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BIOL 2323 Exam 4 (Spring 2024) 1 EXAM 4 GENERAL GENETICS BIOL 2323 –
SPRING 2024 •
Mark your answers to the exam questions on a blue ANSWER SHEET NO. 4521 using a pencil No. 2. It is essential that you enter
your name and university ID
on the form by filling in the bubbles on the sheet!! •
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Read each question and the selection of answers completely and carefully before choosing the best answer
. Mark the correct answers on your exam by circling the corresponding letter. It is recommended that you transfer the results to the answer sheet after you have answered all questions and you are sure that your answers are final. •
Trying to correct entries on the answer sheet may lead to errors when the sheet is read. It you are unable to erase completely, it is much safer to fill in a new sheet. •
You can earn a maximum of 36 points on this exam by answering all 35 questions and the bonus question (on page 11) correctly. GOOD LUCK! ……………………………………………………………………………………………………………….
BIOL 2323 Exam 4 (Spring 2024) 2 The following table may be useful for answering some of the questions: 1. The following is the sequence of a very small eukaryotic protein:
N-Met-Ser-Trp-His-Tyr-C Which of the following is a plausible
mRNA sequence for translating the above protein? (Hint: *think* like a eukaryotic ribosome initiating translation) A. 5’
-ATGAGTTGGCACTAT-
3’
B. 5’
-
ACAGCAUGUCUUGGCAUAUACCGGAUU3’
C. 5’
-GACGCAUAGUUGGCCUUCGAAAUAC-
3’
D. 5’
-UCCACGUGUGACCUUAAACCAAAAAA-
3’
E. None of the above. 2. As hinted at in question above, there can be several possible mRNA sequences that are translated into an identical protein. Why is this the case? A. The genetic code is unambiguous. B. The genetic code is degenerate. C. The genetic code is a triplet. D. The genetic code is nearly universal. E. The genetic code is non-overlapping.
BIOL 2323 Exam 4 (Spring 2024) 3 3. What was the significance of Sir Archibald Garrod
’s research on alkaptonuria (black urine disease)? A. He was the first to sequence DNA to discover that alkaptonuria patients lacked a key metabolic gene. B. He was the first to show that alkaptonuria patients had a hypermorphic mutation leading to excessive production of a metabolite that turns urine black. C. He was the first to connect a heritable disease to a specific metabolic defect. D. He was the first to visualize intragenic recombination. E. He was the first to discover why asparagus makes pee smell funny. 4. The following figure shows a simplified pathway for methionine biosynthesis in E. coli
. If you generated a mutant defective only in producing functional Enzyme 4, which compound would you expect to accumulate
inside
of the cell? A. serine B. O
-acetyl serine C. cysteine D. cystathione E. homocysteine 5. Mutations in the gene encoding isocitrate dehydrogenase (IDH) are associated with several different types of cancers. Wild-type IDH converts isocitrate to alpha-ketoglutarate + CO
2
, while the mutant IDH instead produces 2-
hydroxyglutarate (an “oncometabolite” that promotes cancer progression). This is what type of mutation? A. null. B. hypomorphic. C. hypermorphic. D. neomorphic. E. mighty morphin. 6. Still considering the IDH allele above, which description best matches your predictions for the properties of the disease allele? A. The disease allele is recessive to the wild-type allele, and results from a deletion of the IDH gene. B. The disease allele is recessive to the wild-type allele, and results from a missense mutation in the IDH gene. C. The disease allele is dominant to the wild-type allele, and results from a deletion of the IDH gene. D. The disease allele is dominant to the wild-type allele, and results from a duplication of the IDH gene. E. The disease allele is dominant to the wild-type allele, and results from a single amino acid substitution of IDH. Serine O-
acetyl serine cysteine cystathione homocysteine methionine Enzyme 1 Enzyme 2 Enzy
me 3 Enzyme 4 Enzyme 5
BIOL 2323 Exam 4 (Spring 2024) 4 7. All of the following were key observations in Neurospora crassa
that contributed to Beadle and Tatum devising the “one gene
-
one enzyme” hypothesis except…
A. the observation that UV could induce auxotrophic mutations in haploid conidia (spores).
B. the observation that auxotrophic mutations showed at least dihybrid inheritance patterns when backcrossed to the wild-type parental strain.
