ps 1 answers

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LS4 Problem Set 1, 2010

Complementation testing, fine mapping by recombination, biochemical pathways,reading frames, mutations

Corresponding Lectures: October 30, November 2, 4, 6

Corresponding Reading: Hartwell et al., 224-238, 255-265

Corresponding Quizzes: Nov 2-6 (simple quiz on complementation); Nov 9-13

If you want additional problems, try these from the book: Chapter 7: 16-19, 21, 23,32; Chapter 8: 3-7, 16.

Problem 1. 1. What is the purpose of a complementation test?To determine if two mutations with the same phenotype occur in the same gene or in different genes

2. What is one criterion that must be met in order for a complementation test to be avalid experiment?Mutations must be recessive 3. How do complementation tests differ between animals, bacteria and viruses?Most eukaryotes are diploid or are sometimes diploid as a part of their life cycle(Also accept easier to score phenotypes) 4. Is this an example of complementation or of failure to complement?Complementation

Problem 2. You have 3 turtles, each with a mutation that causes them to have astrange polka dotted pattern on their shells. You don’t know if the mutations are in thesame gene or in different genes. Assume that all 3 mutations are homozygousrecessive.

a) turtle1 X turtle2turtle2 X turtle3

turtle3 X turtle1

b) Turtle3 x turtle2

Complementation, wild type phenotypeTurtle1 x turtle3

No complementation, mutant phenotype



Problem 3. You have performed a mutagenesis screen in Drosophila and have found 5recessive loss-of-function mutants with an essential role in the neural pathway thatallows flies to jump.

You perform a complementation test and get the following results.+ = wildtype, - = non-jumpers

There are three complementation groupsGroup 1: mutations 1, 3, and 4Group 2: mutation 2Group 3: mutations 5 and 6

Problem 4. Four rII- recessive point mutations were discovered. Coinfections of bacteria with allpossible combinations of rII- mutant T4 phages were performed, and the followingresults were obtained.

Fill in the chart

1 2 3 4

1 - + - +

2 - + +

3 - +

4 -

A. How many genes are identified by this analysis?From row 1, we see that mutation 1 fails to complement mutation 3, so they are inthe same gene. Mutations 2 and 4 complement mutation 1, so they are not in thesame gene as mutation 1. 2 and 4 are not necessarily in the same gene, we must

keep looking at the chart to figure that out. All we know so far is that mutations 2and 4 are not in the same gene as mutation 1 and 3.From row 2, we see that mutation 2 complements mutation 4, so they must also bein different genes. B. Which mutants belong to the same complementation groups?1 complementation group (mutations in same gene) – mutations 1 and 3mutation 2 is in a different gene than 4 and (1 and 3)mutation 4 is in a different gene than 2 and (1 and 3)



Problem 5. It’s a very well known fact that most dragons do not breathe fire, and it’s alsowell known that alleles that enable fire breathing are recessive to the wild-type SweetBreath phenotype. However, it’s unclear how many different genes when mutated causethe fire breathing phenotype. To answer this question you use four different strains of dragon and perform a series of crosses. Your strains are the Hungarian Horntail (Strain

A), the Norwegian Ridgeback (Strain B), the Romanian Longhorn (Strain C), and theSwedish Shortsnout (Strain D). All strains are true breeding for a single homozygousrecessive mutation that allows them to breathe fire.

Cross #1 Hungarian Horntail (A) X Norwegian Ridgeback (B) = F1: All Sweet Breath

Cross #2 Romanian Longhorn (C) X Swedish Shortsnout (D) = F1: All Sweet Breath

Cross #3 Swedish Shortsnout (D) X Norwegian Ridgeback (B) = F1: All Firebreathing

Cross #4 Romanian Longhorn (C) X Hungarian Horntail (A) = F1: All Firebreathing

How many genes are involved in creating a firebreathing phenotype? Group thedragons (A, B, C, and D) as to which contain mutations in the different genes.

There are 2 genes, A and C have mutations in the same gene, as do B and D.

Problem 6.

A) Construct a map of the deletions that fits the data in this table1 _________ 2 __________ 3 _______________

4 ______ 5 ___________________

B) Draw a map that includes both the deletion and point mutations.1 _________ 2 __________ 3 _______________ 4 ______ 5 ___________________

c e,d a b

C) Why do d and e point mutants yielded the same result? Which of the following are the

possible explanations for this? Circle all the possible answer.i. They are point mutations in the same nucleotideii. They are not in the same complementation groupiii. They are point mutations in different nucleotides but they both overlap with thesame deletion mutants.iv. It just happened by chance that they yielded the same result



Problem 7. You have four known deletion mutants for a single gene: W, X, Y and Z.Point mutations 1, 2, 3, 4, 5, and 6 are all mutations in the same complementationgroup. Based on the recombination tests shown in the table below, draw map showingthe deletions and point mutations. Please make the map as accurate as possible.

