Sample questions for Exam 3

Sample questions for exam 3 -- from 1997

If this set of questions takes you more than an hour to do, DON'T PANIC! I've included more than one exam's worth of questions here. Also, don't read too much into what's included or not included -- this set should in no way be treated as in indicator of what will or will not be on the final. These are just questions I had available from 1997.

1.
You have a diploid fungus with the genotype:

y+/y, ala+/ala, hyg+/hyg, kno+/kno; where

y is a recessive yellow color
ala is a recessive allele that causes alanine requirement
hyg is a recessive allele conferring resistance to the drug hygromycin
kno is recessive allele causing knobs at the ends of hyphae

All four loci are on the same chromosome . You plate the fungus on hygromycin so that only hyphae that are hygromycin-resistant can grow. 200 hygromycin-resistant colonies are obtained. Further analysis shows that:

none of the hygromycin-resistant colonies requires alanine
40 of the hygromycin-resistant colonies are yellow but don't have knobs
60 are yellow and have knobs

(i)
How did these recessive phenotypes arise? (Only a two-word answer needed!)

Mitotic recombination

(ii)
Where is the centromere in relation to these loci? Show all four loci.

ala cen kno y hyg
--------------o----------------------------

hyg is the furthest from the centromere--the 100 colonies that are hyg-resistant but not knobbed or yellow tell us that there were crossovers between hyg and the other markers. Likewise, the presence of colonies (40) that are hyg-resistant and yellow tells us that yellow is next to hyg; and there are knobbed, yellow, hyg-resistant colonies (60), so these three markers must be on the same side of the centromere, and knob must be the closest to the centromere. The complete absence of ala-requiring, hyg-resistant colonies suggests that ala must be on the other side of the centromere (although it is possible that it is also on the same side, but very tightly linked to the centromere).

(iii)
Where possible, show the relative map distances between the loci (including the centromere). (Show your work!)

CEN-kno = 60/200 = 0.3;
kno-y = 40/200 = 0.2;
y-hyg = 100/200 = 0.5

2. You are a poor, overworked grad student whose project involves looking for mutant flies that have an extended life span. You grow up an age-matched group of mutagenized flies in a bottle. At a certain time after the expected life span has passed, you examine each fly and pick out the ones that are still living.

(i)
Is this a screen or a selection?

Screen __X___

Selection ______ (check one)

Explain briefly:

You have to examine each fly in the collection to decide if it's living or dead, so it's a screen.

(ii)

Mutant m1 m2 m3 m4 m5 m6

m1 - + + - + +

m2 - + + + +

m3 - + + -

m4 - + +

m5 - +

m6 -

After doing the appropriate back-crosses, you have obtained six true-breeding lines of long-lived mutant flies (m1 through m6, each with a single mutation). You do pairwise crosses between these mutant lines, and note the phenotypes of the progeny ("+" = wildtype, "-" = mutant, long-lived) as shown.

Group the mutations into complementation groups. Name the complementation groups A, B, C, etc.; assume that the mutations are all recessive.

Look for mutations that fail to complement (i.e., mutation pairs that give the "-" phenotype). Thus,

A: m1, m4
B: m2
C: m3, m6
D: m5

3. In the course of your research, you isolate two pure breeding strains of mice. One has an average tail length of 1 inch while the other has an average length of 7 inches. When these two strains are crossed, the F1 progeny all have a tail length of 4 inches.

(i)
We have talked about two modes of inheritance in class that could explain these results: incomplete dominance at a single locus, versus quantitative inheritance involving multiple loci . What crosses could you perform that may distinguish between the two?

Mate the pure-breeding 1"-tailed mice with the 7"-tailed pure breeders; cross the 4"-tailed F₁ progeny with each other and examine the tail lengths in the F₂ progeny.

(ii)
What would your experimental results look like for incomplete dominance? (State the expected phenotypes and give the proportions in which they are expected.)

Parents: L₁L₁ (1" tail) x L₇L₇ (7" tail)

F₁: L₁L₇ (4" tails)

F₂: L₁L₁ (1" tails), L₁L₇ (4" tails), L₇L₇ (7" tails) in 1:2:1 ratio

(iii)
What would your experimental results look like for continuous variation due to three unlinked genes? (State the expected phenotypes and give the proportions in which they are expected.

