tested by Meselson and Stahl. They used nitrogen isotopes to label replicating DNA which was then separated into bands on a density gradient. The      three possible models of replication predicted different numbers of bands after one and then two DNA replications. For the conservative model of DNA replication two bands were expected after the first round of replication. The

 

      semiconservative model predicted the presence of _____ bands after the first

 

      DNA replication.         a. 2    b. 4    c. 3   d. 1   e. 8

 

8.One reason that parents in many cultures have for centuries asked that their children

 

   marry their "own kind" was the belief  that the material basis of inheritance was

 

   fluid (BLENDING THEORY). Marriage outside their own kind would adulterate

 

   the purity of the blood lines. Mendel's experiments refuted this fluid based theory of

 

   inheritance.(Not all parents know this.)  The critical observation in Mendel's work

 

   which led to the rejection of BLENDING THEORY was

 

a. chromosome condensation      b. reappearrance of the recessive phenotype in the F2

 

                     c. anaphase I pairing     d. mitosis        e. enteleky

 

9.     Professor Dropkin has been collecting data from couples known to be carriers or

 

      heterozygotes for the STR gene wowzer. The data collected look like this

 

                          CHILDREN's PHENOTYPES

 

                             WW             Ww              ww

 

                              121              249            130

 

            The expected number of heterzygotes is

 

              a. 250  b. 400  c.125   d. 249  e. 15

 

      10.  Using his hypothesis [single locus two allele] the expected number of ww

 

             homozygotes is

 

                                         a. 18    b. 36   c. 41.25   d. 18.33  e. 125

 

      11.  and the expected number of WW homozygotes is

 

                                         a. 125    b. 19   c. 13.75   d. 36.67  e. 18.33

 

      12. The chi-square value for goodness of fit of this data is

 

                                                                       Chi-square = SUM (obs-exp)2/exp 

 

                  a. 0.0   b. 3.84  c. 0.332   d. 55     e. 0.152

 

     13. The hypothesis [single locus, two allele, wowzer recessive] should be

 

               a. accepted as true    b. accepted as consistent with the hypothetical predictions

 

           c. rejected as false        d. rejected as unlikely       e. rejected as true

 

     14. The basis of inferential(testing) statistical analysis is that data are composed of

 

           two elements,

 

                    a. theoretical expectation + error                         b. mean + standard error

 

             c. dominant + recessive           d. alpha + beta                       e. skew I + skew II

 

     15. The chi-square distribution relates

 

            a. genetics to statistics    b. deviation from theoretical value to probability

 

       c. dominants to recessives  d. metaphase  I to telophase           e. bases to sugars

 

     16.  Gwendolynn and Igor have decided to marry and procreate. They are worried

 

           because both of them have siblings sufferring from Winkle Syndrome which is

 

           really bad and expressed in homozygous recessives. The chance that Gwendolynn

 

           is a carrier of the allele for Winkle Syndrome is  

 

                   a. 1/4    b. ½     c. 2/3   d. 1.0  e. 9/12

 

     17. The chance that both Gwendolynn and Igor are carriers is

 

                a.  ½        b.  4/9       c.  1/16      d. 1.0      e. 25%

 

     18. The chance that their first-born will have Winkle Syndrome is

 

                a. ½ X ½     b. 4/9 X 1/4     c. 2/3 X 1/4    d. 9/16   X 1/4

 

     19. The cis-trans or complementation test involves infecting a bacterial

 

          cell with two different mutant phage at the same time.  If

 

          complementation takes  place the two mutants are in _______.

 

a. oppostion   b. separate genes      c. one operon     d.  linkage

 

   20. To map the rII region of T4 bacteriophage, Benzer co-infected E. coli

 

with different rII mutants and measured the frequency of___________

 

from the progeny of mixed infection.

