<> 

<> 

<>8.) Concerning the discussion paper: 

<>

<> 
<>

<>       c)    Why does the RNAi construct for the Drosophila dYARS gene not inhibit the human transgene?      

Bring the homework to class (no email attachments)  and type your answers.  If you must write your answers,  please print and make sure ALL letters  are clear.   

4.)  Suppose we isolate two independent mutations conditioning dwarfing in pea.  Each mutant was propagated over 20 years.  Through examination of millions of plants arising from mutant plants, no evidence for reversion to wildtype was noted. In addition, each yields only tall plants when crossed to a wildtype, tall plant.  A cross was made between a dwarf plant from each mutant class.  This yielded 345 dwarf plants and one tall plant.

        A cross was made between two plants of the following genotypes:

T-Sh2, Sh2, – (no Bt2 transgene), Bt2

3.)  Suppose we isolate two independent mutations conditioning dwarfing in pea.  Each mutant was propagated over 20 years.  Through examination of millions of plants arising from mutant plants, no evidence for reversion to wildtype was noted. In addition, each yields only tall plants when crossed to a wildtype, tall plant.  A cross was made between a dwarf plant from each mutant class.  This yielded 345 dwarf plants and one tall plant.

                 

Below are the scores on the first homework.  People did well!!!                                     

 

	8
	8
	8.5
	8.5
	8.5
	8.5
	8.5
	8.5
	8.7
	9
	9
	9.5
	9.5
	10
	10
	10
	10
	10
	10
	10
	10
	10
	10
average	9.248

                                                    

                                                                                     Name:          ___Key_______________________________

Advanced Genetics Fall 08, Homework 1.

Due:  Tuesday September 9^th.   Bring hard copy to class (no email attachments)

 

Question 1.

 

          Two sets of DNA PCR primers (a total of four primers) were used in a single PCR reaction with DNA from each of three inbreds from a hypothetical organism.  (For purposes of this question, assume that each inbred is homozygous for all genes.). 

 

            Inbred A produced a 1.1 kb fragment and a 2.1 kb fragment.  Inbred B yielded a 1.2 kb and a 2.2 kb fragment while inbred C gave rise to a 1.3 and a 2.5 kb fragment.

 

            The hybrid from the cross of inbreds A and B contained four fragments:  1.1 kb, 1.2 kb, 2.1 kb and 2.2 kb.  The hybrid of the A X B cross was crossed to inbred C.  This yielded the following types of progeny with the following molecular markers.  The various classes occurred in equal frequency.

 

 

class    1.1 kb              1.2 kb              1.3 kb              2.1 kb              2.2 kb              2.5 kb

1          yes                                          yes                   yes                                           yes

2          yes                                          yes                                           yes                   yes

3                                 yes                   yes                    yes                                          yes

4                                 yes                   yes                                           yes                   yes

 

For the following combinations, do the markers segregate following the (a) Law of Segregation, (b) the Law of Independent Assortment or (c) it is impossible to discern the pattern from the data given? 

 

a.)   Marker 1.1 with marker 1.2   Law of Segregation

b.)   Marker 1.1 with marker 1.3    Impossible to discern

c.)   Marker 1.1 with marker 2.1    Law of Independent  Assortment

d.)   Marker 1.2 with marker 2.2    Law of Independent  Assortment

e.)   Marker 1.2 with marker 2.5    Impossible to discern

(A note on PCR.  PCR stands for polymerase chain reaction.   It is used to amplify particular regions of DNA.  Normally a set (two) primers are added to DNA. The DNA is heated to denature the DNA (make it single stranded) and then cooled to allow DNA renaturation.   The primers hybridize or anneal to the single stranded DNA.   The mixture also contains DNA polymerize, deoxyribonucleotides and all the goodies to allow DNA synthesis.  This cycle is repeated many times so that for every single copy of a particular sequence that is present in the DNA, one ends up with many fold more copies of that particular sequence.  For example,  after one cycle there are now two copies of each fragment that originally was present in single copy, after two cycles, there are 4, then 8, 16, 32, 64, etc). 

  So in our case, in inbred A, one set of primers gives rise to a 1.1 Kb fragment and the other set gives rise to a 2.1 kb fragment. There is genetic heterogenity for these fragments across inbreds. For those of you who do PCR, click on the following link.  It is hilarious!!!  http://bio-rad.cnpg.com/lsca/videos/ScientistsForBetterPCR/ )

 

Question 2.

 

 

          A collection of six dwarf pea mutants was tested for allelism.  In the following scenarios, the number of crosses yielding tall plants is given.  For each scenario, list the number of complementation groups identified.

 

a)  15 talls; # complementation groups =  __6_
b)  13 talls; # complementation groups =  __4_
c)  11 talls; # complementation groups =  __3_
d)    8 talls; # complementation groups =  __2_
e)    5 talls; # complementation groups =  __2_
f)     1 tall, ; # complementation groups =  _1__   (intracistronic complementation)
g)   no talls; # complementation groups =  __1_

 

Question 3.

 

          Five mutants are known of the waxy (wx) locus of maize.  Pollen containing the mutant gene do not stain with iodine whereas wild type or Wx pollen do stain with iodine. Staining of pollen produced by plants homozygous for each of the five mutants showed that each of the first three mutants produce wild type pollen at a frequency of ~1 X 10^-6 whereas the latter two mutants do not produce wild type pollen.  Given below is the frequency of wild type pollen produced from heterozygotes involving the stated alleles.

 

            Heterozygote                                 Frequency of wild type pollen

            1 with 2                                               ~1 X 10^-3

            1 with 3                                                ~1 X 10^-6

            1 with 4                                                ~1 X 10^-6

            1 with 5                                                ~1 X 10^-6

            2 with 3                                                ~1 X 10^-3

            2 with 4                                                ~1 X 10^-6

            2 with 5                                                ~1 X 10^-6

            3 with 4                                                ~1 X 10^-6                                

            3 with 5                                                ~1 X 10^-6

            4 with 5                                                   0

 

            Sequencing revealed that two of the mutants are partial deletions covering a significant portion of the gene while the other three are point mutations.  Sequencing also revealed that two of the mutants are actually the same point mutation.

 

Taking all data into account, which mutants are deletions, which are point mutants and which of the two point mutants are, in fact, the same mutation?

Deletions are mutants 4 & 5,  point mutants are 1, 2, and 3.  Mutants 1 and 3 are the same.

 

                                                                                   

End of Homework 1

Advanced Genetics.  Exam 1.  Fall 2007
                        September 20, 2007

 

Total points on this exam:  80 points.

 

Question 1.   In the space provided, define the following terms as used in modern genetics (2 points each):

 

   These terms are defined in your notes, so no need to copy/paste here.  Note that controlling elements (last term) is the term Barbara McClintock used to describe her transposable elements.  By definition, it has nothing to do with regulatory genes.

