BIO 121 Practice Exam Questions And Already Passed Answers.
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Bio 121
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Bio 121
Eukaryotes employ several forms of sexual reproduction
A. Vegetative reproduction that involves mitosis
B. Parthenogenesis that is the production of an organism from an unfertilized egg
C. Hermaphroditism that involves both gametes from a single parent
D. Two of the above
E. All of the above ...
BIO 121 Practice Exam Questions And
Already Passed Answers.
Eukaryotes employ several forms of sexual reproduction
A. Vegetative reproduction that involves mitosis
B. Parthenogenesis that is the production of an organism from an unfertilized egg
C. Hermaphroditism that involves both gametes from a single parent
D. Two of the above
E. All of the above - Answer C
Hermaphroditism is most common among species that: - Answer Are sessile (organisms fixed in place)
A cell has a diploid chromosome number of 2n = 4. We will designate these four as chromosomes "A",
"B", "C", and "D". If meiosis occurred WITH the formation of homologous pairs "AC" and "BD", and the
chromosomes were then distributed randomly between the resulting cells, how many gametes could be
formed? - Answer 4 (AB, AD, BC, CD)
In wonderland, the red queen said to alice, "You have to run faster than that just to stay in place."
According to the material in the course, the biological hypothesis that corresponds to this idea is: -
Answer Genetic variability is needed to keep up with new adaptations of potential predators and
parasites
When Chlamydomonas utilizes sexual reproduction:
A. Adults use mitosis to produce haploid gametes which are morphologically indistinguishable
B. Stressful environmental conditions such as nutrient shortages appear to trigger the onset of this
reproductive mode
C. Syngamy and then meiosis combine to produce haploid adults
D. More than one of the above
E. All of the above - Answer E. All of the above
Some cells do not divide (muscle/nerve cells). This means that during the cell cycle, they remain in:
,A. G0 phase
B. Gi phase
C. Interphase
D. Two of the above
E. All of the above - Answer D. Two of the above
Pairs of homologous chromosomes _____
A. are exact copies of each other
B. Share a single centromere
C. Separate during anaphase II
D. Are present together in the cells produced at the end of meiosis
E. None of the above - Answer E. None of the Above
Cancer kills millions of people worldwide. Characteristics of cancer cells include:
A. Mutations that activate oncogenes so that there is a constant signal to divide
B. Enormous ability to replicate themselves and metastisize
C. Mutation that inactivate tumor suppressor genes that normally inhibit division
D. Two of the above
E. All of the above - Answer E. All of the above
Following meiosis II, an organism's gametes contain 7 chromosomes. In prophase I, this dividing cell
contained ___ chromosomes with __ chromatids each. - Answer 14, 2
The diploid chromosome number for a certain plant species is 10. In a section of tissue from the plant
you observe a dividing cell and notice its chromosomes are in two separate clusters. Closer examination
reveals that each group contains 5 double stranded chromosomes. The cell is in: - Answer Anaphase of
meiosis I
Below is a graph of CO2 concentration in the northern (solid line) and southern hemisphere (dashed
kine). What is most likely driving these oscillations? - Answer In the northern hemisphere,
photosynthesis by plants in summer generates a CO2 decrease
,During Pre-synthesis interphase before meiosis a cell has 1 g of DNA. Are each of the resulting cells
diploid or haploid and how many grams of DNA does each cell contain at the end of meiosis 1 and
meiosis 2? - Answer End of meiosis 1: Haploid with 1 g DNA; End of meiosis @: Haploid with 0.5 g DNA
Genetic differences among siblings can be the results of:
A. Nondisjunction
B. Chromosomes from mom and dad lining up independently during telophase I
C. Crossing over (recombination)
D. Two of the above
E. all of the above - Answer C. Crossing Over
In terms of photosynthesis with which of the following would you agree?
A. the wavelengths of light that are absorbed by pigments are generally the ones that result in high PHS
activity
B. the source of carbon for the dark reaction is CO2
C. Oxygen produced during photosynthesis comes from the cleavage of H2O
D. two of the above
E. all of the above - Answer all of the above
CAM plants have evolved a particular adaptation that allows tham to inhabit hot, dry environments and
this adaptation involves: - Answer The utilization production of PEP corboxylase
In terms of reproduction with which of the following would you agree?
A. the random alignment of homologues in meiosis I can lead to genetic variation
B. babies conceived using IVF exhibit genetic variation among siblings
C. Sexual reproduction is not necessary for the evolution of all organisms
D. two of the above
E. all of the above - Answer E. all of the above
, Imagine Photosystem I is broken but all other parts of photosynthesis function normally. What molecules
accumulate in the Calcin cycle and why? - Answer PGA accumulates because there is not enough
NADPre to supply electrons for the reaction
although c3 and c4 plants both perform the Calvin cycle, differences exist in which of the following ways?
A. primary CO2 acceptor molecules; C3 plants use RuBp and C4 plants use PEP
B. C4 plants use more energy than C3 plants to get CO2 into the Calvin Cycle
C. The level of photorespiration can be extensive in C3 plants, but minimal in C4 plants
D. Two of the above
E. All of the above - Answer All of the above
If photosynthesizing green algae are provided with CO2 synthesized with heavy oxygen (18O), later
analysis will show which of the following compounds does not contain the 18O label - Answer O2
Suppose that you are measuring CO2 exchange by a leaf. During the time period of your measurement, 8
x 10^6 molecules of phosphoglyceraldehyde (PGAL) are made from CO2 in photosynthesis and 6 x 10^6
molecules of pyruvate are used in cellular respiration. How many net molecules of CO2 would you detect
being taken up by the lead in your measurement? - Answer 8 x 10^6 PGAL (3 Carbons) = 24 x 10^6
6 x 10^6 pyruvates (3 Carbons) = 18 x 10^6
Total = 24-18 = 6 x 10^6
Vitamin B1 is an essential vitamin that is required to help convert pyruvate to acetyl-CoA. If an animal
doesn't have enough vitamin B1 to convert pyruvate to acetyl-CoA, what other molecule could be
produced instead? - Answer Lactic Acid
A cell is carrying out cellular respiration of glucose when suddenly ATP synthase is degraded. What
percentage of ATP will be produced in this scenario compared to regular cellular respiration? - Answer
Glucose Net of 2
Krebs Net of 2
=4
4/total of 30 = 13%
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