An Introduction To Genetic Analysis 11th Edition By Griffiths – Test Bank

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Chapter 5

The Genetics of Bacteria and Their Viruses

MULTIPLE-CHOICE QUESTIONS

Section 5.1 (Working with microorganisms)

1. A bacterium that is unable to grow on minimal medium is said to be:

A) allotrophic.

B) autotrophic.

C) auxotrophic.

D) heterotrophic.

E) prototrophic.

Answer: C

2. A gal^– mutant:

A) can make its own galactose.

B) cannot grow without galactose.

C) cannot utilize galactose as a carbon source.

D) can utilize galactose as a carbon source.

E) is resistant to galactose.

Answer: D

3. A Str^R mutant:

A) can grow in the presence of streptomycin.

B) can make its own streptomycin.

C) cannot grow in the presence of streptomycin.

D) cannot grow without streptomycin. 

E) cannot make its own streptomycin.

Answer: A

4. A strain of E. coli has the genotype (and phenotype) leu^– lac^–. This notation means that this strain if bacteria is unable to:

A) metabolize the amino acid leucine and synthesize the sugar lactose.

B) utilize the amino acid leucine and synthesize the sugar lactose.

C) utilize the amino acid leucine and the sugar lactose.

D) synthesize the amino acid leucine and utilize the sugar lactose.

E) synthesize the amino acid leucine and the sugar lactose.

Answer: D

5. A strain of E. coli with the genotype arg^–bio^–gal^–Tet^R is able to grow on minimal medium (containing glucose as a carbon source) supplemented with:

A) the amino acid arginine and the antibiotic tetracycline.

B) the amino acid arginine and the vitamin biotin.

C) the amino acid arginine and the sugar galactose.

D) the vitamin biotin and the sugar galactose.

E) the sugar galactose and the antibiotic tetracycline.

Answer: B

6. A bacterial strain is able to grow on rich medium (which contains all the amino acids, vitamins, and a variety of carbon sources) but not on minimal medium or on minimal medium supplemented with leucine. What could be the genotype(s) of this bacterial strain with respect to the arg and leu loci?

A) arg⁺leu^–

B) arg^–leu⁺

C) arg⁺leu^– or arg^–leu⁺

D) arg^–leu^– or arg⁺leu^–

E) arg^–leu^– or arg^–leu⁺

Answer: E

7. An E. coli culture is growing in (rich) liquid medium. A representative sample of this culture is plated on solid rich medium, and 950 colonies grow on this medium. A sample of bacteria from each of these colonies is then plated on minimal medium, and only 108 of these samples grow into a colony. What proportion of the bacteria in the original liquid culture are auxotrophic?

A) 0.102

B) 0.114

C) 0.796

D) 0.886

E) 0.898

Answer: D

8. An E. coli colony grew on minimal medium supplemented with arginine and leucine. However, bacteria from this colony are unable to grow and form colonies on minimal medium supplemented with arginine and methionine. What is the genotype of the bacteria in this E. coli colony?

A) arg⁺leu⁺met^–

B) arg⁺leu^–met⁺

C) arg⁺leu^–met^–

D) arg^–leu⁺met⁺

E) arg^–leu^–met^–

Answer: B

9. The table below summarizes the results of a replica-plating experiment where a mixture of 347 E. coli strains of different genotypes were tested for their ability to grow on a variety of media.

Medium	Number of colonies that grew
complete medium (contains all amino acids, vitamins, and a variety of carbon sources)	347
minimal medium (glucose as a carbon source)	261
minimal medium + ampicillin (an antibiotic)	51
minimal medium + methionine (an amino acid)	303
minimal medium + arginine (an amino acid)	301
minimal medium + methionine + ampicillin	57
minimal medium + arginine + ampicillin	64

a) Based on the data presented above, how many colonies are wild type (met⁺ arg⁺Amp^S)?

A) 51

B) 210

C) 261

D) 296

E) 347

Answer: B

b) Based on the data presented in the table above, how many colonies are met^– arg⁺Amp^S?

A) 36

B) 44

C) 246

D) 252

E) 296

Answer: A

c) Based on the data presented in the table above, how many colonies are met^–arg^–?