 

a)  Law of Segregation

 

 

b)  Law of Independent Assortment

 

 

c)  Central Dogma of Genetics

 

 

d)  null mutation

 

 

e)  non-reverting mutation

 

 

f)  silent mutation

 

 

g)  frameshift mutation

 

 

h)  missense mutation

 

 

i)  Intracistronic complementation

 

 

j)  Controlling elements

 

 

 

 

 

 

 

 

 

Question 2.  (10 points) One good thing Seymour Benzer did was exploit deletion mutations.  List two advantages and/or uses of deletion mutations.  

 

Because deletions can not revert, wildtypes coming from a cross of two deletions must have been the result of recombination.  Hence the smallest size of the “recon” can be defined in these experiments since the reversion background has been removed.  Because overlapping deletions cannot recombine, but non-overlapping deletions can, a very strict prediction is made concerning the pattern of wildtypes/non-wildtypes in recombination tests if the gene is linear.   Benzer showed that the gene is linear this way.

 

  Secondly because deletions are a finite size, placement of other mutants to a particular area required only crosses to that deletion defining the area.  Hence the number of crosses required went from ~ n²  to n to define the location of a whole series of mutants.

 

            Likely there are other uses/advantages and I graded accordingly.

 

 (Some people just said that deletions don’t revert and define a particular space. The question asks for advantages and uses.  AGAIN, READ THE QUESTION!!! )

 

 

b)

 

 

Question 3.   (10 points)  Two recessive shrunken seed mutants are known in corn.  Self-pollination of the first mutant, m1, yielded 312,587 progeny.  All were shrunken.  Likewise, self-pollination of the second mutant, m2, yielded 352,635 progeny.  All but 20 were shrunken.  Plants from shrunken seed of the two mutants were crossed and 578,439  progeny were examined.  All but 15 were shrunken.  Shrunken F₁ seed were crossed to m1 and 213,000 progeny were examined.  All but 240 were shrunken. 

    For the following statements, state (a) whether the data are compatible with the statement, (b) whether the data are not consistent with the statement or (c) the data do not bear on the statement.  Explain your answer in one sentence and place a letter (a, b, or c) on the line between the parentheses. 

    a) The mutants are not allelic.  (  b )  Data say just the opposite.


    b) The exceptional non-shrunken seeds seen upon selfing of m2 are due to recombination. (   b   )  Again, recombination to produce a wildtype cannot occur in a homozygote.  PLEASE, IF YOU DON’T LEARN ANYTHING ELSE, PLEASE REMEMBER THIS!!!  IT IS NOT THAT DIFFICULT.

    c.) The mutant m2 is a deletion mutation.  (    C  )  It reverts; hence not a deletion
     (Some people talked about pollen/seed contamination being a complication here so I adjusted my grading appropriately. )


    d) The mutant m1 is a deletion mutation. (   a   )  Note the data are compatible with this interpretation; not that they prove this interpretation.  There is a difference.  I also accepted (c) here as well because of this nuance.  (Perhaps the population was not large enough to detect reversion.)



    e.) The majority of the non-shrunken seed seen in the cross of the F₁ with m1 are likely due to recombination   (   a   )

 

 

 

 

 

 

Question 4   (10 points) Concerning the Lac operon, a mutant was found with the following properties: (a) it cannot produce the three enzymes of the Lac operon in the presence or absence of the inducer,  (b)  it is allelic to the i gene, (c) it acts in both cis and trans, and (d) it is dominant to both i- and to i+.   In no more than two sentences and in the space below, provide a molecular explanation. 

 

This is the classic i^s mutant.  It was described in detail in your reading material. The mutant repressor recognizes the operator but cannot recognize the inducer.

 

 

 

 

Question 5  (10 points).  Suppose we obtain a 9 tall to 7 dwarf ratio in an F₂ family for plant height (ie, segregation for two genes).  If we could clone and sequence these genes, which of the following scenarios would you expect:  (i) the genes encode two separate steps in a biochemical pathway, (ii) the two genes encode two separate subunits of one enzyme, (iii) the two genes exhibit significant DNA similarity,   (iv) the segregation data do not allow for any meaningful biochemical predictions.   Circle your answer and explain your rationale below.

 

  The segregation here is classic.   One needs a wildtype allele at each of the two genes to get the phenotype.  Both (i) and (ii) fit the data.  Because the mutant phenotypes here are identical, this has nothing to do with epistatis.

 

 

Question 6. (10 points)  A common observation is that homozygosity for loss-of-function, transposable element-induced alleles in coding regions of genes does not lead to a phenotype differing from the wildtype phenotype.  Provide an explanation.  (10 points)

 

 

I made a big deal of this in lecture because it is very important.  There exists functional redundancy for the vast majority of the genes in an organism, even in the “lower” families.

 

 

Question 7. (10 points)  Kimchi-Sarfaty et al. (Science 315:525) described a “synonymous SNP” that altered the phenotype of the ATP-driven efflux pump MDR1.  Note there are lots of ways to pick up the full 10 points on this question.

 

a)  What is meant by “synonymous SNP”?

 

(2 pt.)  A single nucleotide base change that does not alter the amino acid sequence of the predicted gene product.

 

b) Based upon your reading of the paper by Kimchi-Sarfaty et al. and the commentary by Parmley and Hurst (BioEssays 29:515), briefly describe three different molecular mechanisms by which synonymous SNPs might produce changes in phenotype

 

b-1)  A synonymous SNP that changes a commonly used codon into a rarely used codon (1 pt)

Rare codons potentially change the rate of protein synthesis (1 pt). This can result in aberrant or unusual protein folding (1 pt.) such that the properties of the protein are altered (1 pt).

 

b-2) A synonymous SNP that changes the RNA sequence such that intron splice recognition sites are changed (1 pt).  This can result in failure to splice an intron or in an alternatively spliced intron (1 pt) such that no protein or a different protein is produced (1 pt.)

 

b-3) A synonymous SNP that changes the RNA sequence such that mRNA secondary structures, hairpin loops for example, are stabilized (1 pt). This can impede the translation of the message by the ribosome (1 pt) such that less protein, no protein, or an alternatively folded protein is produced (1 pt).

 

A synonymous SNP that changes the RNA sequence such that the binding patterns of small regulatory RNAs (such as miRNAs) are altered (1 pt). miRNAs generally function as negative regulators of RNA stability or translation (1 pt.) Failure of a miRNA to bind its target could lead to unexpected patterns of gene expression (1 pt).  Alternatively a new miRNA binding site could be created (1 pt.) such that an mRNA is no longer expressed (1 pt).

 

 

1.  (three points) Assume, in our hypothetical situation discussed in class that the Mendel’s dwarf mutant is functionally allelic to the Florida dwarf mutant.  Given this result, we can not formally exclude the possibility that the two mutants are, in fact, the same mutant. Perhaps someone brought Mendel’s mutant to Florida and the mutant was then rediscovered.