Workspace

3 genes; n = 3, so the number of additive alleles = 6

6" difference between 0 additive alleles and 6 additive alleles, so each additive allele contributes 1"

# of phenotypes = 2n + 1 = 7

Distribution = (a+b)²ⁿ where a = additive, b = non-additive

= (a+b)⁶

Expected phenotypes:

7", 6", 5", 4", 3", 2", 1" tails

Proportions:

from (a+b)⁶ --

1:6:15:20:15:6:1 distribution of tail lengths in the order shown for "expected phenotypes"

4. Genes X and Y are structural genes required for the breakdown of altrose (a sugar) by a certain bacterium, and are transcribed as a polycistronic mRNA. X and Y proteins are present at low level in the absence of altrose, and at a 30-fold higher level in the presence of altrose. A mutation in a third gene (Z) causes expression of X and Y at constitutively high levels.

(i)
Assuming that gene Z makes a protein that regulates X and Y gene transcription, suggest two possible models (explanaations) for the induction of X and Y transcription by altrose. Your models should explain the function of Z, the effect of altrose, and the nature of the mutation in Z.

Model 1: Positive regulation. Altrose binds to Z protein and activates it; activated Z promotes transcription of X and Y. The mutated Z must be a gain-of-function mutation that causes activation of X and Y transcription even in the absence of inducer (altrose).

Model 2: Negative regulation. Z protein inhibits transcription of X and Y. Altrose binds to Z and blocks this inhibition. The mutation in Z must be loss-of-function; the mutated Z protein no longer binds to the promoter/operator of X-Y.

(ii)
Outline a genetic experiment that you would do to distinguish between your two models, and state the outcome you would expect for each model.

Experiment: Make a Z/z- partial diploid with everything else wildtype and look at the response to altrose.

Prediction for Model 1: The z- mutation is a dominant gain-of-function mutation, so the partial diploid will have constitutively high levels of X and Y protein, just like the z- single mutation.

Prediction for Model 2: The mutation is a recessive loss-of-function mutation, so the partial diploid will behave just like wildtype--low level of X and Y expression in the absence of altrose, elevated expression in the presence of altrose.

(iii)
A frame-shift mutation in X has an x- phenotype (no detectable X protein, normal level of Y protein) but a frameshift mutation in Y has an x-y- phenotype (no detectable level of either X or Y protein). Which gene is closer to the promoter, X or Y?

Y is closer to the promoter. The mRNA is polycistronic, i.e., the coding sequences for both genes are present on the same mRNA, and are sequentially translated 5' to 3'. A frame-shift in the second open reading frame (ORF) affects the second protein but not the first--it has been translated already. In contrast, a frameshift in the first ORF can terminate translation, preventing the ribosome from even getting to the second ORF.

5. In a Caucasian population of 9000 individuals, the frequency of the allele for male pattern baldness is 0.3.

(i)
What are the expected genotypes and their frequencies in the population? (Assume Hardy-Weinberg conditions.)

Frequency of baldness allele = p = 0.3

Frequency of wildtype allele = q = __1-0.3 = 0.7___

Expected genotypes and frequencies:

B^bB^b = p² = 0.3² = 0.09

B^bB^h = 2pq = 2 x 0.3 x 0.7 = 0.42

B^hB^h = q² = 0.7² = 0.49

(ii)
The baldness allele is recessive in females (only B^bB^b females are bald) but dominant in males (B^bB^b and B^bB^h males are bald). What are the expected phenotype frequencies in males and females?

Males

Bald = B^bB^h and B^bB^b = 2pq + p² = 0.42 + 0.09 = 0.51

Non-bald = B^hB^h = q² = 0.49 (= 1 - 0.51)

Females

Bald = B^bB^b = p² = 0.09

Non-bald = 1 - 0.09 = 0.91

(iii)
A group of 1000 new individuals joins the population. If p = 0.1 in the incoming group, what will be the frequency of heterozygotes in the population after one generation? (Assume random mating.)

Number of B^b alleles after immigration = (0.3 x 9000) + (0.1 x 1000) = (2700 + 100) = 2800.

New value of p = frequency of B^b in the expanded population = 2800/10000 = 0.28

New value of q = 1-0.28 = 0.72.

Frequency of heterozygotes in the next generation = 2pq = 2 x 0.28 x 0.72 =0.403

6.
Claws on the forelegs of Great Northwestern Arboreal frogs are pigmented, while claws on the hind legs are unpigmented. You are interested in the regulation of this pigmentation, and you have narrowed it down to the two models shown below. In both models, Gene B is needed for pigment production; in Model 1, Gene A is an activator of B and only functions in forelegs, while in Model 2, Gene A is a repressor of B and acts only in hind legs.