A) 2

B) 4

C) 36

D) 82

E) 86

Answer: B

Section 5.2 (Bacterial conjugation)

10. Joshua Lederberg and Edward Tatum discovered a sex-like process in bacteria using:

A) high-powered microscopy in combination with antibiotic resistant genes. 

B) populations of cells with complementing auxotrophs, and the measurement of the generation of prototrophs during microbial mixing.

C) mutations in bacteria that inhibit the mating process.

D) comparison of bacterial behavior to yeast behavior.

E) chromosome sequencing and analysis.

Answer: B

11. Bernard Davis tested the “cross-feeding” interpretation of some data that showed the phenotype of one microbe as capable of being changed by another microbe.  His contribution can be summarized as:

A) using one strain of microbes as food for another, and then evaluating the impact upon microbial phenotype.

B) combining microbes with complementing auxotrophs in a single tube and then analyzing the cells of gene exchange.

C) placing a barrier between bacterial strains with complementing auxotrophs to assess whether contact between cells was required for genetic exchange. 

D) using fungi to feed bacterial strains with unusual auxotrophic mutations.

E) Davis was not a significant contributor to this research question.

Answer: C

12. An F⁺ microbial strain would best be described as:

A) any bacterium that readily takes up naked DNA through its membrane.

B) incapable of conjugation due to inhibitory genes on the F plasmid.

C) a bacterial strain with specific auxotrophic characteristics.

D) a bacterial strain that carries a phage with membrane fusion competency.

E) a bacterium with a plasmid that confers mating competency (or fertility).

Answer: E

13. An Hfr microbial strain would best be described as:

A) any bacterium that readily takes up naked DNA through its membrane.

B) a bacterial strain with specific auxotrophic characteristics.

C) a bacterium containing a chromosomally integrated plasmid conferring mating competency.

D) a bacterium carrying an Hfr phage with membrane fusion competency.

E) a bacterium that lacks the histidine frequenase gene.

Answer: C

14. If a met^–thr^– Hfr strain is mated with an F^– of genotype leu^–thi^–, prototrophic recombinants can be detected by plating the mixture on minimal:

A) medium.

B) medium supplemented with leucine and methionine.

C) medium supplemented with leucine and thiamine.

D) medium supplemented with methionine, threonine, leucine, and thiamine.

E) medium supplemented with threonine and thiamine.

Answer: A

15. An Hfr strain of E. coli with the genotype gly⁺azi^RStr^S is mated with an F^– strain of E. coli of genotype gly^–azi^SStr^R. Gly refers to the amino acid glycine, azi refers to sodium azide, and Str refers to the antibiotic streptomycin. Conjugation occurs and the progeny are screened on a selective medium to detect recombinants. If you wanted to select for the F^– recombinant genotype gly⁺azi^RStr^R, you should use a minimal medium containing:

A) glycine.

B) glycine, sodium azide, and streptomycin.

C) glycine and streptomycin.

D) sodium azide. 

E) streptomycin and sodium azide.

Answer: E

 16. Two bacterial strains were obtained with the following genotypes:

	Hfr:  ala–leu+aziSStrS
	F–:    ala+leu–aziRStrR

After an uninterrupted conjugation, you want to select F^– recombinants that are ala⁺leu⁺. Which of the following media would you use for this selection?

A) complete medium containing streptomycin

B) minimal medium containing streptomycin

C) minimal medium containing leucine and streptomycin

D) minimal medium containing sodium azide and leucine

E) minimal medium containing alanine, leucine, and streptomycin

Answer: B

17. Which of the following are characteristics of an E. coli strain of genotype F^–gal⁺lac^–pur⁺ile^–str^r?

A) It can act as a donor in conjugation experiments.

B) It can form a pilus.

C) It can grow on minimal media where lactose is the only carbon source.

D) It is auxotrophic for lactose.

E) It is susceptible to streptomycin.

Answer: D

 

18. A cross is made between an Hfr strain that is Str^Sa⁺b⁺d⁺ in genotype and an F^– strain that is Str^Ra^–b^–d^– in genotype. Interrupted-mating studies show that b⁺ enters the recipient strain last, and that the Str locus is very far away from b⁺, so it never enters the recipient strain. The b⁺ recombinants are then tested for the presence of the a⁺ and d⁺ alleles. The following data were obtained:

	a+	b+	d+	326
	a–	b+	d+	2
	a+	b+	d–	14
	a–	b+	d–	58
		400

a) What is the gene order?

