Bio 4564/6564                        Advanced Genetics                      Maymester, 2018

 

                                                             Test I

 

Name__________________                                            Student ID: _______________

           

 

 

Section 1 (40 pts.)

Unless stated otherwise each question in this section is worth 2 pts. Please circle the letters that precede the most appropriate answer(s) to the question asked.  Circling more than one letter in answering any given question will count against you. 

 

1.  Mendel’s 2nd law states that:

 

            A.        During gamete formation the segregation of one gene pair is independent of all other gene pairs

            B.        Two members of a gene pair segregate from each other into the gametes, whereby one half of the gametes carries one of the traits, the other half carries the other  

            C.        The union of one gamete from each parent to form a resultant zygote is random with respect to that particular characteristic.

            D.        Like all things biological, genes come in pairs.

            E.        When fertilization occurs, the zygote gets one half of its chromosomal set from each parent, thus restoring the pair.

 

 

2. Which of the following methods was not used by Mendel in his study of the genetics of
the garden pea?

 

 

            A.        maintenance of “true-breeding” lines.

            B.        cross-pollination.

            C.        basic microscopy.

            D.        production of hybrind plants.

            E.        quantitative analysis

 

 

3. When a given trait is the result of multigene action, one of the genes may mask the
expression of one or all other genes. This phenomenon is termed

 

            A.        epigenesis.

            B.        non-dysjunction.

            C.        overdominance or heterosis.

            D.        epistasis.

            E.        incomplete dysgenesis

 

 

4.    Given that non-albino mice that lack the dominant agouti allele at the B locus are black.

An agouti mouse that is heterozygous at the agouti and albino loci (Aa Bb) is mated to an albino mouse that is heterozygous at the agouti locus (aa Bb).

 

     What percent of the progeny do you expect to be agouti?

            A.        0

            B.        12.5

            C.        37.5

            D.        50

            E.        100

 

 

5. What percent of the progeny would you expect to be albino?

 

            A.        0

            B.        12.5

            C.        37.5

            D.        50

            E.        100

 

 

6. The complete phenotype of an organism is dependent on

 

            A.        genotype.

            B.        penetrance.

            C.        expressivity.

            D.        polygenes.

            E.        All of the above

 

 

 

 

7. Which of the following statements about Mendelian genetics need not be true?

 

            A.        Alternative forms of genes are called alleles.

            B.        A locus is a gene’s location on its chromosome.

            C.        Only two alleles can exist for a given gene.

D.        A genotype is a description of the alleles that represent an individual’s genes.

E.        Individuals with the same phenotype can have different genotypes.

 

 

 

8. Given that, in dogs, the two traits of erect ears and large canines are due to dominant alleles; and consequently -droopy ears and small canines are due to recessive alleles. A dog homozygous dominant for both traits is mated to a droopy-eared, small canine dog

If the two genes are unlinked, the expected F1 phenotypic ratios should be

 

            A.        9:3:3:1.

            B.        1:1.

            C.        16:0.

            D.        1:2:1.

            E.        None of the above

 

 

9. (a)  If NCO = 1 - (SCO + DCO) then in yeast formation of haploids from meiosis:

 

A.        NCO = 1 - 2(4NPD) + (T-2NPD)

            B.        NCO = 1 - (T-2PD) + 2(4PD)

            C.        m = 1 - (T-2NPD) + 2(4NPD)

D.        NCO = 1 - (T-2NPD) + 2(4NPD)

E.        NPD = 1 - (T-2PD) + 2(4PD)

 

 

 

10.  “B form DNA: is said to be a right handed double helix in which the bases are perpendicular to the helical axis, with ~11 base pairs per turn, contains clearly defined major and minor grooves, C2'-endo pucker preferred, and where the bases adopt an anti- configuration.”

 

   Please indicate (1) which part of the previous statement is NOT true by underlining the phrase that is incorrect and then (2) choosing the appropriate (correct?) statement below.

 

            A.        B  form DNA is a right -handed helix.

            B.        B  form DNA is a left handed helix.

            C.        B  form DNA has a clearly defined major and minor groove.

D.        The bases in B form DNA adopt an syn- configuration 

            E.        There are ~10 bases per turn in B form DNA

 

 

11. The ABO blood groups in humans are determined by a multiple allelic system in which IA and IB are codominant and are dominant to iO. If an infant born to a type A mother is found to be type AB, possible genotypes for the father are

 

            A.        O or A.

            B.        B or AB

            C.        B only.

            D.        O, A, or B.

            E.        impossible to determine.

 

 

12.   A centiMorgan defines:

 

A.    the number of cents in a genetic unit, the “Morgan”

B.    the genetic distance between one chiasmatic event and the next

C.   the number of chiasmata that occur in 100 recombinant individuals

D.   the distance between two genes as it relates to the frequency of crossover events occurring between them

E.    a physical map unit that is less than a microMorgan, but more than a milliMorgan

 

13.   One of the following describes a Type I restriction enzyme.  Please choose the most appropriate description.

 

A.    A restriction-modification enzyme with three functionally distinct subunits

B.    A restriction-modification enzyme with two subunits

C.  An enzyme that binds and cuts DNA within a palindromic DNA sequence

D.  An enzyme that binds and cuts DNA upstream of a palindromic DNA sequence

E.    An enzyme that copies a DNA palindrome.

 

 

14. In discerning the correct interpretation for phenotypic variations in the yeast asci if two genes are unlinked then

 

