Bio 4564/6564                        Advanced Genetics                      Spring, 2022

 

                                                             Test I

Section 1 (38 pts.)

 

Unless stated otherwise each question in this section is worth 2 pts. As directed to do so in the lecture, please give your answer to each of the questions in this section next to the appropriate question number in your Microsoft word written answer that you will submit on-line. Choosing more than one letter in answering any given question will count against you. 

 

1.  Mendel’s 1st law states that:

 

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

B.        When two different unit factors governing the same phenotypical trait occur in the same organism, one of the factors is dominant over the other one, which is called the recessive trait

            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.

            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       

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

 

 

2. Phenylketonuria PKU, is a relatively rare genetic disorder that causes the amino acid, phenylalanine, to build up in the body. It is a recessive trait. The buildup of the amino acid in the body is because the gene responsible for breaking it down is defective.  If a couple who are both carriers for the trait have three children, what is the likelihood (probability) that all three are phenotypically “normal”?

 

            A.        3/64

            B.        6/64

            C.        18/64

            D.        27/64

            E.        64/64

 

 

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

 

            A.        epigenesis.

            B.        non-disjunction.

            C.        overdominance or heterosis.

            D.        epistasis.

            E.        incomplete dysgenesis

F.        epitomyosis

 

4. (4 pts)    Given that non-albino mice, which 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).

 

(b)     What percent of the progeny would you expect to be albino?

 

            A.        0

            B.        12.5

            C.        37.5

            D.        50

            E.        100

 

(a)   What percent of the progeny would you expect to be agouti?

 

A.        0

            B.        12.5

            C.        37.5

            D.        50

            E.        100

 

 

5.         Recessive suppression” is an extension of Mendelian genetics where:

 

A.        each gene suppresses the other

B.        recessive suppression gives rise to a 15:1 ratio of one trait over the other when two strains, heterozygous for each are mated.

            C.        a masking of one of the phenotypes when the suppressor allele is suppressed

D.        recessive suppression gives rise to a 9:7 ratio of one trait over the other when two strains, heterozygous for each are mated

E.        recessive suppression gives rise to a 13:3 ratio of one trait over the other when two strains, heterozygous for each are mated

F          recessive suppression is a condition when one of the alleles is lethal only when the suppressor is present in the recessive state.

 

 

6. Which of the following statements about Mendelian genetics needs 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.

 

7. Given that, in dogs, the two traits of long tails and large canines are due to dominant alleles; and conversely -short tails and small canines are due to the corresponding recessive alleles. A dog that is homozygous dominant for both traits is mated to a short tailed, 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

 

 

8. In a suburb of London (England) called Lower-East Hempstead, a recessive allele at a unique locus on the X chromosome of the men in this community gives rise to a very rare phenomenon called “purple-eye”.   Assuming that this is a closed community, which one of the following answers would best describe the inferred F1 and/or subsequent generational offspring of a purple-eyed male and a phenotypically normal eye-coloured female, if all their F1 offspring were female.

 

      A       All F1 offspring will have purple eye.

      B       All the F2 male offspring will have purple-eye.

      C       All the F1 male offspring would not survive.

      D       Half the F1 female offspring would have purple-eye

      E       One quarter of the couple’s grandchildren could be assumed to have purple-eye

 

 

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

 

            A.        OO or AB.

            D.        AO, AA, or AB

            C.        BB only.

            B.        BB or AB

            E.        impossible to determine.

 

 

10.   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

F.    the recombinatory frequency of a single chiasma occurring between two unlinked genes located on the same chromosome

 

11.   One of the following sentences describes a Type III restriction enzyme.  Please choose the most appropriate description for this type of enzyme.

 

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 normally binds DNA and cuts DNA downstream of a non-palindromic DNA sequence

E.    An enzyme that copies and restricts any palindromic DNA sequence.

 

 

12. In discerning the correct interpretation for phenotypic variations in the yeast asci if two genes are linked then:

 

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

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

            C.        the number of non parental ditypes will equal the number of tetratypes

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

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

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

 

13.      The term “alpha complementation” potentially refers to:

 

A.             the ability of one form of a lacZ gene to complement its counterpart in a heterozygous cross

B.             the flexibility in genetic outcomes from numerous, related crosses of the lacZ gene in E. coli

C.            the presence of specific male traits being exhibited in the male offspring following a heterozygous cross

D.            the phenotypic expression of multiple genes, when expressed, having a synergisitic or “greater effect” than the sum of each individual gene.

E.             The ability of part of a protein from one source to interact with the rest of the protein (from another genetic source) to restore function.

 

 

14.      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

 

15. (4 pts)     Two plasmid vectors are considered 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?