A) a d b Str

B) b a d Str

C) b d a Str

D) d a b Str

E) d b a Str

Answer: D

b) What is the map distance between markers d and b?

A) 4 map units

B) 18 map units

C) 19 map units

D) 36 map units

E) 38 map units

Answer: C

19. An interrupted conjugation experiment is conducted using the following strains: Hfr Str^Sx⁺y⁺z⁺ and F^– Str^Rx^–y^–z^–. Samples of exconjugants are tested at regular intervals for their ability to grow on minimal medium (MM) containing streptomycin plus compound X, Y, or Z. The table below summarizes the number of colonies obtained on the different media when conjugation was interrupted after 10, 15, 20, or 25 minutes.

	Media used
Time of interruption	MM+Str+X	MM+Str+Y	MM+Str+Z
10 min	0	0	0
15 min	0	10	0
20 min	15	250	0
25 min	250	250	10

Based on these results, what is the order of entry of the four markers in question?

A) Str x y z

B) Str y x z

C) Str z x y

D) y x z Str

E) z x y Str

Answer: B

 

20. From one F⁺ strain, three distinct Hfr strains were derived. The first three markers transferred during an Hfr X F^– cross (different for each of the three Hfr) are:

Hfr 1: . . . D A F→

Hfr 2: . . . E B F→

Hfr 3: . . . E C D→

Which of the following must be the order of the genes on the bacterial chromosomal circle? (A is shown at both ends to represent circularity. Assume that the Hfr picks up all intermediates between any two represented genes.)

A) A D C E B F A

B) A B C D F E A

C) A C D F E B A

D) A E F B C D A

E) A F B D E C A

Answer: A

Section 5.3 (Bacterial transformation)

21. To demonstrate transformation of bacteria, one could:

A) extract DNA from an auxotroph and add it to prototrophic cells.

B) extract DNA from arg^– cells and add it to arg⁺ cells.

C) extract DNA from arg⁺ cells and add it to arg^– cells.

D) extract DNA from Str^S cells and add it to Str^R cells.

E) mix the DNA from arg⁺ and arg^– cells to allow recombination.

Answer: C

22. To demonstrate linkage of two markers A and B by transformation, one needs to demonstrate that the frequency of transformation by:

A) A and B is greater than the product of their individual transformation frequencies.

B) A and B is greater than the sum of their individual transformation frequencies.

C) A and B is less than the product of their individual transformation frequencies.

D) A and B is less than the sum of their individual transformation frequencies.

E) A is equal to the frequency of transformation by B.

Answer: A

23. If two markers are closely linked, they will show:

A) a high frequency of recombination and a high frequency of co-transformation.

B) a high frequency of recombination and a low frequency of co-transformation.

C) a low frequency of recombination and a high frequency of co-transformation.

D) a low frequency of recombination and a high frequency of co-transformation.

E) the same frequency of recombination and frequency of co-transformation.

Answer: C

Sections 5.4 and 5.5 (Bacteriophage genetics, Transduction)

24. The most commonly used phage characters in the study of phage inheritance are:

A) phage morphology and host morphology.

B) phage morphology and host range.

C) plaque morphology and host morphology.

D) plaque morphology and host range.

E) plaque morphology and phage morphology.

Answer: D

25. A permissive bacterial strain is simultaneously infected with two strains of T4-like phage. One phage strain has the genotype a^–b⁺, while the other is a⁺b^–. If the map distance between markers a and b is about 10 map units, what proportion of the progeny phages are expected to have the genotype a⁺b⁺?

A) about 1/20

B) about 1/10

C) about 4/10

D) about 4/5

E) about 9/10

Answer: A

26. A double infection experiment is conducted to determine the distance between two markers in the bacteriophage RB49, a phage similar to T2 and T4. A mutation in the first marker (F13^–) results in the inability to grow on E. coli strain B178, while a mutation in the second marker (P6^–) prevents the phage from growing at high temperature. A permissive E. coli strain is infected with both F13^– and P6^– phages, and the lysate is analyzed. 1002 plaques were formed on a lawn of permissive E. coli at 37ºC, and out of these 1002, 49 also formed plaques on E. coli B178 at high temperature. What is the predicted frequency of phages with genotype F13⁺P6^– in the lysate?