(i)
What phenotype would you expect for each of the following strains in each model? (Upper case alleles = wildtype; lower case = null mutations)

Model Strain Phenotype

Model 1 aaBB No pigmented claws

Model 1 AAbb No pigmented claws

Model 2 aaBB All claws pigmented (fore and hind legs)

Model 2 AAbb No pigmented claws

(ii)
Unfortunately for you, before you can score the single mutant phenotypes, the lab pet (python) gets into the frog cages and snacks on your entire collection of single mutants, leaving behind only a cage of AaBb males and and a cage of AaBb females. However, being an accomplished geneticist, you know that you can still distinguish between the two models by mating the survivors and examining the phenotypes of the progeny. What phenotypes of progeny would you predict, and in what proportions, for Model 1 vs. Model 2? (Assume that a and b are null alleles of unlinked genes.)

This is just a dihybrid cross giving 9:3:3:1 ratio of progeny phenotypes. Depending on whether model 1 is correct or model 2 is correct, different ratios will be obtained as follows:

Proportion Genotype Phenotype

Model 1 Model 2

9/16 A_B_ Wildtype Wildtype

3/16 A_bb No pigmented claws No pigmented claws

3/16 aaB_ No pigmented claws All claws pigmented

1/16 aabb No pigmented claws No pigmented claws

Overall :
9 normal: 7 unpigmented

(complementation)

9 normal: 3 all pigmented : 4 unpigmented

(recessive epistasis)

(9:7 ratio) (9:3:4 ratio)

[In Model 1, A and B are both required for pigmentation, so a homozygous recessive of either one gives only unpigmented claws. In Model 2, gene A is needed to repress B in hind legs--so in the aaB_ progeny, pigment is made in forelegs and hindlegs.

.. Color

1.	You have a diploid fungus with the genotype: y+/y, ala+/ala, hyg+/hyg, kno+/kno; where y is a recessive yellow color ala is a recessive allele that causes alanine requirement hyg is a recessive allele conferring resistance to the drug hygromycin kno is recessive allele causing knobs at the ends of hyphae All four loci are on the same chromosome . You plate the fungus on hygromycin so that only hyphae that are hygromycin-resistant can grow. 200 hygromycin-resistant colonies are obtained. Further analysis shows that: none of the hygromycin-resistant colonies requires alanine 40 of the hygromycin-resistant colonies are yellow but don't have knobs 60 are yellow and have knobs
	(i)	How did these recessive phenotypes arise? (Only a two-word answer needed!) Mitotic recombination
	(ii)	Where is the centromere in relation to these loci? Show all four loci. `ala cen kno y hyg` `--------------o----------------------------` hyg is the furthest from the centromere--the 100 colonies that are hyg-resistant but not knobbed or yellow tell us that there were crossovers between hyg and the other markers. Likewise, the presence of colonies (40) that are hyg-resistant and yellow tells us that yellow is next to hyg; and there are knobbed, yellow, hyg-resistant colonies (60), so these three markers must be on the same side of the centromere, and knob must be the closest to the centromere. The complete absence of ala-requiring, hyg-resistant colonies suggests that ala must be on the other side of the centromere (although it is possible that it is also on the same side, but very tightly linked to the centromere).
	(iii)	Where possible, show the relative map distances between the loci (including the centromere). (Show your work!) `CEN-kno = 60/200 = 0.3;` `kno-y = 40/200 = 0.2;` `y-hyg = 100/200 = 0.5`

3.	In the course of your research, you isolate two pure breeding strains of mice. One has an average tail length of 1 inch while the other has an average length of 7 inches. When these two strains are crossed, the F1 progeny all have a tail length of 4 inches.
	(i)	We have talked about two modes of inheritance in class that could explain these results: incomplete dominance at a single locus, versus quantitative inheritance involving multiple loci . What crosses could you perform that may distinguish between the two? Mate the pure-breeding 1"-tailed mice with the 7"-tailed pure breeders; cross the 4"-tailed F₁ progeny with each other and examine the tail lengths in the F₂ progeny.
	(ii)	What would your experimental results look like for incomplete dominance? (State the expected phenotypes and give the proportions in which they are expected.) `Parents: L₁L₁ (1" tail) x L₇L₇ (7" tail)` `F₁: L₁L₇ (4" tails)` `F₂: L₁L₁ (1" tails), L₁L₇ (4" tails), L₇L₇ (7" tails) in 1:2:1 ratio`
	(iii)	What would your experimental results look like for continuous variation due to three unlinked genes? (State the expected phenotypes and give the proportions in which they are expected. Workspace `3 genes; n = 3, so the number of additive alleles = 6` `6" difference between 0 additive alleles and 6 additive alleles, so each additive allele contributes 1"` `# of phenotypes = 2n + 1 = 7` `Distribution = (a+b)²ⁿ where a = additive, b = non-additive` `= (a+b)⁶` Expected phenotypes: `7", 6", 5", 4", 3", 2", 1" tails` Proportions: `from (a+b)⁶ --` `1:6:15:20:15:6:1 distribution of tail lengths in the order shown for "expected phenotypes"`