  <>            In the space below, outline two fundamentally different, genetic experiments to test whether in fact the two mutants are the same.  Your tools are (1) seed of the two mutants, (2) the ability to make crosses, grow plants and score plant height, (3) the genetic map location of the dwarf  locus and (4) seed stocks of various pea mutants genetically linked to dwarf.  Molecular techniques are not available.  For each experiment, be explicit concerning the possible outcomes and how those outcomes would be interpreted relative to the question in hand.

 
 See homework 1  for the brief answer.  A full answer will give numbers of  progeny to be screened, etc







2.) A series of 10 mutants for plant height  was isolated.  Each mutant is recessive and the mutant phenotype is dwarf.
Non reciprocal crosses were made among the 10 mutants and the following results were obtained.  A "+" means that the
hybrid is tall  whereas a "--" signifies dwarfness.  At least 100 progeny were examined for each cross and all plants of that cross exhibited the same phenotype.
 

      m1 m2 m3 m4 m5 m6 m7 m8 m9
m1
m2  -
m3  -   -
m4  -   -     -
m5  -   -     -    -
m6  -   -     -    -     -
m7   -   -    -    -     -     -
m8   -   -    -   -      +    -    -
m9   -   -    -    -      -    -    -   -
m10 -   -    -    -      -    -    -   -    -
 
 
 

 For each statement below, state whether (a) the data are compatible with the interpretation given, (b) the data are incompatible
with the interpretation given or (c) the data do not bear on the interpretation given.  Explain your answer in the space provided.

 (A)  The phenotype given by the cross of m5 by m8 is likely due to non-allelic non-complementation.  (b)  we see complementation when it is not expected; hence incompatible with  this.

(B) The phenotype given by the cross of m5 by m8 is likely due to the fact that one mutation is polar and is located in a gene upstream from the gene harboring the second mutation.  (c)  Data do not bear on this.  I also accepted  (b)  as well

 

 

 (C) The phenotype given by the cross of m5 by m8 is likely due to intracistronic complementation. (a)  this is exactly what intracistronic complementation looks like.

D).   The  phenotype resulting from the cross of m5 with  m8  but not seen in any of the other crosses is due to the fact
that  both mutants are deletions covering two genes.  (b)  we see complementation when it is not expected by this model; hence incompatible with  this.

(E).   The  phenotype resulting from the cross of m5 with  m8  but not seen in any of the other crosses is due to recombination.  (b)  Since we did not put the two mutants in a heterozygous condition, there was no place for recombination to occur.  Besides all 100 progeny for this cross were wildtype; inconsistenet with recombination.   I became quite sensitized to people telling me that recombination could occur; it would just be a low frequency event.  I took off for this.  Recombination cannot account for this since the two mutants being examined in these plants came from homozygous parents.
 

(F).   The  phenotype resulting from the cross of m5 with  m8  but not seen in any of the other crosses is due to reversion. (b)  If it were 1 or 2 of the 100 scored progeny, then it is possible; not all 100 though.  To get all 100 wildtypes note that BOTH alleles of one of the parents would have to revert.  Think about it. 
 
 
 

 3.)   In a hypothetical (i.e. made up) mutant hunt in E. coli, a mutant was found that expresses all three structural genes of the Lac operon in the absence of the inducer.  The mutation does not map to any of the known genes of the Lac operon.  The mutation was shown to be recessive to the wildtype allele.   Appropriate  tests showed that the wildtype allele of this gene blocked expression of the Lac operon in both the cis and trans arrangement in the absence of the inducer.
 

    Provide a molecular model for the mode of action of this newly described E. coli gene.

  There are several possiibilites:  the wildtype gene could encode a protein that modifies the repressor.  Without modification, the repressor cannot bind to the operator.  It could also be another repressor and both repressors are need to block transcription.  You may have come up with a different/better answer.

4 .)  Assume that a new transposable element system, call Mu-pea  has been described in the pea plant.  Assume that the element has been cloned, clones are available  and its sequence is known.  Assume that in a large population of a cross between a tall plant containing an active  Mu-pea element and  Mendel's dwarf mutant,  a dwarf plant was isolated.   Assuming that the new mutatation  is due to transposition of Mu into the dwarf gene,  outline, in detail, the crosses and molecular experiments you would do to clone the Mendel's  dwarf gene. 

Basically,  a succinct rewriting of the  crosses and experiments in the "Have mutant, want gene" of the notes will be suffice.  Note that the mutant has already been isolated.  Virtually everyone went back and made the mutant.  I don't get it.  Yes, the notes were quite helpful here but it is not asking too much to apply the notes appropriately.  While I didn't take off for this it really irritated me.  Read the question!!!

End of Homework 2,  Fall 07

 

Key

Homework 1 Fall 2007 (modified  8:30 am  August  30)

First, some general comments.  Scores were the following.  Actual score is listed in the first column and the number of homeworks with that score are given to the right.  The average was a little over 6.  In general, I was not pleased with the performance of the class in general.  Question 3 was good in that it allowed people to put their thoughts down on paper.  This exposed some areas that need some work and attention.   I will give the answers below and some points that students need to work on. 

Score	number
10	3
8.5	1
8	3
7	5
6	4
4.5	1
4	2
3.5	1
2.5	1
0	2

 

There are four questions of unequal weight.

1.) (two points) In maize, mutations at two separate genes (as defined functionally) affect a single enzyme that converts ATP and glucose-1-phosphate to ADP-glucose and pyrophosphate. Mutation at either gene gives rise to a loss of ADP-glucose synthesis, in turn, a loss of starch synthesis and, in turn, a shrunken seed. Provide an explanation in the space below as to how two non-allelic genes can control a single enzyme. 

 This was not a good question.  I take the blame.  The actual answer is that the enzyme is a heterotetramer made up of two identical small subunits and two identical large subunits.  Since the enzyme requires a good large subunit and a good small subunit, we need two good structural genes for activity. 

  There are of course other possibilities.  One person said there could be a gene encoding a positive acting regulatory protein needed for transcription and the other was the structural gene.  That works just fine.  One person invoked post-translational modification.  That was fine.  Most simply said that the enzyme was the product of a pathway and nothing more.  I felt  I had to accept this answer given the way the question was written. I didn't really want to, but I felt like I had to.  So, bottom line, this question will go away in the futrue.

 

 

 

2.)   (three points) In celebration of the 100^th anniversary of the University of Florida, the progeny from a cross between the alligator Albert and Alberta was analyzed via molecular markers. DNA from the two parents and 16 offspring was analyzed via PCR analysis. One set of primers was used for DNA amplification from each individual.  Resulting DNA was electrophoresed on agarose gels, separating the DNA by size.  In total 8 DNA fragments were found in the two parents and in the progeny.  Fragments were assignment numbers based on size (1 is the largest and 8 is the smallest). In the table below, a plus means that the animal contained the particular molecular marker. 
 