A) 5%

B) 45%

C) 47.5%

D) 90%

E) 95%

Answer: B

27. A bacterial chromosome has four markers, A, B, C, and D, evenly spaced throughout the circle. A generalized transducing phage will:

A) always transduce at least two of the four markers.

B) never transduce more than one marker.

C) transduce any of the markers by different transduction events.

D) transduce only one specific marker such as A.

E) pick up all the markers in one particle.

Answer: C

 

28. If two bacterial genes are very closely linked (less than one map unit apart), then their frequency of cotransduction will be: 

A) less than 1%.

B) more than 1%, but no more than 25%.

C) between 25% and 50%.

D) between 50% and 75%.

E) very high, almost 100%.

Answer: E

29. The three bacterial markers a, b, and c are linked, but their order is unknown. A cotransduction experiment is performed and the following results are obtained.

– Markers a and b are co-transduced at a frequency of 2%.

– Markers a and c are co-transduced at a frequency of 20%.

– Markers b and c are co-transduced at a frequency of 65%.

What can be concluded about the arrangement of the three markers on the bacterial chromosome?

A) The order is a-b-c, and b is closer to a than to c.

B) The order is a-b-c, and b is closer to c than to a.

C) The order is a-c-b, and c is closer to b than to a.

D) The order is b-a-c, and a is closer to b than to c.

E) The order is b-c-a, and c is closer to a than to b.
Answer: C

MATCHING QUESTIONS

Section 5.2 (Conjugation)

30. The following table contains statements pertaining to Hfr, F⁺, F′, and/or F^– strains of bacteria. Indicate in respective boxes if the statements apply to each category by inserting a check mark (√).

Statement	Hfr	F+	F′	F–
a) Contains genes that regulate conjugation and the transfer of DNA from one cell to another.
b) Always acts as the recipient in conjugation.
c) F factor is integrated into the host chromosome.
d) Cannot form a pilus.
e) F factor is in the plasmid form carrying a segment of chromosomal DNA.

Answer:

Statement	Hfr	F+	F′	F–
a) Contains genes that regulate conjugation and transfer of DNA from one cell to another.	√	√	√
b) Always acts as the recipient in conjugation.				√
c) F factor is integrated into the host chromosome.	√
d) Cannot form a pilus.				√
e) F factor is in the plasmid form carrying a segment of chromosomal DNA.			√

Sections 5.4 and 5.5 (Bacteriophage genetics, Transduction)

31. The following table contains statements that may apply to conjugation, transformation, and/or transduction.  For each blank space, insert a check mark if that statement applies to that mode of DNA exchange. 

	Transformation	Conjugation	Transduction
Bacteriophages are required for gene transfer.
DNA is acquired directly from the environment.
DNA is transferred via an F plasmid.
The movement of DNA is unidirectional from F+ to F–.
The DNA is replicated via rolling circle replication prior to the transfer.
The transferred DNA must recombine and replace the DNA in the chromosome in order to be maintained.
It can be used to study the dominance relationship between alleles via the creation of partial diploids.

Answer:

	Transformation	Conjugation	Transduction
Bacteriophages are required for gene transfer.			√
DNA is acquired directly from the environment.	√
DNA is transferred via an F plasmid.		√
The movement of DNA is unidirectional from F+ to F–.		√
The DNA is replicated via rolling circle replication prior to the transfer.		√
The transferred DNA must recombine and replace the DNA in the chromosome in order to be maintained.	√	(√) – Hfr = yes; F′ = no	√
It can be used to study the dominance relationship between alleles via the creation of partial diploids.		√(F′)

OPEN-ENDED QUESTIONS

Section 5.1 (Working with microorganisms)

32. Auxotrophic mutations have been extremely useful to study a number of processes in E. coli.

Briefly define the term “auxotrophic.” 

Suppose you were given two unlabeled tubes, one that contains a methionine auxotroph and the other that contains a leucine auxotroph. Design an experiment to determine which auxotroph is found in each tube.

