Cell Biology Exam 3

Photosynthesis, Cell Cycle, Cell Division

What is the summary reaction for photosynthesis (light and carbon reaction?)

Overall equation (combining light and carbon reactions):

6CO2+6H2O+light energy→C6H12O6+6O26 CO_2 + 6 H_2O + light\ energy \rightarrow C_6H_{12}O_6 + 6 O_26CO2​+6H2​O+light energy→C6​H12​O6​+6O2​

• Light reactions (in thylakoid membranes):
  Use light to split water and produce ATP, NADPH, and O₂.

  2H2O+2NADP++3ADP+3Pi+light→O2+2NADPH+3ATP2 H_2O + 2 NADP^+ + 3 ADP + 3 P_i + light \rightarrow O_2 + 2 NADPH + 3 ATP2H2​O+2NADP++3ADP+3Pi​+light→O2​+2NADPH+3ATP

• Carbon (dark) reactions (in stroma):
  Use ATP and NADPH to fix CO₂ into carbohydrates.

How is the energy from sunlight converted into chemical energy (ATP)?

Energy conversion occurs through photosystems and the electron transport chain (ETC) in the thylakoid membrane.

1. Photon absorption:

   • Light energy excites electrons in chlorophyll molecules of Photosystem II (PSII) (P680).

   • The excited electrons are transferred to a primary electron acceptor.

2. Water splitting (photolysis):

   • To replace the lost electrons, PSII splits water into 2 e⁻, 2 H⁺, and ½ O₂.

   • Electrons go to PSII, protons contribute to the gradient, O₂ diffuses out.

3. Electron transport:

   • Electrons travel down the ETC (via plastoquinone → cytochrome b6f → plastocyanin).

   • As they move, protons are pumped into the thylakoid lumen, creating a proton gradient.

4. ATP synthesis (photophosphorylation):

   • Protons flow back through ATP synthase to the stroma.

   • The energy drives the conversion of ADP + Pi → ATP.

5. Photosystem I (PSI):

   • Light excites electrons again (P700).

   • Electrons are passed to ferredoxin (Fd) and finally to NADP⁺ reductase, forming NADPH.

Thus, sunlight → excited electrons → proton gradient → ATP + NADPH.

Describe three things that are accomplished by oxidation of H20 and why they’re useful.

1. Supplies electrons to replace those lost by PSII → keeps electron flow continuous.

2. Releases O₂ → provides atmospheric oxygen for respiration and life.

3. Contributes protons (H⁺) to the thylakoid lumen → helps build the proton gradient used to synthesize ATP.

All three are essential for maintaining photosynthesis and life.

What is the final electron acceptor in the light reactions?

→ NADP⁺, which is reduced to NADPH by NADP⁺ reductase at the end of the light reactions.

Describe cyclic electron flow. How does it compare to non-cyclic electron flow in terms of the following :

a. ATP production

b. NADP+ reduction

c. H2O Oxidation

• Involves Photosystem I only.

• Electrons from ferredoxin are cycled back to cytochrome b6f instead of reducing NADP⁺.

• Generates ATP only, no NADPH, and no O₂ is produced.

Feature	Noncyclic flow	Cyclic flow
Photosystems	PSII + PSI	PSI only
ATP production	Yes	Yes (extra)
NADPH production	Yes	No
H₂O oxidation	Yes	No
O₂ release	Yes	No

Cyclic flow helps balance the ATP:NADPH ratio, since the Calvin cycle consumes more ATP than NADPH.

What is the first product of carbon fixation? Describe the experiment used to discover it.

• The first stable product is 3-phosphoglycerate (3-PGA), a 3-carbon compound.

• Discovery: Melvin Calvin and colleagues used ¹⁴CO₂ labeling in Chlorella algae. By exposing cells to light and quenching reactions at different times, they traced radioactive carbon into intermediates — 3-PGA was the first labeled compound.

What are the three different phases of the Calvin-Benson cycle? Describe what takes place during each stage (input/output).