 Animal                       Molecular Marker
                           1              2              3             4           5           6          7            8
   father                +                                                         +                       +            +
   mother                             +              +            +                        +

offspring
    1                      +            +              +                                                   +
    2                                    +              +                           +                      +
    3                      +                            +            +                                     +
    4                                                    +            +             +                      +
    5                      +            +             +                                                                  +
    6                                    +              +                           +                                    +
    7                       +                           +            +                                                   +
    8                                                    +            +             +                                    +
    9                       +           +                                                        +        +
   10                                   +                                           +           +        +
   11                      +                                         +                          +        +
   12                                                                 +             +           +        +
   13                      +            +                                                       +                        +
   14                                    +                                          +           +                        +
   15                      +                                         +                          +                        +
   16                                                                 +             +           +                        +
 

    For the following pairs of markers, which are segregating in accordance with the Law of Segregation, which are segregating in accordance with the Law of Independent Assortment and for which combinations can you not discern the pattern of inheritance from the data given?
 

The trick here is that one can only interpret markers coming from one parent.  Note that marker 5 and 1 both come from the father and note that when a progeny gets marker 5 it does NOT get marker 1.  So these two follow the Law of Segregation.  Note that marker 5 and marker 7 can both be found in a progeny and neither can be found in a particular offspring.  These markers follow the Law of Independent Assortment.  The same holds true for marker 5 and marker 8.

   Since markers 2, 3 4 and 6 are derived from the maternal parent, we can not surmise a relationship with marker 5.  People either got this or they didn't .  Most did.

Marker 5 versus 1.
Marker 5 versus 2.
Marker 5 versus 3.
Marker 5 versus 4.
Marker 5 versus 6.
Marker 5 versus 7.
Marker 5 versus 8.

 

 

3.  (three points) Assume, in our hypothetical situation discussed in class that the Mendel’s dwarf mutant is functionally allelic to the Florida dwarf mutant.  Given this result, we can not formally exclude the possibility that the two mutants are, in fact, the same mutant. Perhaps someone brought Mendel’s mutant to Florida and the mutant was then rediscovered.

  <>            In the space below, outline two fundamentally different, genetic experiments to test whether in fact the two mutants are the same.  Your tools are (1) seed of the two mutants, (2) the ability to make crosses, grow plants and score plant height, (3) the genetic map location of the dwarf  locus and (4) seed stocks of various pea mutants genetically linked to dwarf.  Molecular techniques are not available.  For each experiment, be explicit concerning the possible outcomes and how those outcomes would be interpreted relative to the question in hand.

 This question really exposed some facts.  While I gave full credit to some people, I really didn't see a perfect answer. 

    The two fundamentally different genetic experiments here are (1) measurement of reversion and (2) measurement of recombination between the two mutants.  Remember our discussions of the lozenge locus of Drosophila.  All one had to do was apply the tools of that study to this question. 

<>    So (1) reversion.   This is simply done by growing lots of progeny of each homozygous mutant and screening progeny for tall plants. If one mutant reverts one will see tall plants.  If it doesn't you won't see tall plants.  So, you would know that the two mutants are different if one reverts and the other one does not.  Of course,  the two mutants might behave in exactly the same way (both revert or both don't revert) even if they are different.  So,  a negative result here doesn't  mean that the mutants are the same. 

     No one really addressed the issue of how many progeny one would grow out to screen or safeguards to detect pollen contamination or seed contamination in producing the large populations that are grown out.  We will address this on the examination. 

 So (2)  recombination.  If the two mutants are different, one might be able to generate wildltype revertants in a heterozygote involving the Florida and the Austria allele. On the other hand, they could be different  and one might still get a negative result here; for example the overlapping deletions as Benzer described.  Recombination cannot occur between them to generate a wildtype because they overlap.  Yet, of course they are different since they delete different segments of the gene.

Experimental approach:   Cross the two mutants, take the hybrid and either self it or cross it onto one of the parents.   Then look for wildtype progeny.  If one finds wildtype progeny and test (1) above showed that neither of the mutations can revert, then we have evidence that the two mutants are different.  If, however, both mutants were shown capable of reverting in test 1, then the wildtype coming from the heterozygote here could be due to reversion  and the two mutants could, in fact, be the same.  If however the frequency of wildtypes coming from the heterozygote is significantly higher than the frequency of reversion measured above and/or a large proportion of the wildtypes coming from the heterozygote are associated with exchange of outside markers, then we have good evidence for recombination (and, in turn, the mutants are indeed different). 

    There were a number of points I found disturbing.

    (1)  People suggested tests of function even though the question stated that the two mutants were allelic in a functional sense.  Either people are not readng the question or don't understand tests of function (or perhaps simply trying to fill space).   I took off for this even if the rest of the answer was OK.   One person told me that if the test of function gave rise to a mutant individual, then the two mutants were in fact the same mutation.  That hurts.

   (2)  I think some people don't appreciate the fact that recombination to produce a wildtype gene requires that the organism producing the recombinant gamete must be heterozygous.  And that one doesn't see the wildtype gamete until one screens the next generation of organisms.  For example, some people said that they would makes lots and lots of crosses between the two mutants and screen the progeny for recombination. If wildtypes are seen in the progeny of this cross, they must be due to reversion (or stray pollen, missplaced seed etc).  They are not due to recombination.  One sees the results of recombination between two alleles in the NEXT generation following heterozygosity;  be it the F2 or the backcross to one of the parents.

(3)  People are not precise and did not walk me through the results of their experiments.  A perfect answer would address the results of both the reversion experiment and the recombination experiment in concert; not independently.  Quantitative data were not listed.  How many progeny need to be screened in both experiments.  How does one safeguard against pollen contamination, seed contamination.  There were facts built into the question for you to work with, but nobody did. 

 

 

4.  (two points) A large hypothetical collection of pea dwarf mutants was assembled.  All possible non-reciprocal crosses were made, resulting in 105 crosses.  Of these crosses 91 resulted in all mutant progeny while the remaining crosses produced progeny that were all wildtype.  How many complementation groups are represented in this collection of mutants?   Do you find this result surprising?  Why or why not

<><>

People seem to get this OK.  There are 15 mutants here.  Complementation occurs only when one mutant is crossed with each of the 14 others. Hence there are two complementation groups.  One group has 14 mutants while the other has one.

  There were a couple of reasons why one might find this surprising.  Some people were not surprised.  I didn't  grade your answer here.

  Some people explained the results above, said there were two genes and 14 complementation groups.  The number of complementation groups does not equal the number of crosses giving  complementation.  Complementation group is equatable with a gene.  I took off 1/2 credit for this.

End of Homework 1,  Fall 07

 

 

 

 

Advanced Genetics  Exam 1.  Fall 2006 -  Key
                         

  <>1.  (20 points) Suppose we isolate two independent mutations conditioning wrinkled seed in pea.   Each yields only round seed when crossed to a wildtype plant. In a large grow-out, over 1 million seed derived from self-pollination of each mutant were scored and all were found to be mutant. A cross was made between a series of plants from each mutant class.  This yielded  F₁ seed composed of  447, 435 wrinkled seed  and one round seed.  

    a)  The round seed is most likely due to (i) a rare cross over event occurring in one of the parents,   (ii) a rare crossover event occurring in the F₁ plant, (iii) rare intracistronic complementation occurring between the two mutants, (iv) epistasis or (v) none of the above.    Circle your answer and explain your reasoning below.  (You may circle more than one answer)   Neither mutant can revert, nor could recombination give rise to this since the gametes here come from the two parents.  While intracistronic complementation is rare, if it occurs between two alleles it always occurs, so not (i).   The phenotype is not like one of the parents so not epistasis.  (v) is the correct answer.