Answer:

a) It is a cell that is unable to synthesize essential nutrients and therefore cannot grow unless the appropriate nutrient(s) are added to the medium.

b) Experiment: Grow each sample on 1) minimal media (negative control), 2) complete media (positive control), 3) minimal + methionine, and 4) minimal + leucine and observe growth on the various media.

Growth = +      no growth = –

Medium	WT/prototroph met+leu+	Contents: tube A	Contents: tube B
minimal	+	–	–
complete	+	+	+
minimal + methionine	+	+	–
minimal + leucine	+	–	+

Therefore, tube A contains the met auxotroph, whereas tube B contains the leu auxotroph.

Section 5.2 (Bacterial conjugation)

33. An E. coli F^– strain has the following genotype: S^–U^–R^–I^–E^–C^–. Three different Hfr strains, all carrying S⁺U⁺R⁺I⁺E⁺ and C⁺ markers and all having lambda phages integrated at the same specific site in their chromosomes are mated with the F strain in separate matings. The interrupted-mating results are given below. The numbers indicate time (in minutes) when different donor markers appeared in F^– cells after conjugation began. Assume that the E. coli map consists of 100 minutes, and gene C is mapped at 80 minutes.

	Markers	Hfr 1	Hfr 2	Hfr 3
	R	10	20	−
	I	40	−	5
	U	25	5	−
	E	−	60	−
	C	−	45	−
	S	55	−	20

The positions of Hfr 2, the gene C, and the polarity of Hfr 2 are given in the accompanying map.

a) Give the gene order, starting from C and going clockwise.

b) What is the location of the origin of Hfr 1? Hfr 3?

c) What is the location of the I marker? The S marker?

d) What is the location of the lambda phage in these strains?

Answer:

a) CRUISE (if counterclockwise, CESIUR)

b) 95 minutes (Hfr 1); 30 minutes (Hfr 3) 

c) 35 minutes (I); 50 minutes (S)

d) between S and E (50 to 65 minutes)

34. At time zero, an Hfr strain (strain 1) was mixed with an F^– strain, and at various times after mixing, samples were removed and agitated to separate conjugating cells. The cross may be written as:

	Hfr 1:	a+  b+ c+d+e+f+g+h+StrS
	F:	a–b–c–d–e–f–g–h–StrR

No order is implied in the above listing. The samples were then plated onto selective media to measure the frequency of h⁺Str^R recombinants that had received certain genes from the Hfr cell. The following graph shows the percentage of recombinants against time in minutes.

a) Draw a linear map of the Hfr chromosome indicating:

(1) the point of insertion (origin).

(2) the order of the genes a⁺, b⁺, e⁺, g⁺, and h⁺. 

(3) the shortest distance between consecutive genes on the chromosome in minutes.

b) An additional four Hfr strains (strains 2−5) were obtained. Each carried the wild-type alleles of the genes listed below. They were mated individually to the same F^– strain shown above (carrying the recessive alleles of all genes). With interrupted mating, the times of first appearance in minutes of individual Hfr markers in exconjugants were determined, as shown in the following chart.

	Minutes
	Hfr marker	Hfr 1	Hfr 2	Hfr 3	Hfr 4	Hfr 5
	a+		2			20
	b+				11
	c+		12		28	10
	d+		42				5
	e+					35	50
	f+				88		35
	g+
	h+

Draw the genetic map showing: 

(1) the positions of genes a through h.

(2) the origins of the Hfrs (including strain 1).

(3) the orientation of the origins.

(4) the shortest distance between consecutive genes or origins.

Answer:

a)

b)

Numbers on the outside of the circle are distances between genes. Numbers on the inside are distances between origins and adjacent genes. The above map is not to scale.

35. Strain A of E. coli is auxotrophic for methionine, while strain B is auxotrophic for lysine.  Both strains are F^–.  You want to produce a strain that is prototrophic.  Design an experiment to produce the desired prototroph.  Include in your experimental design your screening method for detecting the prototroph.