C. the observation that auxotrophic mutants required a single type of nutrient. D. the observation that auxotrophic mutants were blocked at a single biochemical steps. E. All of the above were key observations. 8. The tertiary structure of a protein…
A. is determined by its primary amino acid sequence. B. refers to the amino acid sequence of a protein. C. can include α
-
helices or β
-pleated sheets. D. refers to the oligomeric state of a protein. E. can include stem loops and hairpins. 9. Your friend Logan has discovered a bacterium that biosynthesizes Adamantium and requires it for growth. Being something of a scientist yourself, you perform a feeding experiment with four known intermediates. Based on the results of the feeding experiment below, what is the biochemical pathway for Adamantium biosynthesis. (Note: +/- refers to growth or no growth, respectively) A. Compound A →
Compound B →
Compound C →
Compound D →
Adamantium B. Compound C →
Compound D →
Compound B →
Compound A →
Adamantium C. Compound D →
Compound A →
Compound B →
Compound C →
Adamantium D. Compound D →
Compound C →
Compound B →
Compound A →
Adamantium E. Compound B →
Compound A →
Compound D →
Compound C →
Adamantium 10. Based on the results of the feeding experiment above, which enzyme is responsible the reaction that ultimately produces Adamantium? (Note: Mutant 1 corresponds to a defect in Enzyme 1, etc.) A. Enzyme 1 B. Enzyme 2 C. Enzyme 3 D. Enzyme 4 E. Wild Type
BIOL 2323 Exam 4 (Spring 2024) 5 11. Which of the following events signals the termination of translation? A. The ribosome runs out of tRNAs. B. The polypeptide folds into a protein. C. The ribosome reaches the end of the mRNA. D. The ribosome reaches a start codon. E. The ribosome reaches a stop codon. 12. You have isolated 2 new rII -
T4 phage mutants and wish to determine whether they are located in either the rII A gene or rII B gene. Fortunately, you have previously characterized a mutant with a mutation in the rII A gene (Mutant A1). You performed complementation analysis by co-
infecting mutant A1 with either Mutant 2 or Mutant 3. The results are shown below. Note that lysis only occurs if wild-type rII +
phage are recovered from co-infection. Which genes are likely mutated in Mutants 2 and 3, respectively? A. gene A, gene A B. gene B, gene B C. gene A, gene B D. gene B, gene A E. gene C, gene C 13. If you were designing an experiment to calculate the recombination frequency between two different T4 rII
-
mutations, which scheme would you use to most easily
identify rare recombinant phage? A. A selection for phage able to infect the E. coli K12 strain. B. A selection for phage unable to infect the E. coli K12 strain. C. A screen for large plaques on the E. coli B strain. D. A screen for normal plaques on the E. coli B strain. E. A screen based on microscopic inspection of individual phage. 14. When studying proflavin-induced mutants, Crick and colleagues observed that combining 3 single base-pair deletions or 3 single base-pair insertions restored the phage to Wild Type. However, wild-type phage were only recovered if the 3 deletions or insertions mapped very close together. What is the most likely
explanation for this observation? A. Proflavin induced mutations naturally cluster next to each other. B. Proflavin induced mutations always occur in threes. C. Cellular repair machinery prevents insertions or deletions from occurring at great distances. D. Recombination reduces the distance between insertions or deletions. E. The further the mutations are separated from each other, the more of the open reading frame remains frameshifted. Co-infected Mutants Results (+/- lysis) A1 + 2 + A1 + 3 -
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Two rough black guinea pigs when bred together have two offspring. One of them is rough white and
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O A. Cells can function equally as well at all temperatures because enzymes are effective at all temperatures.
OB.
Cells only function within a specific temperature range because enzymes only function within a specific temperature
range.
OC.
Cells can function eaually as well at all pH's because enzvmes are effective at all pH's.
Session Score: 100% (4/4)
Session Timer: 44:44
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A Test - Unit 2B F x
E Practice -5. Ec x
E Alya Alhashim x
E Alya Alhashim x
E Alya Alhashim x
A examlogin.com/#/virtual_form/7c8d0e46-30d8-11ec-8b04-a772c9f594f9
Morris Bye Element..
Library Search
Media services / Ele.
E What's Going On in.
6 Bases housing U.S.
4 of 6
+
100%
18. A breeding pair of rabbits escaped from their cage behind a farmer's barn. The farmer observed the rabbits and.kept
data on their population size for ten years. During this time, the rabbits reproduced and their offspring reproduced many
times. The fenced, undisturbed two acres where the ribbits live support populations of predators, such as hawks and
snakes, as well as a limited supply of grass and water. At the end of the ten years, the farmer graphed the data he
collected. Which of the following graphs would its population growth look like? Don't pick E, you're welcome.