  <>    b)  Mutant  F₁  plants mentioned above were grown and crossed to one of the two mutant parents, resulting in 358,556 progeny.   Of these all but 97 were mutant.  The cause of the 97 wildtype seeds is most likely due to (i) a rare cross over event occurring in one of the parents,   (ii) a rare crossover event occurring in the F₁ plant, (iii) reversion of a mutant allele(s) (iv) rare intracistronic complementation occurring between the two mutants,  (v) epistasis or (vi) none of the above.    Circle your answer and explain your reasoning below.  (You may circle more than one answer)  Answer is (ii).  Neither mutant can revert, crossing over to give wildtypes does not occur in either parent.  Rest is obvious.






    c)  Based on all the observations in a) and b) above, the two mutants (i) are in separate functional units, (ii) are in the same functional unit , (iii) could, in fact, be the  same mutation, or (iv) none of the above.   Circle your answer and explain your reasoning below.  (You may circle more than one answer).

 They are in the same functional unit since virtually all progeny are mutant; they are not the same since recombination can generate wildtype.

 

2.  (10 points) As we know, an allelism test is normally used to determine whether two mutants are in the same functional unit (a structural gene plus any cis-dominant regulatory information).  While the allelism test can be used with confidence, three phenomena, all rare in occurrence,   lead to the wrong conclusion being drawn from this test.  What are they?

1.)  Non-allelic non-complementation
2.) Polar mutants
3) Intracistronic complmentation. 

 

          

 

            d) Of these three, which is clearly evident if the cis test (along with the trans test) is also performed? 

 1.)  Non-allelic non-complementation

 

3.  (10 points) A hypothetical mutant search for suppressors of O^c was performed in  E. coli.   A strain containing the O^c mutant was mutagenized and the resulting bacteria were screened for mutants that did not produce the enzymes of the Lac operon in the absence of the inducer.  Two non-allelic suppressors were obtained.  One mutant (termed mutant 1) produced the three enzymes of the Lac operon in the presence of the inducer whereas the other mutant (mutant 2) did not.  Subsequent analysis showed the suppressor mutants were actually alleles of the i gene and the P gene of the Lac operon.   Which is which and in the space below, explain your answer.  

  Mutant 1 could be an i mutant  that produces a protein that now recognizes the mutant operator allele.  Mutant 2 is a non-functional p mutant
<>4.  (10 points) Why are virtually all loss-of-function mutants recessive?
<Because  there is a huge buffering in   gene product such that 50%  of the gene product  produces the wildtype phenotype.

5.  (10 points) Associate the phenomena below with the people to the right

 

i. Transposable elements __McClintock _                                        a. Benzer

 

ii. Lac operon _Jacob and Monod

 

__                                                        b. Crick

 

iii. DNA structure_Watson and Crick_                                                   c. Jacob and Monod

 

iv. Dominance__Mendel__                                                       d. Mendel                                            

 

v. Recon __Benzer__                                                              e. Robertson

 

vi. Maize Mu  _Robertson___                                                      f. Watson and Crick

 

vii. Independent Assortment  __Mendel __                                g. McClintock

 

viii Cistron   _Benzer___

 

ix. Law of segregation  __Mendel ___

 

x. Central dogma of genetics  _Crick___

 

6.  (10 points) In the space provided, what is meant by forward genetics and what is meant by reverse genetics? 

 Forward:  have mutant, want gene:  reverse:  have gene want mutant.

 

7.  (10 points) In a hypothetical population of a swamp animal with the scientific name, A.  mississippiensis and termed here “Fighting Gators”,  tongue color is normally black. However, two variants were found in this population, brown and colorless.  Appropriate genetic crosses showed that these variants were caused by a recessive allele. A cross between a Fighting Gator with a colorless tongue and one with a brown tongue yields all progeny with black tongues.  A double mutant was made between the two mutants.  Tongues of such animals were colorless. 

 

Which of the following is likely true?   

 a.)  In terms of the biochemical pathway, the product of the gene whose recessive allele leads to colorless tongue comes before the product of the gene whose recessive allele conditions brown tongue.    not always true.

 

b.)   In terms of the biochemical pathway, the product of the gene whose recessive allele leads to colorless tongue comes after the product of the gene whose recessive allele conditions brown tongue.   not always true.

c.)  The gene whose recessive allele conditions colorless tongue is epistatic to the gene whose recessive allele conditions brown tongue.

     True.

d.)  The gene whose recessive allele conditions brown tongue is epistatic to the gene whose recessive allele conditions colorless tongue.

 False

e)  None of the above.

Homework 2,  Fall 2006.  Each question is worth two points.

Question 1.  A cistron is defined genetically as a relationship between two mutants.  In the space below, biochemically what does a cistron correspond to?

 

Question 2)  Assume we have a polar mutation in the z gene of the Lac operon.  Assume we genetically introduce a non-sense suppressor.  We then note that functions of the downstream y and a genes are restored, but beta-galactosidase arising from the z gene is not enhanced.  Provide an explanation in the space below.

 

<>Question 3) (two points)  In a hypothetical (i.e. made up) experiment involving swine (pigs), molecular markers were used to analyze the DNA of the parents and offspring of a cross.  DNA was extracted, cut with a restriction enzyme, electrophoresed and blotted to a nylon support.  A cDNA probe, different from either of the ones used in the experiment of Homework 1, was used to probe the nylon membrane.  In each of the parents and in each of the 16 offspring, four DNA fragments were found to hybridize to the probes. In total, 8 fragments hybridized to the probes.  In the table below, these are referred to as fragments 1 to 8.   A plus means that the animal contained the particular marker.  

 Animal                       Molecular Marker
                           1              2              3             4           5           6          7              8
   father                +                                                         +                       +            +
   mother                             +              +            +                         +
offspring
      1                    +            +                             +            +                                      
      2                    +                             +                           +           +         
      3                                   +                            +                                    +               +
       4                                                   +                                        +       +               +
     
      5                    +                             +                           +           +

      6                    +            +                             +            +          

      7                                                   +                                         +       +               +

      8                    +            +                             +            +

      9                    +                             +                           +           +   

    10                                                    +                                        +       +               +

    11                      +                          +                           +           +

    12                                     +                            +                                    +               +

    13                      +            +                             +            +          

    14                                     +                            +                                    +               +

    15                                     +                            +                                    +               +

    16                                                    +                                         +       +               +

 
    For the following pairs of markers, which are segregating in accordance with the Law of Segregation, which are segregating in accordance with the Law of Independent Assortment and for which combinations can you not discern the pattern of inheritance from the data given?
 