Phase	Description	Inputs	Outputs
1. Carbon fixation	CO₂ combines with ribulose-1,5-bisphosphate (RuBP) via Rubisco, forming 3-PGA	CO₂, RuBP	3-PGA
2. Reduction	3-PGA is phosphorylated by ATP and reduced by NADPH to form glyceraldehyde-3-phosphate (G3P)	3-PGA, ATP, NADPH	G3P, ADP, NADP⁺
3. Regeneration	Some G3P regenerates RuBP using ATP	G3P, ATP	RuBP

• Net reaction for 3 CO₂ fixed:

  3CO2+9ATP+6NADPH→1G3P+9ADP+8Pi+6NADP+3 CO_2 + 9 ATP + 6 NADPH \rightarrow 1 G3P + 9 ADP + 8 Pi + 6 NADP^+3CO2​+9ATP+6NADPH→1G3P+9ADP+8Pi+6NADP+

What is the precursor molecule used for fixation of CO2 (molecule x)? What is the name of the enzyme (shorthand) which catalyzes this reaction? What are its limitations?

• Precursor (molecule X): Ribulose-1,5-bisphosphate (RuBP)

• Enzyme: Rubisco (Ribulose bisphosphate carboxylase/oxygenase)

• Limitations:

  • Rubisco can bind O₂ instead of CO₂, leading to photorespiration (wastes energy).

  • It has a low affinity for CO₂ and is relatively slow.

Define photorespiration and why it’s detrimental to the cell. How do C4 plants handle the dark reactions in order to limit photorespiration? How does this differ from C3 plants.

• Occurs when Rubisco fixes O₂ instead of CO₂, producing phosphoglycolate, which cannot enter the Calvin cycle.

• The cell must recycle it through energy-consuming reactions that release CO₂.

• Detrimental because:

  • Consumes ATP and reducing power.

  • Releases CO₂, undoing fixation.

  • Does not produce sugars.

Why do C3 plants typically require temperate, high CO2 climates?

• C4 plants (e.g., maize, sugarcane) spatially separate CO₂ fixation and the Calvin cycle:

  • In mesophyll cells, CO₂ is fixed by PEP carboxylase into 4-carbon oxaloacetate (OAA → malate).

  • Malate diffuses into bundle sheath cells, where CO₂ is released and refixed by Rubisco.

  • This maintains high CO₂ concentration near Rubisco → minimizes oxygenase activity.

C3 plants fix CO₂ directly with Rubisco in mesophyll cells — no CO₂ concentration mechanism → more photorespiration, especially in hot, dry climates.

Define the following terms: Diploid, Haploid, Germ Line, Somatic Cell, Zygote

Term	Definition
Diploid (2n)	A cell that contains two sets of chromosomes, one from each parent. Most body (somatic) cells are diploid.
Haploid (n)	A cell that contains one set of chromosomes. Gametes (sperm and egg) are haploid.
Germ Line	The cell lineage that gives rise to gametes (sperm or egg). These cells can undergo meiosis.
Somatic Cell	Any body cell that is not part of the germ line. Somatic cells divide by mitosis and are diploid.
Zygote	The diploid cell formed when two haploid gametes fuse during fertilization (n + n → 2n). The zygote develops into a new organism.

For a species with the ploidy number 2n = 12: How many unique chromosomes are present? How many copies are there of each chromosome?

Question	Answer
How many unique chromosomes?	There are 6 unique chromosomes (since 2n = 12 → n = 6).
How many copies of each chromosome?	2 copies of each unique chromosome — one maternal and one paternal homolog.

Can a haploid cell undergo both mitosis and meiosis? What about diploid cells? Explain.

Cell Type	Mitosis	Meiosis	Explanation
Haploid (n)	✅ Yes	❌ No	Haploid cells can undergo mitosis because their DNA can replicate and divide equally, but they cannot undergo meiosis — they only have one set of chromosomes, so they can’t pair homologs.
Diploid (2n)	✅ Yes	✅ Yes	Diploid cells can undergo mitosis (to make identical diploid cells) or meiosis (to make haploid gametes).

Compare and contrast mitosis and meiosis, addressing each of the following: How parent cells compare to daughter cells Ploidy at the beginning Number of daughter cells and their ploidy

Feature	Mitosis	Meiosis
Parent vs. Daughter Cells	Daughter cells are genetically identical to the parent	Daughter cells are genetically unique
Starting Ploidy	Diploid (2n)	Diploid (2n)
Number of Divisions	One	Two (Meiosis I and Meiosis II)
Number of Daughter Cells	2	4
Daughter Cell Ploidy	Diploid (2n)	Haploid (n)
Purpose	Growth, repair, asexual reproduction	Gamete formation, sexual reproduction
Crossing Over	No	Yes (in Prophase I)
Homolog Pairing	No	Yes (synapsis in Prophase I)

Draw a representation of what takes place during Meiosis I and Meiosis II. Your diagram should focus specifically on Metaphase and Anaphase.