4.	Genes X and Y are structural genes required for the breakdown of altrose (a sugar) by a certain bacterium, and are transcribed as a polycistronic mRNA. X and Y proteins are present at low level in the absence of altrose, and at a 30-fold higher level in the presence of altrose. A mutation in a third gene (Z) causes expression of X and Y at constitutively high levels.
	(i)	Assuming that gene Z makes a protein that regulates X and Y gene transcription, suggest two possible models (explanaations) for the induction of X and Y transcription by altrose. Your models should explain the function of Z, the effect of altrose, and the nature of the mutation in Z. Model 1: Positive regulation. Altrose binds to Z protein and activates it; activated Z promotes transcription of X and Y. The mutated Z must be a gain-of-function mutation that causes activation of X and Y transcription even in the absence of inducer (altrose). Model 2: Negative regulation. Z protein inhibits transcription of X and Y. Altrose binds to Z and blocks this inhibition. The mutation in Z must be loss-of-function; the mutated Z protein no longer binds to the promoter/operator of X-Y.
	(ii)	Outline a genetic experiment that you would do to distinguish between your two models, and state the outcome you would expect for each model. Experiment: Make a Z/z- partial diploid with everything else wildtype and look at the response to altrose. Prediction for Model 1: The z- mutation is a dominant gain-of-function mutation, so the partial diploid will have constitutively high levels of X and Y protein, just like the z- single mutation. Prediction for Model 2: The mutation is a recessive loss-of-function mutation, so the partial diploid will behave just like wildtype--low level of X and Y expression in the absence of altrose, elevated expression in the presence of altrose.
	(iii)	A frame-shift mutation in X has an x- phenotype (no detectable X protein, normal level of Y protein) but a frameshift mutation in Y has an x-y- phenotype (no detectable level of either X or Y protein). Which gene is closer to the promoter, X or Y? Y is closer to the promoter. The mRNA is polycistronic, i.e., the coding sequences for both genes are present on the same mRNA, and are sequentially translated 5' to 3'. A frame-shift in the second open reading frame (ORF) affects the second protein but not the first--it has been translated already. In contrast, a frameshift in the first ORF can terminate translation, preventing the ribosome from even getting to the second ORF.

5.	In a Caucasian population of 9000 individuals, the frequency of the allele for male pattern baldness is 0.3.
	(i)	What are the expected genotypes and their frequencies in the population? (Assume Hardy-Weinberg conditions.) Frequency of baldness allele = p = 0.3 Frequency of wildtype allele = q = __`1-0.3 = 0.7`___ Expected genotypes and frequencies: `B^bB^b = p² = 0.3² = 0.09` `B^bB^h = 2pq = 2 x 0.3 x 0.7 = 0.42` `B^hB^h = q² = 0.7² = 0.49`
	(ii)	The baldness allele is recessive in females (only B^bB^b females are bald) but dominant in males (B^bB^b and B^bB^h males are bald). What are the expected phenotype frequencies in males and females? Males Bald = `B^bB^h and B^bB^b = 2pq + p² = 0.42 + 0.09 = 0.51` Non-bald = `B^hB^h = q² = 0.49 (= 1 - 0.51)` Females Bald = `B^bB^b = p² = 0.09` Non-bald = `1 - 0.09 = 0.91`
	(iii)	A group of 1000 new individuals joins the population. If p = 0.1 in the incoming group, what will be the frequency of heterozygotes in the population after one generation? (Assume random mating.) `Number of B^b alleles after immigration = (0.3 x 9000) + (0.1 x 1000) = (2700 + 100) = 2800.` `New value of p = frequency of B^b in the expanded population = 2800/10000 = 0.28` `New value of q = 1-0.28 = 0.72.` `Frequency of heterozygotes in the next generation = 2pq = 2 x 0.28 x 0.72 =0.403`