Marker 6 versus 1. 
Marker 6 versus 2    
Marker 6 versus 3.  
Marker 6 versus 4. 
Marker 6 versus 5. 
Marker 6 versus 7.  
Marker 6 versus 8.  

  <>

 <>4.)  Two phenotypically-identical and recessive mutants were crossed in a diploid organism.  Resulting progeny were mutant in phenotype. To validate the positive allelism test, the phenotypically mutant hybrid was crossed to each of the two parents.  Unexpectedly,  each of the two resulting backcross progenies segregated three mutant to one wildtype. 

<> <> Provide in the space below a possible explanation to account for the unexpected backcross data.    <>

5.)  Assume you are furnished heterozygous seed which contain two alleles of your favorite gene.  One allele is due to a cloned transposable element while the other allele is of unknown origin.  In the space provided below, describe how you might clone your gene of interest.

 

End of home work 2

Homework 1,  Fall 2006  

 

<>Question 1. (two points)   Aminoacyl-tRNA synthetases are the enzymes that recognize both the specific tRNA and cognate amino acid to “activate” or load the tRNA with an amino acid before translation.  For proper reading of the genetic code, aminoacyl-tRNA synthetases must recognize the correct tRNA and the correct amino acid.  Suppose we isolate and engineer the gene that encodes the enzyme that normally recognizes leucine and alter the binding site such that the enzyme now recognizes both leucine and isoleucine with equal efficiency. Would you expect this mutant to behave as a dominant or recessive mutation? Explain your answer. (Rationale for your position rather than your position is important here)

   

 

 

 

            Assuming we can turn this mutant gene on and off at will in Drosophila (technology exists to do this), describe what the phenotype of this mutant gene might be. Be as specific as the data allow.  (Rationale for your position rather than your position is important here) 

 

 

 

Question 2. (three points)    As discussed in lecture, assume you perform all non-reciprocal crosses in a diploid organism among a set of phenotypically-indistinguishable, recessive mutants.   This totals 10 crosses.  In every cross, all resulting individuals have the same phenotype. How many complementation groups would be represented in this collection of mutants if you observed the following numbers of crosses yielding only mutant individuals?

 Some of these are straightforward, some of very difficult, perhaps impossible.

 # crosses yielding mutants                  # complementation groups

   0                                                              

  1                                                               

  2                                                               

  3                                                                 

  4                                                                

  5

  6                                                                 

  7

  8

  9                                                                   

10                                                                    1

 

 

 

 

Question 3 (three points).  In the hypothetical case below, a gene is defined by a series of alleles: a wildtype allele designated M1 and a series of recessive alleles termed m1-a, m1-b, m1-c, m1-d and m1-e. Classification is based solely on allelism tests.   In a series of large scale experiments, it was found that homozygotes involving each of the recessive alleles, m1-a, m1-b, m1-d and m1-e produced wildtype gametes at a frequency of approximately 1 in 100,000 to 1 in 1,000,000.  The mutant m1-c was found not to revert.

            Certain heterozygotes of mutant alleles did not produce wildtype progeny above the frequency of reversion.  These heterozygotes were the following: a with c, a with d, b with c and c with d.   <>

Crosses listed below did yield wildtypes at frequencies 10 to 100 fold above reversion.  Crosses with outside marker arrangements are listed.  Outside markers are designated the Z gene and the Y gene, each with contrasting alleles. Also given is the outside marker arrangement of the most prevalent crossover recombinational class.  

Heterozygote                                                    Prevalent crossover outside marker class of wildtypes

 

Y       m1-a             Z                                           y    Z

y        m1-b             z

 

Y       m1-a             Z                                           y    Z

y        m1-e             z

 

 

Y       m1-b             Z                                           Y   z

y        m1-d             z

 

 

Y       m1-b             Z                                          y    Z

y        m1-e             z

 

Y       m1-d             Z                                          y    Z

y        m1-e             z

 

Y       m1-c             Z                                          y    Z

y        m1-e             z

 

 

Question a) Based on the data above, order the position of the mutant sites within the M gene.  Be as specific as the data allow. 

Question b)  After performing all the crosses and analyses, data from another laboratory suggested that two of the five mutants above might, in fact, be the same mutation. Are the data above consistent with that and, if so, which two mutants might, in fact, be the same lesion? 

 

 

 

 

Question 4) (two points)  In a hypothetical (i.e. made up) experiment involving swine (pigs), molecular markers were used to analyze the DNA of the parents and offspring of a cross.  DNA was extracted, cut with a restriction enzyme, electrophoresed and blotted to a nylon support.  Two cDNA probes from two different genes were labeled, mixed and then used to probe the nylon membrane.  In each of the parents and in each of the 16 offspring, four DNA fragments were found to hybridize to the probes. In total, 8 fragments hybridized to the probes.  In the table  below, these are referred to as fragments 1 to 8.   A plus means that the animal contained the particular marker.
 





 Animal                       Molecular Marker
                           1              2              3             4           5           6          7            8
   father                +                                                         +                       +            +
   mother                             +              +            +                         +
offspring
    1                      +            +              +                                                   +
    2                                    +              +                           +                      +
    3                      +                            +            +                                     +
    4                                                     +           +             +                      +
    5                       +            +              +                                                                +
    6                                     +              +                          +                                    +
    7                        +                           +            +                                                  +
    8                                                     +            +            +                                    +
    9                        +          +                                                        +        +
   10                                   +                                            +          +        +
   11                      +                                          +                         +        +
   12                                                                  +             +          +        +
   13                       +           +                                                       +                       +
   14                                    +                                            +         +                        +
   15                       +                                          +                        +                        +
   16                                                                    +            +         +                        +
 
    For the following pairs of markers, which are segregating in accordance with the Law of Segregation, which are segregating in accordance with the Law of Independent Assortment and for which combinations can you not discern the pattern of inheritance from the data given?
 
Marker 6 versus 1.   
Marker 6 versus 2    
Marker 6 versus 3.
Marker 6 versus 4.  
Marker 6 versus 5. 
Marker 6 versus 7.
Marker 6 versus 8. 

 

1.)  As we all know, Mendel is considered the father of Genetics. One fact that Mendel did not discover was that recombination can occur within a gene (a gene as defined as a unit of function).  In the space provided:

     a.)  Why do you think Mendel did not observe this (5 points)?