_____________*____*___*__*__________*___*___________

X______ Y_________________

Z_________________

W______

Problem 8.A) Take those strains that did not grow in minimal media and try to grow them inminimal media + lysine. If they now grow, you know that they are auxotrophic for lysine.B) Complementation tests can reveal how many different genes there are.C)

1 3 2 4B > D > A > C > lysine

Problem 9.A)

Compounds supplementing minimal mediaMutant Strains

Ornithine Citrulline Arg-Succ Arginine

arg-F + + +

arg-G + +

arg-H +

arg-F; arg-G + +

arg-F, arg-H +

arg-G, arg-H +arg-F, arg-G, arg-H +

B) You have mixed up the vials of your double and triple mutant strains (which havebeen growing in complete media) and discover that there is one vial that you forgot tolabel. You find that spores from that vial grow on minimal media supplemented witharginine, but not on media supplemented with ornithine, citrulline or arginino-succinate.List all of the possible strains that this could be.1. arg-F, arg-H2. arg-G, arg-H3. arg-F, arg-G, arg-H

C) You cross this mystery strain to the Arg-F single mutant to create a diploid. Thisdiploid grows on minimal media. What was the mystery strain?The strain is arg-G, arg-H

D) If you cross the mystery strain to the Arg-G mutant, which compounds will theresulting diploid be able to grow on?

Arginino-Succinate and Arginine



Problem 10. If you insert an extra nucleotide at position A, what will be the phenotypeof an animal homozygous for this mutation? (Please do not invoke any obscurephenomena or concepts we have not talked about in class—these questions are just asstraight-forward as they appear!)Defective tail, but otherwise wild-type

If you insert a nucleotide at position C, what will be the homozygous phenotype?Wild-type

If you insert a nucleotide at position D, what will be the homozygous phenotype?Infertile

If you insert a nucleotide at position C, and take one away from position D, what will bethe homozygous phenotype?Infertile

If you insert a nucleotide at positions A and C, what will be the homozygous phenotype?Defective tail

B) If you insert a nucleotide at position A, how might you be able to suppress the mutantphenotype at position B?Remove a nucleotide at position B

If you add two nucleotides at position B, name two distinct alterations you could makeat position A that would suppress the phenotype.1. Add a nucleotide at position A 2. Remove two nucleotides at position A

C) You find that you can suppress a mutation at position D by removing a nucleotide atposition E, but not by removing one at position F. What does this tell you about thestructure of the Fertilin gene?

The region between E and F is critical for function, but the region between D and Eis not.



Problem 11.“The cylons were created by man. They rebelled. They evolved. There are many copies. Andthey have a plan…”

Cylon models affectionately known as “skinjobs” are part machine, part organic – they look and

feel human. In order to increase our knowledge about cylons (and perhaps spot sleeper agentshiding among us), Dr. Gaius Baltar has given you DNA samples of known cylons. You run aseries of experiments to learn about cylon gene expression.

1) You discover the cylons do not have a triplet genetic code. Four nucleotides make up acodon in the cylon genome. How many possible cylon codons are there?256

2) Stop codons have the same first three letters as in humans; the fourth is a wobble position(i.e. stop codons = UAAN, UAGN, UGAN). How many stop codons are in the following DNAsequence?

5’ – GTACGTGCCTTATTCGATCGATC – 3’3’ – CATGCACGGAATAAGCTAGCTAG – 5’

One stop codon

3) The following mRNA sequence encodes the polypeptide His-His-Glu-Lys:

CAUGCACGGAAUAAGC

A mutation is induced that encodes the polypeptide His-His-STOP.

What kind of mutation is responsible?

Insertion of U at position 9 or substitution G > U at position 9



Problem 12. While working in the lab, you find six deletion mutants of Neurosporacrassa that are auxotrophic for lysine. The A and α spores for each mutant are matedwith each other to form a dikaryon (a diploid cell) and plated on minimal media.