Answer:

Since both strains are F^–, a conjugation experiment cannot be conducted.  Instead, a transformation or transduction experiment should be designed.  To do a transformation experiment, take DNA from Strain A and mix it with Strain B bacteria (or vice versa).  The goal is to combine the functional met⁺ gene from strain B and the functional lys⁺ gene from strain A via transformation and recombination.  Some of the transformants will contain the desired combination of met⁺ and lys⁺.  Next, grow the transformed bacteria on minimal media.  Any bacteria that can grow on minimal media are prototrophs.

36. An extraterrestrial scientist is trying to obtain a genetic map of a newly discovered bacterium using standard techniques such as interrupted conjugation and Hfr mapping. She is particularly interested in three loci (ata, tic, and lan) encoding proteins that are required for the synthesis of atanine, ticovine, and lancytrine (three essential extraterrestrial amino acids), respectively. With the help of interrupted-conjugation experiments, the scientist determines that ata is the last marker to enter the recipient bacterium.

She now sets up a cross between a wild-type Hfr strain that is tertacycline sensitive and an ata^–tic^–lan^– F^– strain that is tetracycline resistant. 5000 exconjugants are recovered on minimal medium plus tetracycline, ticovine, and lancytrine, and subsequently replicated onto a variety of media.

The results are as follows:

– minimal medium and tetracycline: 4704 colonies grow;

– minimal medium plus lancytrine and tetracycline: 4708 colonies grow;

– minimal medium plus tetracycline and ticovine: 4900 colonies grow.

Draw a genetic map of these three loci (including all the map distances).

Why is the scientist using tetracycline on her plates (what is the purpose)?

Why is she recovering the exconjugants on plates that lack atanine?

Answer:

a) ata⁺lan⁺tic⁺ 4704

ata⁺lan^–tic⁺ 4

ata⁺lan⁺tic^– 196

ata⁺lan^–tic^– 5000 – 4794 – 4 – 196 = 96

 

lan is in the middle, ata to lan  2 m.u.; lan to tic  4 m.u.

b) to kill off the Hfr cells

c) to select for the last marker (ensuring that all three markers have entered the recipient)

Section 5.3 (Bacterial transformation)

37. In a transformation experiment, DNA of a p⁺q⁺ strain was used to transform a recipient strain that is auxotrophic for these markers. The number of each class of transformants is shown below: 

Class	Genotype	Number
1	p+q+	405
2	p+q–	300
3	p–q+	195

a) If the total number of transformants isolated was 900, what is the co-transformation frequency for the p and q loci?

b) Use diagrams to show how each class of transformants were obtained starting with the recipient auxotroph.

Answer:

a) co-transformation frequency = (p⁺q⁺)/p⁺ = 405/(405 + 300) =  0.574

                                                                               

b)

 

38. A bacterium with the genotype met⁺his⁺lys^–arg⁺ has been transformed with DNA from a second bacterium that has the genotype met⁺his^–lys⁺arg^–.

 

Following the transformation, scientists streaked a sample of a transformed bacterium onto several different plates that contain different nutritional supplements in order to determine the transformant’s genotype.  Using the results of the experiments (shown below), determine what crossovers occurred to produce the bacterium.  On the above diagram, draw the crossovers that are necessary to produce a bacterium that fits the observed growth profile.

Answer:

The bacteria must be met⁺his^–lys^–arg⁺, so a crossover must have occurred between the met and his genes and a second one between the his and lys genes, as shown below.

Sections 5.4 and 5.5 (Bacteriophage genetics, Transduction)

39. Two labs calculated cotransduction frequencies for the genes A and B. Lab 1 calculated 0.63, and lab 2 calculated 0.47. Which lab reported the genes to be closer together?

Answer: Lab 1. A higher cotransduction frequency indicates a smaller distance between the markers because fewer crossovers occurred to break up the two genes.

40. In a generalized transduction experiment, donor E. coli cells have the genotype a⁺c⁺b^–, and recipient cells have the genotype a^–c^–b⁺. Pl-mediated transductants for a⁺ were selected, and their total genotypes were determined, with the following results:

	Genotype	Number of progeny
	a+	c+	b–	95
	a+	c–	b–	205
	a+	c+	b+	5
	a+	c–	b+	195
		500

a) What is the order of the three genes?

b) What is the cotransduction frequency for:

	a and c?
	a and b?
	b and c?