    Mendel worked with only one mutant allele per gene.  Intragenic recombination is most easily seen as wildtypes arising from a heteroallelic combination of two mutant alleles. 
     b.)  Which of Mendel’s two Laws does intragenic recombination violate (5 points)? 
    Law of Segregation

2.) Define the following types of mutations in the space provided (2 points each):

    a.)polar: affects gene down stream, but not upstream from the mutant gene.
 
    b.) frame shift: shifts the reading frame such that new codons are made and old ones are destroyed.

     c.) nonsense: changes a codon such that it encodes chain termination of protein synthesis.

     d.) missense:   changes from one amino acid to another amino acid

     e.) null:  No residual gene function remains.

    f.) leaky: Partial gene function remains


3.  A hypothetical mutant search for suppressors of O c  was performed in  E. coli.   A strain containing the  O c mutant was mutagenized and the resulting bacteria were screened for mutants that did not produce the enzymes of the Lac operon in the absence of the inducer.  Two non-allelic suppressors were obtained.  One mutant (termed mutant 1) produced the three enzymes of the Lac operon in the presence of the inducer whereas the other mutant (mutant 2) did not.  Subsequent analysis showed the suppressor mutants were actually alleles of the i gene and the P gene of the Lac operon. 


    a.)Which do you think  is which (2 points)?  Mutant 1 is in the i gene while the second mutation is in the promoter.


    b.)Explain your answer in the space provided  (4 points). Mutant 2 does not make enzyme in the presence or absence of the inducer.  This is the classic phenotype of p mutants, and accordingly, straightforward.  Mutant 1 is the challenge.  Here one has to invoke the synthesis of a mutant repressor that now recognizes the   Oc operator and shuts down transcription in the absence of the inducer.






 4.)  A series of 10 mutants for anthocyanin synthesis was isolated.  Each mutant is recessive and the mutant phenotype is colorless.  Non-reciprocal crosses were made among the 10 mutants and the following results were obtained. In each case, 100 progeny were scored from each cross and all 100 had the same phenotype. A "+" means that the hybrid synthesized anthocyanin whereas a "--" signifies no pigment synthesis. 


         m1    m2      m3      m4      m5     m6     m7      m8      m9       m10
m1                                       
m2    +                                   
m3    +       +                               
m4    +       +        +                           
m5    +       +        +        +                       
m6    +       +        +        +        -                   
m7    +       +        +        +        +         +               
m8    +       +        +        +        -          -       +           
m9    +       +        +        +        +         +        +        +       
m10   +      +        +        +        +         +        +        +        +   


    For each statement below, state whether (a) the data are compatible with the interpretation given, (b) the data are incompatible with the interpretation given or (c) the data do not bear on the interpretation given.  Explain your answer in the space provided.

 (A) The phenotype given by the cross of m1 with m2 shows that the mutants are allelic (3 points).  (b), the data say just the opposite.  The phenotype is wildtype, non-allelic.




 (B) The phenotype given by the cross of m1 with m2 shows that the mutants are not allelic (3 points).   (a), straightforward.

 



(C) The phenotype given by the cross of m1 with m2 shows that both  mutants are point mutants (3 points).   C) The data don't bear either way on this interpretation.  



(D).   The  phenotype resulting from the cross of m5  with  m8  is due to the fact that  both mutants are deletions covering two genes (3 points).  I don't know if I love this question or hate it. The data fit with how two deletions covering two genes would behave.  Yet, there is no pattern in the data to suggest such a complicated interpretation.  One could not publish this as evidence for two genes.  The bottom line is that  I gave full credit for either a, b, or c  depending totally on the stated rationale.  Hence, that makes it a good question.   Welcome to the reality of biology!!!




(E).   The  phenotype resulting from the cross of m5 with  m8  is due to recombination (3 points).
    (b), there was no place where recombination could occur to generate non-parental types.  Besides, there is no reason to invoke recombination, since only parental types are recovered.


(F).   The  phenotype resulting from the cross of m5 with  m8  is due to reversion (3 points).

(b) Again, we get parental types.  No need to invoke reversion.  And reversion would not affect 100 of 100 progeny.



(G) The data set, in total, defines _8_  complementation groups (3 points).  








5.)  Suppose we isolate two independent mutations conditioning wrinkled seed  in pea.   Each yields only round seed  when crossed to a wildtype plant. In a large grow-out, over 1 million seed derived from self-pollination of each mutant were scored and all were found to be mutant. A cross was made between a series of plants from each mutant class.  This yielded  F1 seed composed of  447, 435 wrinkled seed  and one round seed.  

    a)  The round seed  is most likely due to  (i) pollen or seed contamination,  (ii) a rare cross over event occurring in one of the parents,  (iii) a rare crossover event occurring in the F1 plant, (iv) rare intracistronic complementation occurring between  the two mutants or (v) none of the above.    Circle your answer and explain your reasoning below (5 points).  (You may circle more than one answer)

Recombination in either parent will not produce a wildtype, recombination in the F1 is seen in the phenotype of the progeny, while intracistronic complementation is rare (only seen in crosses involving only a few alleles), when it occurs, it will occur every time the two alleles are crossed.


    b)  Mutant  F1  plants mentioned above were grown and crossed to one of the two mutant parents, resulting in 445,233 progeny.   Of these all but 111 were mutant.  The cause of the 111 wildtype seeds is most likely due to (i) recombination occurring in the F1 plant , (ii) reversion of a mutant allele(s),   (iv) interallelic (intracistronic) complementation or (v) none of the above. Circle your answer and explain your reasoning below (5 points).  (You may circle more than one answer) 

The alleles don't revert.  See above for interallelic complementation






    c)  Based on all the observations in a) and b) above, the two mutants (i) are in separate functional units, (ii) are in the same functional unit , (iii) could, in fact, be the  same mutation, or (iv) none of the above.   Circle your answer and explain your reasoning below (5 points).  (You may circle more than one answer). 


The phenotype of the hybrid is mutant, hence ii.  They are not the same mutation, otherwise we wouldn't see wildtypes via recombination.







6) In the space below, how did the development/advent/invention of TAIL-PCR advance the use of transposable elements in cloning genes (6  points)?

 Because the one primer is degenerate, the technique allows isolation of transposable element-gene junction sequences without having to know the sequence of the gene. 

7) In the paper discussed in class by Lolle et al., what was the evidence the investigators pointed to  for a critical role of RNA in the high reversion frequency (10 points)?

The template must be exact and it is not DNA. Hence, it must be RNA.   It is a rather weak argument, but there does not seem to be a better one.

 END OF EXAM                                                                                                                                

Name:

1)  (modified 9/12, 12:50 pm) A recessive mutant of the T gene was recently (hypothetically) isolated. The mutant is termed "t".  The T gene encodes a homotetrameric enzyme  (four identical subunits).  Biochemical evidence showed that  any enzyme molecule containing at least one subunit from the "t" mutant was enzymatically inactive (dead).  This is a diploid organism and the enzyme is expressed in diploid tissue.  Assuming that the wildtype "T" allele and the mutant "t" allele are expressed in equal amounts and assuming that the encoded subunits  (T and t) can polymerize at random to form a tetramer, what is the  amount of enzyme activity in a heterozygote containing this allele and a wildtype allele that conditions a wildtype phenotype?  Show your answer and rationale below.  (3 points)

  Only enzymes composed of four wildtype subunits are active.  These occur with a frequency of 0.5 to the fourth power or 6.25%











2)  The following is hypothetical. A subset of a large collection of starch mutants of maize was studied.  All are recessive and all produce a  "shrunken" kernel phenotype in homozygous condition.  One complementation group on chromosome 3 contains 12 allelic recessive mutations.  Another collection of 22 allelic mutations resides on chromosome 4.  Unexpectedly, a cross of a chromosome 3 homozygous recessive mutant with a homozygous chromosome 4 recessive mutant gives rise to all mutant seed.  Provide, in the space below,  an explanation for this unexpected observation.  (1 point)

    This is due to non-allelic non complementation.