 

a. control     b.  complementation      c. recombination      d.linkage

 

21. In phage crosses or mixed infections recombination frequency is

 

    calculated as the sum of recombinant plaques divided by the total

 

    number of plaques. To estimate very small recombination frequencies

 

    Benzer devised a system in which only recombinant phage could grow

 

    and form plaques. The total number of plaques ( recombinant and

 

    nonrecombinant ) was estimated by plating a greater dilution on a

 

    permissive bacterial strain. If there are 187 recombinant plaques and

 

    the total number of plaques is 10,463, the recombinant frequency is

 

a.     0.23    b. 0.018     c. 0.32    d.  0.0098    e.   0.103

 

22. Genetic recombination, both independent assortment of nonhomologous

 

    chromosomes and crossing over between homologues, generates amazing

 

    amounts of genetic diversity and allows the constant generation of novel

 

    multilocus genotypes. T. H. Morgan hypothesized a chromosomal

 

    structure in which genes were linearly arranged along the chromosome

 

    and each gene had a specific location or locus. He also felt that crossovers

 

    took place at random along the chromosomal length. Alfred H. Sturtevant

 

    an undergraduate research assistant in Morgan’s lab realized in 1911 that

 

    this meant recombination could be used to map loci along  chromosomes

 

    based on recombination frequency. In a cross of  D. melanogaster flies

 

    Sturtevant recovered recombinants and nonrecombinants out of a total of

 

    8000 progeny. His progeny classes were:  3500 Zh, 3500 zH, 500 ZH and

 

    500 zh. He estimated the distance between Z and H as

 

                  a. 0.125        b. 22            c. 13.5 %             d. 0.083      e. 64

 

23. . Recombination mapping using the method of Lod scores compares two

 

      hypotheses:  _____ and _____.

 

a. homozygous and heterozygous                        b. haploid  and  diploid

 

c. prokaryotic and eukaryotic           d. linked  and unlinked

 

                             e. chromosomal  or  mitochondrial

 

24. In Lod score analysis the recombination distance between the marker

 

     locus and the locus affecting the phenotype is not known. Lod score

 

     analysis

 

      a. requires the analyst to guess at the recombination distance.

 

      b. requires review of the literature to find the best estimate of

 

          recombination rate.

 

      c. requires performance of an arcsin transform on the preliminary data.

 

      d. suggests c2 complementation testing.

 

 e. is only possible using mitochondrial markers.

 

25.    If you can draw in the pedigree and make up the missing parts of this

 

        question, you are SAFE.    

 

                                           3,9                   4,6

 

 

 

 

 

 

 

           4,9         6,9             4,9          3,4            4,9               4,9          3,6

 

     The affected mother I2 has a disorder, Labatt’s syndrome, which is

 

thought to be inherited as a single locus dominant. A group of human

 

geneticists are attempting to map this  disorder.  After collecting this pedigree from a family in Coral Gables, they first hypothesized that the marker locus with 3,4,6 and 9 alleles was seen in the affected father’s chromosomes in _________  and  ___________ phase configurations.

 

                     a.   3 L   and   9 l                   b.    3 l    and    9  L

 

c.        4  L    and   6  l              d.     3  N    and      4  Z

 

26.Under this hypothesized phase configuration there are _________ non

 

crossover and  ________  crossover progeny.

 

     a.   7 and  0         b.    6  and  1      c.   5  and 2       d. 4 and  3    e.  7 and 1

 

27. The  probability of  observing this particular family under the hypothesis  

 

      of independent  assortment is        a.  ½      b.  ¼     c.  ¼7     d.  ½7    e.  0

 

28. If we assume that the rate of recombination is r the probability    

 

     of observing this particular family under the hypothesis of linkage is

 

          a.  (r)6(1/10)   b.  ½6(1/10)     c.  (1-r)6(r)         d.  ½(1/10)       e.  

 

29. The ratio of the probability under linkage and independent assortment is

 

 equal to greater than 1.0.  This means that the hypothesis of _______

 

 is supported by the data in this pedigree.

 

a. dominance  b. recessiveness  c. linkage  d. independent asortment 

 

30. Mutations in the I or repressor gene which produce polypeptides unable

 

      to bind to the operator are termed ____________ mutants, because they

 

      are always __ .

 

  a. constitutive   off        b.  repressible   off       c.    repressible      on

 

             d.  constitutive    on                         e.  lagomorph     furry

 

31.Operons which are repressed by end-product are usually biosynthetic

 

pathway controls like the tryptophan operon. Mutations in the  trpR

 

gene may prevent formation of the repressor-corepressor complex

 

(tryptophan-repressor). This leads to a state in which RNA _________

 

is always able to initiate transcription.