 

A.             Compatible,        do

B.             Multicopied,        do

C.        Incompatible       do

            D.        Compatible,        cannot

            E.        Compatible,        do not

            F.        Incompatible       do not

 

 

16. (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 ultimate consequences of such an event, ie monosomy, trisomy, normal diploid or “other” -with a brief explanation?

 

Trisomy or monosomy for these particular chromosome… as the daughter cells are viable cells and would normally be diploid cells

IF THE ANSWER was “MEIOTIC Nondisunction”..then this would change -for partial credit.

 

 

Section 2 (52 pts.)

 

Please answer the following questions using short, concise answers and as this is effectively a “take home examination”, feel free to expand upon your answers.  Either way, SHOW YOUR WORKING OUT, whenever possible. If you DO NOT, I cannot give any partial credit.

 

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

 

The allele wng of Drosophila (the fruit fly)  gives rise to “wingless”, while the wings of a wng+ fly are completely normal.  The allele b of a completely independent gene gives rise to “small bristles” on the thorax and b+ gives large bristles (the wild-type phenotype).  Finally, 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 that is being test crossed?

 

            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 and which two genes are closer?

 

A.             b   vg   cn  

B.        vg  b   cn 

C.        b   cn   vg

D.        cn  vg   b

     Do not have enough information

 

2.2a. (5 pts.)   Given that RF = ? x (1-e-m), and you remember what “?” equals, the empirically defined recombination frequency between two genes = 16.5%.  what is the value of the “ultimate mapping unit”? SHOW ALL WORK -Credit given for showing work -if wrong!

 

            A.        1 – 0.36

            B.        1 + 0.54

C.        0.67

D.        0.8

E.        0.4

F.        0.2

 

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

 

 

m

e-m

0.000

1.000

0.100

0.905

0.200

0.819

0.300

0.741

0.400

0.67

0.500

0.607

0.600

0.549

0.700

0.497

0.800

0.450

0.900

0.407

1.000

0.368

 

 

2.2b. (5 pts.)   Now, calculate the more accurately determined recombination frequency in centi Morgans, and briefly explain your answer.

 

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. (6 pts) A friend comes to you for some “unofficial” advice in your capacity as an “advanced geneticist”. She tells you that her brother can’t discriminate between reds and greens very well, but both of her parents are able to do so. She then tells you that her boyfriend appears to have a normal colour spectrum, but she would like to know (if they were ever to have children) what the probability would be that any of their children would exhibit red/green colour blindness. Please give your answer with a brief explanation as to how you arrived at that answer.

__X-linked, The friends mother has to be a carrier, thus the friend has a 50 : 50 shot at also being a _

 

_carrier so her male offspring would have a 50:50 shot at getting __________

 

__the X with the red/green colour blindness gene. 

 

   Only boys would show the trait (as the husband has no history)_

 

__so either 0% or 50% of her offspring (none or half  the boys) would have a probability of showing red/green colour blindness ___

 

 

 

2.4. (5 pts.) Give a brief explanation as to two of the roles OR consequences of DNA methylation in prokaryotes that were discussed in the lectures

 

 

Control of replication initiation at the origin of replication,

 

Definition of self-identity, if the methylation is attached to specific bases…

 

 

 

2.5. (6 pts.)    If DCO = 4NPD , and SCO = T – 2NPD

 

What am I talking about?

Explain what all the “alphabet soup” means.  Then explain  to me what NCO equals?

Hint: it is not “Non Commissioned Officer”.

 

___________You are talking about ditypes in yeast where the two genes in question_____________

 

___________are “linked” on the same chromosome________________________________________

 

___________NPD = Non parental ditype, DCO Double cross over____________________________

 

___________T = Tetratype , and ______________________________________________________

 

___________ NCO (No cross over) = PD - NPD _________________________________________

 

 

2.6. (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 ___________________________________________

 

 

 

2.7. (6 pts.) Briefly differentiate between B and Z form DNA, giving a specific example of some of the respective characteristic differences between the two.

 Indicate at least one for Z-form DNA and why this variant might be present in the cell.

 

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B Form is the typical structure of DNA in the cell, while Z-form DNA is sometimes found in “overwound “ or “stressed” DNA.

EXTRA credit if more detail with respect to function or induction of Z-DNA.

 

2.8. (4 pts.)  List  FOUR  discrete proteins or protein complexes that may be found in or around the DNA replication fork of a EUKaryotic cell,

and briefly explain their role in DNA replication.

Be sure to mention if and why they effect “leading strand” replication, “lagging strand” replication or both.  

 

If they are having trouble differentiating between PROKS and EUKS… give some credit for what they do get correct in Prokaryotes! ALSO SSB proteins in both!!!!!

 

Diagram

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