A)
E)
Time
Time
D
Time
Time
I am finished
azis uoueindo
azis uouejndo
Population size
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Session 2
siology
Questions 44 and 45 are open-response questions.
to dis.
BE SURE TO ANSWER AND LABEL ALL PARTS OF EACH QUESTION.
Show all your work (diagrams, tables, or computations) in your Student Answer Booklet.
If you do the work in your head, explain in writing how you did the work.
Write your answer to question 44 in the space provided in your Student Answer Booklet.
44
The graph below shows the birth rate and the death rate of a mouse population over a
three-year period. The immigration rate and the emigration rate of the population are equal.
Birth Rate and Death Rate
of a Mouse Population
ogg td e
odr
Year 1
Year 2
Year 3
Time
Key
------ Birth rate
Death rate
a. Describe what happens to the size of the population for each year shown on the graph.
Explain your answers.
b. Identify three factors that could affect the death rate of the mouse population, and explain
why each factor affects the death rate.
400
This Way
slife.
Birth and Death Rates
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+ Bio Questions - Google Sheets
b Biology Question | bartleby
b My Questions | bartleby
+ Smartwings Tracking sheet June- x
A https://www.bartleby.com/questions-and-answers/biology-question/282fe512-981d-404b-8986-3fce76bb4d9c
由
* Question Completion Status:
Question 2
name the rythym
Junctional rythm
Moving to the next question prevents changes to this answer.
Ques
Waiting for nk6xemh85d.execute-api.us-east-1.amazonaws.com...
梦 $
02:02
O Type here to search
a
(a ) ENG
06-11-2020
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Task 4: A man is unsure if he is the biological father of all 4 of his children. DNA testing was
done to confirm.
mother
child 2
father child 4
child 1 child 3
1 2 3 4 5 6
Based on these DNA fingerprints, are any of the children not related to the father? If so, which
ones? Explain how you arrived at your conclusion.
d山TO
OCT
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Directions: Answers must be precise and concise. You can use other sources
of information related to the topic. Write down your responses on a separate
answer sheet.
1. What is the role of recombinant DNA technology in improving life?
Elaborate on your answer.
2. How will knowledge of recombinant DNA technology be useful to resolve
the current challenges and issues we are facing today? Cite an example.
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DGM-StudentWS-CL (2).pdf
d Central Dogma and Genetic Med X H
Users/ACER/Downloads/CDGM-StudentWS-CL%20(2).pdf
A Read aloud
V Draw
F Highlight
O Era
Biolnteractive
Central Dogma and Genetic Medicine
Click & Learn
Student Worksheet
OVERVIEW
This worksheet complements the Central Dogma and Genetic Medicine Click & Learn.
PROCEDURE
As you proceed through the Click & Learn, follow the instructions below and answer the questions in the spaces
provided.
1. Let's review! The central dogma of molecular biology refers to the process of gene expression, Write the
definition of gene expression in your own words.
2. Click on the "Central Dogma" menu tab at the top of the screen.
The table below outlines the steps in eukaryotic gene expression. Click on each tab or scroll through the page
and briefly summarize each step below.
Gene
Molecules Involved
Summary
Expression
What molecules and proteins
are involved in this step?
What happens during this step?
Steps
CDGM-StudentWs-C.pdf
Open file
...
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ce Final Part 2 (page 9 of 25)
arkanacollege.edu/mod/quiz/attempt.php?attempt-502922&cmid%3353028page=8
BIology for Non-Sclence Majors I (BIOL-1308-TH-TRA, 2022-FA) Lecture Final Exam-Fail 20Z1 7 Lecture Final Part 2
- TEXARKANA
COLLEGE
My Courses
Diogy ior NOn-Science iMiajuis i (DIOL-1308-TH-TRA, 2022
Question 9
Not yet answered Points out of 1.00
P Flag question
Which sequence correctly lists the different levels of biological organization, from the smallest
and simplest to the largest and most complex?
Select one:
O a. cells-organs-tissues-organ systems-organism
O b. cells-tissues-organs-organ systems-organism
Oc. tissues-organs-organ systems-organisms-cells
O d. tissues-cells-organs-organ systems-organism
ime left 0:40:08
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