Exam 2, Fall 2014

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name:_______________________ student ID:_____________________

Genetics L311 exam 2

October 24, 2014

Directions: Please read each question carefully. Answer questions as concisely as possible. Excessively long answers, particularly if they include any inaccuracies, may result in deduction of points. You may use the back of the pages as work sheets, but please write your answer in the space allotted. However, you must show all your work. Clearly define your genetic symbols.

We will not make guesses as to what a particular symbol is intended to mean. Also, don’t assume that strains are true-breeding unless this is stated in the question. Finally, show all your work. Good luck. page 2 page 3 page 4 page 5 page 6

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(20 points possible)

(23 points possible)

(24 points possible)

(24 points possible)

( 9 points possible) total _______ (of 100 points possible)

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name:_______________________ student ID:_____________________

1.

Short answers (2 points each, 20 points total)

A.

Locus refers to a position on a chromosome, of a gene for example.

B.

Traits that appear in both genders but that are more severe in one gender than the other are said to be sex influenced .

C.

The tendency of two genes located on the same chromosome to co-segregate is called linkage .

D.

Base analogs are compounds that can be incorporated into the DNA during replication and can cause mis-pairing, thereby inducing mutation (looking for the class of compounds not a specific example).

E.

Genetic anticipation refers to the tendency of some disorders, such as Huntingon’s

Chorea and Fragile X syndrome, to become more severe and with earlier age of onset in subsequent generations.

F.

Telomeres are specialized structures found at the ends of many eukaryotic chromosomes.

G.

Acquired characteristics is the belief that traits influenced by the environment can be passed on to subsequent generations.

For the following, please provide a brief definition of the term given:

H.

Prion: Infectious protein capable of producing disease, apparently by interacting with and causing misfolding of endogenous protein.

I.

Photolyase: An enzyme that directly reverses the covalent bonds of pyramidine dimers formed by exposure to UV light. It requires visible light to function.

J.

Position effect variegation: Produced when a gene has been shifted to the vicinity of a boundary between heterochromatin and euchromatin. Depending on the precise boundary, the gene may be expressed or silenced, producing cells that appear normal or show a mutantlike phenotype.

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name:_______________________ student ID:_____________________

2.

In a fly the genes px , sp , and cn are all linked. Flies heterozygous for these three loci were crossed to the total recessive condition and the following progeny were obtained. px sp cn 1,461 px sp + 3,497 px + cn px + +

+ sp cn

+ sp +

1

11

9

0

+ + cn 3,482

+ + + 1,539

10,000

A.

What is the order of the genes (2 points)? sp px cn ( or cn px sp )

B.

What is the map distance between the three linked genes (6 points)? px – sp: (9+11+1+0)/10,000 = 0.21 cM sp – cn: 30.22 cM px – cn: (1461+1539+1+0)/10,000 = 30.01 cM

C.

Assume for the rest of the problem (warning this does not necessarily represent the correct answer for a and b) that the order and distance are the following:

px 30mu sp 20mu cn

What is the probability of obtaining a phenotypic px sp + offspring from a cross between a heterozygous fly( px sp cn/+ + +) and a px sp cn recessive fly (3 points)? sp – cn : (SCO + DCO))/total X 100 = 20

SCO/total + DCO/total = 0.20 frequency of SCO + frequency of DCO = 0.20, and frequency of DCO = 0.3(0.2) = 0.06 freq. SCO = 0.20 – 0.06 = 0.14 but only half of the cross overs between sp and cn produce the px sp + chromosome to 0.14/2 = 0.07

3. A rare condition is controlled by a paternally imprinted gene. A man who is homozygous mutant for this gene marries a woman who is homozygous wild-type. All who marry in are wild type (12 points).

0 What is the probability that his sons will show the trait?

What is the probability that his daughters will show the trait? 0

What is the probability that his son’s son will show the trait? 0

What is the probability that his daughter’s son will show the trait? ½

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name:_______________________ student ID:_____________________

4.

When traveling in a cave near Bloomington, you came across a species of bat, T. nicholasae . Two recessive mutations are known, one causing glow-in-the-dark wings ( g ) and the other causing hairy feet ( h ). When trying to create a monster bat for a reality TV show, you cross heterozygous parents, both with the mutations in trans, hoping to produce bats with hairy feet and glowing wings. Since the mutations are only 4 mu apart you assume no double crossovers.

A. What are the genotypes of the parents (4 points)? g +/+ h and g +/+ h

B. What percentage of offspring will have glowing wings and hairy feet (4 points)?

0.02(0.02) = 0.0004 or 0.04%

C. What percentage of offspring will have the same genotype as their parents (4 points)?

0.48(0.48) + 0.48(0.48) = 0.46 or 46%

5.