Answer:

a) The donor is a⁺ c⁺ b^–, the recipient is a^– c^– b⁺, and the rare class is a⁺b⁺c⁺. Since the b allele is not transferred but the other two are, b must be the middle gene. The order is a b c.

b) cotransduction frequency for a and c = (95 + 5)/500 = 0.2 or 20% and for a and b = (5 + 195)/500 = 0.4 or 40%.

The frequency for b and c cannot be calculated because they are not selected markers.

41. A transduction experiment was conducted using a prototrophic donor and a recipient that was auxotrophic for lys, arg, and trp.  Selection was done for trp⁺ transductants. The trp⁺ transductants could be categorized into the following classes:

Genotype	Number
lys+arg–trp+	10
lys–arg+trp+	13
lys–arg–trp+	82
lys+arg+trp+	1

a) Which marker is closest to trp? Explain your reasoning.

b) What is the order of the three markers?  Explain your reasoning.

Answer:

a) arg.  arg was co-transduced with trp at a higher frequency than was lys.

b) arg trp lys (i.e., trp is in the middle).  We can predict this because the frequency of triple transductants is much lower than either of the double transductants. If arg⁺ was in the middle, then the frequency of lys⁺arg⁺trp⁺ would be higher than lys⁺arg^–trp⁺ (because fewer crossovers would be needed). This would be true, also, if lys⁺ was in the middle.

42. A new bacteriophage of E. coli was isolated from the steam tunnels of New York City. Six mutant strains (a, b, c, d, e, f) were derived from it, each having a different single point mutation for the genes A, B, C, D, E, and F. Strain BDE was mated with strain bde. The order of genes given below does not necessarily reflect the order of the genes on the chromosome. The following progeny were observed:

BDE	376	BdE	21	bDE	99	bdE	5
BDe	4	Bde	92	bDe	20	bde	383

a) What are the genotypes of the double crossovers, showing gene order?

b) Give the percent recombination between each pair of genes.

c) Draw a map showing the positions of the B, D, and E genes, with map distances between genes, where appropriate.

d) 

	(1) Give the formula you would use to calculate the coefficient of coincidence.
	(2) Calculate the coefficient of coincidence.

e) 

	(1) Give the formula you would use to calculate interference.
	(2) Calculate interference and specify whether the value is positive or negative.

Answer:

a) BeD and bEd

b) b, d: 25%; b, e: 20%; d, e: 5%

c)	20	5
	B−−−E−−−D         (Distance from B to E: 20 m.u.; distance from E to D: 5 m.u.)

d) 

	(1) c.o.c. = observed double crossovers/expected DCOs
	(2) c.o.c. = 0.009/(0.2 x 0.005) = 0.9

e) 

	(1) I = 1 − c.o.c.
	(2) I = 1 − 0.9 = 0.1

43. A mixed infection of an E. coli strain was performed using the h^–r⁺ and h⁺r^– genotypes of the T2 phage. Progeny phage were collected after lysis and plated on two E.coli strains to obtain plaque characters important for detection of parental and recombinant types.

h⁺  can only infect one strain of E.coli

h^–   can infect both strains

r⁺   shows slower lysis, forms small plaques

r^–    shows rapid lysis

In the table below, for each observed phenotype, give the genotype that would produce that phenotype, and indicate if the phenotype is parental or recombinant type. Calculate the map distance between the h and r genes using the information provided. 

Plaque phenotype	Number of plaques	Plaque genotype	Parental/recombinant
cloudy, large	600
clear, small	480
cloudy, small	180
clear, large	180

Answer:

RF% = (recombinants/total) × 100 = map distance

         

Plaque phenotype	Number of plaques	Plaque genotype	Parental/recombinant
cloudy, large	600	h+r–	parental
clear, small	480	h–r+	parental
cloudy, small	180	h+r+	recombinant
clear, large	180	h–r–	recombinant

map distance = (360/1080) × 100 = 33.33 m.u.

                                                             

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minimal

minimal + methionine + histidine + lysine +arginine

minimal + histidine + lysine

minimal + histidine

minimal + lysine

met⁺

his⁺

lys^–

arg⁺

met⁺    his^–    lys⁺   arg ^–