 

                a. DNA trihybrids            b. DNA dihybrids

 

c. polymerase                         d.  furcase                     e.   polypeptide

 

32.Eukaryotic genes are, for the most part, split by introns so that the bases

 

which actually encode polypeptides, exons,  are separated.  This allows

 

_______________________ of transcribed mRNA to produce different polypeptides.       a. operonic processing      b. translational processing

 

          c. differential splicing            d. continuous looping

 

                              e. ubiquitin portage

 

33. In a two locus model A=2, a=1, B=2 and b=1.  In a cross between

 

two double heterozygotes, progeny with scores ranging from four to eight are seen.  The genotypes producing these extreme scores are ________

 

and ________ .    a. AABB and aabb          b.  AaBb   and Aabb

 

       c. AAbb    and   aaBB                 d.  aabb     and  AaBb

 

34.Assume that the model in question 20 applies in the real world and that

 

there is very little environmental variation. A breeder selects only those individuals with scores of 8 for breeding. Which alleles would you expect to decline in frequency ?

 

                a.    a and A             b.  b and  B            c. A  and  B

 

                                          d.   a   and   b            e.  B  and  c’

 

35.Heritability, h2, is a measure of genetic _________ among individuals

 

in a population; it is not the percentage of genetic determination in an

 

____________ .           a. polymorphism     population

 

b. determination    individual                   c.  control   individual

 

            d. differences   individual                e. variability individual

 

36. The assumption used in proving the Hardy-Weinberg Equilibrium which

 

     allows algebraic equality of allele and genotype frequencies across the

 

     generations without sampling error is

 

      a. random mating          b. allelic independence    c.  no mutation

 

                         d. no selection      e. infinite population size

 

37. The frequency of Bird Brain Syndrome carriers is thirty-two percent, and

 

     the frequency of affected homozygotes is four percent. What is the

 

     frequency of the normal allele?

 

      a. 0.8       b. 0.2     c. 0.3    d. 1.0     e. 0.9

 

38. Natural selection was described as “survival of the fittest” or “nature red

 

     in tooth and claw”.  It is, in fact, nothing more than

 

       a. differences in reproductive rates among genotypes.

 

       b. differential survival.                    c. differential mutation.

 

       d. differences in heritability among genotypes.   e. b and d

 

39. In forensic case work analysis the Hardy-Weinberg is used to

 

      calculate genotype frequencies from allele frequencies. If an STR

 

      profile is  10,22 and the frequencies of the alleles are p10 = 0.15 and

 

      p22 = 0.14, the expected genotype frequency is _____________ .

 

a. 0.05       b.  0.042           c. 0.01           d. 0.00005        e. 0.25

 

40. Ralph is accused of fathering Beatrice’s child. Ralph says that he

 

 is not the father; indeed he denies ever having known Beatrice.

 

 Confident that the child, Waldo has a father the assistant state attorney

 

 asks the forensic lab to conduct one of those new fangled STR

 

 parentage tests on Ralph. Here are the results:

 

     Locus           D4S4730           D5S227           D18S38        D22S1376

 

     Beatrice             12,17               9.2,17              11,14            22,30

 

     Child                  11,17              9.2,21               14,18            21,30

 

     Ralph                   9,11                 9,21                10,18            18,21

 

 For D18S38 the paternity index(ratio of alleged father and random

 

     man) is                   a.    1/ p17            b. 0.5/p18         c.  1/p11

 

                    d.   0.5/p11                       e. .05                f. 9:3:3:1

 

41.  RNAi comes in many known flavors and we will almost certainly be

 

       treated to many new flavors. The degradation of message [siRNA] or

 

       the down regulation of translation [miRNA] is a function of

 

     a. RNA translation.      b. mtDNA.       c. copy number variation.

 

               d. allozyme mobility                e. percentage of base pairing.

 

42. Hypothalamic brain-derived neurotropic factor BDNF is a key regulator

 

      of weight and energy balance. In Nature Medicine [April 2009] obesity

 

      and diabetes were controlled in a mouse model system by using an

 

      adeno-associated virus vector to introduce BDNF to induce weight loss

 

      and ____________ targeting BDNF to prevent too much weight loss.

 

      a. auto brain-derived neurotropic polypeptide      b. Drosophila

 

      c. miRNA                 d. splicing factor         e. both b and d

 

 43. Human groups may be broadly classified into bioeographic regions;

 

      these correspond to racial groups. Ninety percent of all polymorphisms

 

      are found in all five biogeographic groups, but only ____ percent of all

 

      alleles are found only in specific racial groups.

 

       a. 10       b. 30      c. 50      d. 70  e. 5    

Explanation / Answer

The Answer to #7 is D