A) Assuming you made no errors while performing the crosses, fill in the expectedoutcomes of yet to be completed complementation tests. (+ = growth on minimal media,

- = no growth on minimal media)

1 2 3 4 5 6

1 - - + + + -

2 - + + + -

3 - + - +

4 - + +

5 - +

6 -

How many genes are important for lysine production? _____ 3 _______

B) You find four new mutants in the same complementation group that are auxotrophicfor lysine and appear to be deletions, and cross these mutants pair wise to makediploids that are heterozygous for both deletions. You induce recombination in thesemitotically dividing diploid cells by exposing them to UV light, and find that now somecells can grow on minimal media. Construct a map of the four deletions based onresults in the table below. (+ = growth on minimal media, - = no growth on minimalmedia) (4 points)

I II III IV

I - + - +

II + - - -

III - - - +

IV + - + -

IV______

II_________________

III_____________

I______

Problem continues on the next page



C) You were also able to find 6 point mutants (a-f) in the same gene that were alsolysine auxotrophs. You crossed them with the deletions found in part B and plated thediploids on minimal media. The results are shown in the chart below. Construct a mapwith both deletions and point mutations.

a b c d e f

I + + - + + +

II - - + + - +

III - + - - + -

IV + - + + + +

B E A D,F C

IV______

II_________________

III_________________

I______

D) The relative locations of point mutants d and f cannot be distinguished from theexperiments you’ve done. How could you do an experiment, involving recombination, todistinguish the order of d and f on the chromosome? Describe this briefly.(Hint: you’ll have to use two different diploids in this experiment)

Mate point mutations d and f separately to another point mutation and look for therecombination between the new point mutation and d and the new point mutation and f.If the recombination frequencies are the same, d and f are mutations in the samenucleotide. If the recombination frequencies are different, then d and f are mutations in

different nucleotides of within deletion III and you can tell the relative order because theone with the higher recombination frequency is further away from the other mutation.This would also work if you looked for recombination between d and f and another deletion.



Problem 13. You discover a new compound that yeast require for survival. To findgenes involved in the synthesis of this compound (compound X) you have screened for mutants that cannot grow on minimal media, but can grown on minimal media +compound X. You find 10 mutants and perform a complementation tests. The results areshown below (+ = growth, - = no growth).

1 2 3 4 5 6 7 8 9 101 - + + - + + + + + +

2 - - + + + - + + +

3 - + + + - + + +

4 - + + + + + +

5 - - + + - +

6 - + + - +

7 - + + +

8 - + -

9 - +

10 -

A) How many genes are in your mutant collection?

4 genes

B) Which mutations are in the same genes?1, 42, 3, 75, 6, 98, 10

You know that the earliest precursor for compound X is compound Y. There are 5 other compounds that could potentially be involved in the pathway (L, M, N, O, P). Todetermine if, and in what order, these compounds are involved in the pathway you growyour mutants from above on minimal media + one of the compounds. The results arebelow.

L M N O P X Y

1 - - + - - + -

2 - + + - - + -

3 - + + - - + -

4 - - + - - + -

5 - + + - + + -

6 - + + - + + -

7 - + + - - + -8 - - - - - + -

9 - + + - + + -

10 - - - - - + -

Problem continues on the next page



C) Given the results reported in the chart on the previous page, draw the biosyntheticpathway for compound X indicating which mutations correspond to which genes in thepathway.

5,6,9 2,3,7 1,4 8,10

Y P M N X

D) You have constructed several double mutant strains. Indicate whether or not thesedouble mutants would grow on minimal media + the following compounds. (3 points)

L M N O P X Y

1, 2 - - + - - + -

1, 4 - - + - - + -

5, 10 - - - - - + -



Problem 14. On a planet in a galaxy far away you discover life. Proteins on this planetare constructed from 200 unique amino acids; however, DNA on this planet is made upof the same 4 nucleotides as ours.

A) What is the minimum number of bases per codon necessary for this system?

4 (4x4x4x4=256, which is more than enough to make 200 amino acids)

B) Assuming the number of base pairs per codon you calculated in part A is true for theorganisms on this planet, how many possible reading frames are there in doublestranded DNA? In other words, if you don’t know where a start codon is, how manypossible ways could you potentially read a double stranded piece of DNA? (3 points)

8 (4 possible reading frames per strand)

Coding sequence of Gene X

C) Again assuming the number of base pairs per codon in part A is accurate, how manybase pairs would you need to add at position B in the gene above to rescue the insertionof a base pair at position A?

3 (1+3=4, restoring the frame)

D) You find that the precise sequence of the region between C and D is vitally importantfor the function of protein X encoded by this gene. If a base pair is deleted at A, list all of the positions where insertions of nucleotides could rescue the function of the protein B and C, although full credit was given for the answer A, B, and C or “anywhere betweenA and C”.

E) Again assuming the codon length in Part A, if you generated deletions in Gene X,which would be more likely to destroy the function of Protein X? Circle the best answer below.

1. A 12 base pair deletion between A and B

2. A 9 base pair deletion between B and C (This would throw themessasge out-of-frame before the important part of the gene)

3. A 5 base pair deletion upstream of the start codon

4. An 11 base pair deletion between E and the end of the gene

A B C D E

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