You are attempting to determine if two genes in zombie mole rats, pl, which causes pale skin and mn , which causes animals to moan eerily, are linked. You cross heterozygotes with homozygous recessive to get the following results.

118 pale, moaning

101 pale

103 moaning

120 wild type

A. Please state the null hypothesis that you will test (2 points).

The null hypothesis is that pl and mn are not linked.

B. Please calculate chi square. Using the table at the right, what is the probability that your null hypothesis is incorrect? What can you conclude (10 points)?

(238 – 221) 2 /221 + (204 – 221) 2 /221

= 1.31 + 1.31 = 2.62

Probability that null hypothesis is incorrect is about 90%. Because this doesn’t reach the 95% threshold we cannot conclude that the genes are linked.

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name:_______________________ student ID:_____________________

6. When on vacation, you discovered a species of giant spiders that were unusually colored. You found two true-breeding strains, one green striped and one yellow striped. When you cross true breeding yellow striped with true breeding green striped, you find that all the F1’s are green. When you perform crosses amongst the F1’s, these are the following results:

481 yellow striped aaB– and aabb

358 blue striped A–bb

1082 green striped A–B–

A. What are the genotypes of the F2 progeny (8 points)?

B. What is the ratio observed (2 points)? 9:3:4

C. If you cross a blue F2 chosen at random with a homozygous recessive, what frequency of the offspring will be yellow (4 points)?

A–bb X aabb

1/3 AABB X aabb => Aabb

2/3 Aabb X aabb => Aabb

1/3(0) + 2/3(1/2) = 1/3 and aabb

7.

You come across two true breeding strains of purple people eaters on an expedition through

Antarctica. One strain has smooth scales and pink spots on its belly. The other strain has jagged scales and yellow spots on its belly. The traits are controlled by two different genes. You become attached to the purple people eaters and decide to study them more. You cross the two strains and get jagged scales and pink spots. Surprisingly, one of the offspring has two small patches next to one another that differ from the rest of its skin. One has smooth scales with pink spots and the other has jagged scales with yellow spots.

A. What are the genotypes of the parents and the offspring (6 points)?

Jagged scales, yellow spots: + y/+ y

Smooth scales, pink spots: s +/s +

F1: s +/+ y

B. If these genes are linked, are they in cis or trans in the F1 offspring (2 points)? trans

C. Draw out the event that could have produced this individual. Please include the genes (8 points). s

+ s s

+ +

+ + y y

+ y s

+ s s

+

+ y

+ y

+ smooth, pink spots

+ y jagged, yellow spots

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name:_______________________ student ID:_____________________

8.

Using the following DNA sequence, consider each of the following cases separately. Use the chart below to determine which of the amino acids corresponds to each codon.

First

Position

U

C

A

U

Phe

Phe

Leu

Leu

Leu

Leu

Leu

Leu

Ile

Ile

Ile

Second Position

C

Ser

Ser

Ser

Ser

Pro

Pro

Pro

Pro

Thr

Thr

Thr

A

Tyr

Tyr

Stop

Stop

His

His

Gln

Gln

Asn

Asn

Lys

G

Cys

Cys

Stop

Trp

Arg

Arg

Arg

Arg

Ser

Ser

Arg

Third

Position

U

U

C

A

U

C

A

G

C

A

G

Starting with the ATG on the top strand, left hand side, please give the sequence of the resulting protein.

Please name the type of mutation that results from the change. Please name a mutagen other than ionizing radiation that is likely to produce the mutation shown. You need not provide the original protein sequence.

The arrow indicates the position of the mutation. Please do not use the same mutagen more than once (9 pts).

G

Met

Val

Val

Val

Val

Thr

Ala

Ala

Ala

Ala

Lys

Asp

Asp

Glu

Glu

Arg

Gly

Gly

Gly

Gly

G

U

C

A

G

DNA Sequence

5’-ATGATAAGAGCATGA-3’

3’-TACTATTCTCGTACT-5’

(This is the original sequence.)

5’-ATGATAGGAGCATGA-3’

3’-TACTATCCTCGTACT-5’

5’-ATGATAAAGCATGA-3’

3’-TACTATTTCGTACT-5’

5’-ATGCTAAGAGCATGA-3’

3’-TACGATTCTCGTACT-5’

Resulting Protein Type of Possible

Mutation Mutagen don’t need to provide this not applicable not applicable

Met-Ile-Gly-Ala-stop Transition or missense

5 BrU or 2 AP

Met-Ile-Lys-His-Asp/Glu Frameshift Acridine orange

Met-Leu-Arg-Ala-stop Transversion or missense

UV light

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