3)  The following is hypothetical. Two corn seeds isolated from the middle east were brought illegally (just kidding)  through customs and back to the states.  These were grown, leaf samples were taken for DNA extraction and the resulting plants were crossed.  Resulting plants were grown and DNA was extracted and digested.  A Southern blot of DNA from the parents and progeny  was probed with  full length cDNA clone of a recently described enzyme WMD vanishase.  Eight fragments of varying size were detected in total.  The largest fragment, slowest in mobility was given the number 1; the smallest, number 8.  One of the parental plants contained bands 1,3, 7 and 8 while the other parent exhibited bands 2, 4, 5 and 6.   Among the 200 progeny  screened, four patterns were seen.   These occurred in approximately equal frequency.  These were 1,2,5 and 7; 1,4,6 and 7; 2,3,5 and 8; and 3,4,6 and 8. 

Provide a defensible interpretation of the data.  (2 points)   Either there is one gene with an internal restriction site or two very closely duplicate genes.   Note that fragments 1 and 7 are always inherited together, fragments 3 and 8, fragments 2 and 5 and fragments 4 and 6.  The simplest explanation is that there is one gene.  This gene contains one internal site for the restriction enzyme used.  The restriction sites neighboring the internal sites are polymorphic.

    The other explanation is that there are two duplicate copies of this gene and the two copies are quite closely linked.  This is not too far fetched since lots of duplications involve slippage of paired homologous chromosomes, and one ends up with closely linked duplicate genes.



4.)  Benzer described a class of  r_II  mutants that each of which lacked function of both the A cistron and the B cistron.   Provide an explanation for this class of mutants in the space below.  (1 point)  Because the A and B cistrons are closely linked, some single deletions take out both genes.





5.)  Assume that Investigator Genejock is attempting to clone the maize gene, shrunken-4.    She successfully isolated shrunken seed from a cross of a stock containing the functional, Sh4 allele in homozygous condition and an active  Robertson Mu element  with a stock containing a recessive sh4 allele in homozygous condition.  Describe, in the space below, the genetic experiments that would be done to allow  detection of cosegregation of the  presumed sh4-TE allele with a  Mu-hybridizing fragment on a Southern blot.  (3 points)

  This is basically a repeat of the notes, starting at Cross 2 (Have mutant, want gene) of the newly created mutant to wildtype and then the five crosses to the original mutant.   I don't normally like  "repeat" type answers, but the importance of this point "forced" me into writing this.























 

End of Homework 2

  1.)  In the year 2056 a space ship returned from Mars with a mutant form of  E. coli.    In the presence of lactose this mutant did not produce the three enzymes of the Lac  operon.  Mapping showed that  the mutant gene did not map to i, P or O.     Cis - trans tests showed that the mutant was recessive  and that the wildtype allele of this gene acted in both orientations.  Sequencing of the wildtype gene showed that it contained an open reading frame.   In the space below, provide a model for how the wildtype gene might control expression of the Lac operon.  

 The gene encodes something needed for expression.  It could be a transcription factor, something that binds to the inducer that then allows binding to the repressor, allows the inducer to enter the cell, etc.  There are a bunch of things that could work.


2.)    A hypothetical  (i.e. made up)  major protein (Pro⁺) is known in corn which is needed to make a plump seed.  The protein can be seen after electrophoresis.            Two electrophoretic variants of this protein are known in wildtype; one migrates  4.2 cm whereas, under the same conditions, the other form migrates 6.8 cm.  Maize lines homozygous for each variant were mutagenized with a mutagen that creates primarily small deletions.  Non-reverting mutants of each variant were found.  These mutants gave rise to non-plump seed and lacked a detectable protein.   A heterozygote, flanked by heterozygous and closely linked outside markers (A/a & B/b), between one variant from each allele was made and crossed to one of the parents:



                                     A                             pro-4.2                                     B              
                                    a                             pro-6.8                                      b    


        Among  the progeny, rare plump kernels were found.  Genetic characterization showed that there was only one class of recombinant outside markers in these rare wildtype progeny .  Genetically these were  a  Pro ⁺   B.    Electrophoretic characterization showed that all of these rare progeny had a wildtype protein migrating at  4.2  cm.    In the space below, draw the Pro locus with placement of the electrophoretic site and the mutant sites in two pro alleles.



                                    A        X                   pro-4.2                                     B              
                                    a                           X pro-6.8                                      b    


Order, from left to right, is knock-out mutantion in the pro-4.2 allele,  knock-out in the other allele and then the electrophoretic  heteroallelic site.  Note that there could be some distance between the second and third sites mentioned above.


   
3.)  Returning to Homework 1,   hopefully you figured out that it was impossible to determine whether markers 1 and 4 were allelic or non-allelic from the data given.  In the space below, describe a genetic  experiment to determine whether these are alleles.  An acceptable answer will involve a specific listing of progeny to cross and a prediction of the Southern banding pattern seen in their offspring.

  Cross progeny with both markers with one with no markers.  If, in the resulting progeny, one get only one or the other marker, then they are allelic. If one gets progeny with both markers or no markers, then non-allelic.



4.) There exist (really) two genes in corn on chromosome 3 termed A1 and Sh2.  The former gene (named for its mutant phenotype of no pigment or anthocyaninless)  and the latter gene (named for its mutant phenotype, shrunken) are 0.2 map units (centimorgans) apart.  Suppose you are assigned the job of cloning the A1 gene via transposon tagging.  Given (1) a stock with an active Robertson's Mutator and containing functional alleles at the two genes mentioned above and (2) a stock that contains in homozygous condition mutant alleles of both sh2 and a1, describe in no more than one page how you would go about this.   Be as specific as one page allows you.


This was basically a repeat of the notes with the addition of a closely linked outside marker to aid in identifying the newly induced pigment allele.  The marker was good news/bad news.  While I was glad to see that some of you used it (I didn't take off if you didn't), everyone who used it lost points, as described.  Cross 1 was straightforward to identify and isolate the mutant.  Cross 2 to a wildtype was fine.  At this point, some people wanted to self to distinguish between the stock with the a1-Ref allele (it would segregate 25% shrunken colorless seed) versus the one with the newly created allele (no segregating shrunken seed), they then wanted to make DNA from the pigmented seed and the colorless seed, do the Southern and look for a band in the mutant that was not in the wildtype.  What was forgotten was that 2/3 of the wildtype seed in a self would also contain the mutant allele and hence have the