            A.        the number of tetratypes is equal to the number of double crossover events

            B.        the number of parental ditypes underestimates the number of recombinants

            C.        the number of non parental ditypes equals the number of tetratypes

D.        the number of parental ditypes equals the number of tetratypes

E.        the number of non parental ditypes would normally approximate half the outcomes.

F.         the number of parental ditypes will be greater than the number of non parental ditypes

 

15. 5.  The term “Monoecious” refers to:

 

A.             the number of chromosomes in a cell

B.             the presence of both male and female structures exhibited by the same individual organism

C.            the presence of only male or only female structures exhibited by the same individual organism

D.            the number of membranes present in a Gram positive bacterium

E.             a genetic unit of hereditary

 

 

16. Complete the phrase, “An episome is…” correctly

 

A.    a small chromosome

B.    a fragment of DNA that is outside of the chromosome

C.   a small, rapidly replicating plasmid

D.   a non-chromosomal vector that has recombined with the chromosome.

E.    A non- chromosomal vector that can be passed to another cell through conjugation.

 

17. (4 pts)     Two plasmid vectors  are said to be ____________ if they are able to co-exist in the same cell, because they ___________ compete for cellular function

 

 

Which combination of words should fill in the blanks to make the statement correct?

 

A.             Compatible,        do not

B.             Multicopied,        do

C.        Incompatible       do

            D.        Compatible,        cannot

            E.        Compatible,        do not

            F.         Incompatible       do not

 

 

 

 

18.  (4 pts)  What does the figure below depict?

 

 

 

             

 

 

 

 

 

 

 

 

A.    Meiotic nondisjunction phase I

B.    Meiotic disjuntion

C.   Meiotic nondisjunction phase II

D.   Sex linkage in Drosophylla

E.    Formation of a male Drosophila (XO)   

F.    Mitotic nondisjunction

 

 

 

What would be the consequences?

 

___triploid or haploid offspring for the chromosome in question_______________________________

 

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Section 2 (50 pts.)

 

Please answer the following questions using short, concise answers.

 

 

2.1.    Please read the following information and then answer the attendant questions below.

 

The allele b gives the Drosophila fly “small bristles” on the thorax and b+ gives large bristles (the wild-type phenotype).  The allele vg of a completely separate gene vg gives “vestigial wings”, while the wings of a vg+ fly are normal.  The allele cn of a third independent gene gives “cinnabar eyes”, and cn+ gives the normal red eye.  A female which is heterozygous for these three genes is test-crossed, yielding one thousand progeny, that were classified as follows.

 

69 vestigial winged flies, cinnabar,

49 vestigial winged flies

67 small bristles

5 wild-type for each locus

6 small bristles, vestigial wings and cinnabar

385 cinnabar;

43 small bristles, cinnabar

380 small bristles, vestigial wings

 

 

2.1a (5 pts.)    What are the parental chromosomes in the triple heterozygote?

 

            A.        b+ vg+ cn+ and  b  vg  cn

B.        b+ vg+ cn and  b  vg  cn+

C.        b+ vg cn+ and  b  vg+  cn

D.        b+ vg+ cn and  b  vg+  cn+

E.             Do not have enough information.

 

 

2.1b (5 pts.)    What would be the gene order that would account for the distribution of recombinants?

 

A.             b   vg   cn  

B.        vg  b   cn 

C.        b   cn   vg

D.        cn  vg   b

E.         Do not have enough information.

 

 

 

2.2a. (5 pts.)   Given that RF =0.5 x (1-e-m)  and that the empirically defined recombination frequency between two genes = 16.5%, what is the value of the “ultimate mapping unit”?

 

            A.        1 – 0.55

            B.        1 + 0.33

C.        0.67

D.        0.8

E        0.4

 

 

m

e-m

0.000

1.000

0.100

0.905

0.200

0.819

0.300

0.741

0.400

0.670

0.500

0.607

0.600

0.549

0.700

0.497

0.800

0.450

0.900

0.407

1.000

0.368

 

An example of an empirically defined RF as being 0.165 (16.5%).  plugging this value into the equation  => 0.165 = 0.5 x (1 - e-m )  => e-m =   1 - 0. 330 = 0.67, then to get a value for "m" go to the log table

 

 

2.2b. (5 pts.)   Now, calculate the more accurately determined recombination frequency in centi Morgans?

 

The more accurately determined distance between two loci that do experience crossing over (in cM or map units)   =  50 x m cM.  .......50 * 0.4 = 20cM which is greater than the original value 16.5 cM, that was empirically deduced. 

 

 

 

2.3. (5 pts.) Explain why the maximum discernable distance between two linked genes (according to Morgan) is 50cM.

 

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2.4a. (10 pts.) Give two similarities between a YAC and a bacterial plasmid and three differences -giving a brief explanation as to the importance of each of your choices.

 

 

 

                             

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2.5. (5 pts.) Briefly explain the phenomenon of “ring pucker” and how it might affect DNA structure.

 

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_______________C2’ endo = B-form DNA ______________________________________________

 

_______________C3’ endo = A-form DNA ______________________________________________

 

_______________C2’ endo, C3’ endo  = Z-form DNA ___________________________________________

 

 

 

 

2.6. (6 pts.) Briefly differentiate between A and Z form DNA, giving a specific example of when each has been shown to be present.

 

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2.6. (4 pts.)  Briefly relate two difference between the replicase “holoenzyme” in prokaryotes and the replicase in eukaryotes.