(Insert Image)

List the pros and cons of: Asexual reproduction Sexual reproduction

Type	Pros	Cons
Asexual Reproduction	- Fast and energy-efficient - Only one parent needed - Produces many offspring quickly	- No genetic diversity - All offspring are clones (vulnerable to disease/environmental change)
Sexual Reproduction	- Creates genetic variation (important for evolution and adaptation)	- Requires two parents - Slower and more energy costly - Only half the population (females) produce offspring

Describe three ways that meiosis contribute to genetic variation within a population. Make sure to discuss how each contributes variation.

Mechanism	Description	How It Creates Variation
1. Crossing Over (Prophase I)	Homologous chromosomes exchange segments of DNA.	Produces new combinations of alleles on chromosomes (recombinant chromosomes).
2. Independent Assortment (Metaphase I)	Homologous chromosome pairs align independently of other pairs.	Each gamete gets a random mix of maternal and paternal chromosomes.
3. Random Fertilization	Any sperm can fertilize any egg.	Enormous number of possible zygote combinations (genetic lottery).

How does non-disjunction contribute to aneuploidy? Why does this generally result in miscarriage or congenital diseases?

What is non-disjunction?

Non-disjunction occurs when chromosomes fail to separate properly during meiosis I or meiosis II.

• Meiosis I error: homologous chromosomes don’t separate.

• Meiosis II error: sister chromatids don’t separate.

How it leads to aneuploidy

As a result, gametes receive an abnormal number of chromosomes:

• Some gametes have n + 1 chromosomes.

• Others have n – 1 chromosomes.

When such a gamete fuses with a normal one:

• Trisomy (2n + 1): extra chromosome (e.g., Down syndrome = trisomy 21)

• Monosomy (2n – 1): missing chromosome (e.g., Turner syndrome = XO)

Why this causes miscarriage or disease

• Embryonic development requires precise gene dosage.

• Too many or too few gene copies disrupt developmental signaling and metabolic balance.

• Most aneuploid embryos die early (miscarriage).

• Some (like trisomy 21, 18, 13, XO) survive but have congenital abnormalities due to gene imbalance.

How does apoptosis differ from necrosis? What are some typical reasons that apoptosis might take place? List at least three examples.

Feature	Apoptosis	Necrosis
Definition	Programmed, controlled cell death	Accidental, uncontrolled cell death
Process	Cell shrinks, DNA fragments, membrane blebs, contents packaged into vesicles (apoptotic bodies) for phagocytosis	Cell swells and bursts, releasing contents and causing inflammation
Outcome	Non-inflammatory, part of normal physiology	Inflammatory, damaging to surrounding tissue

Typical reasons for apoptosis

1. DNA damage that cannot be repaired.

2. Developmental shaping (e.g., removing webbing between fingers in embryos).

3. Elimination of infected or cancerous cells.

4. Loss of survival signals or growth factors.

5. Immune system regulation (removal of self-reactive lymphocytes).

Describe the role of caspases in apoptosis and how they are regulated (ie, how are they kept inactive and how are they activated?)

Role

Caspases are cysteine-aspartate proteases that execute apoptosis by cleaving target proteins to dismantle the cell.

• Initiator caspases (e.g., Caspase-8, -9): activated first, cleave and activate effector caspases.

• Executioner caspases (e.g., Caspase-3, -7): degrade structural and regulatory proteins → DNA fragmentation, cytoskeleton collapse, membrane blebbing.

Regulation

• Inactive form: Caspases are made as zymogens (procaspases) that are harmless until cleaved.

• Activation: Triggered by cleavage at specific aspartate sites — usually by an upstream caspase or apoptosome complex.

• Inhibitors: IAPs (Inhibitor of Apoptosis Proteins) can bind and block active caspases; mitochondrial proteins (like Smac/DIABLO) can relieve this inhibition.

Describe in detail how the following mechanisms regulates apoptosis: Bax/Bak, Fas and Fas ligand, Survival factors

1. Bax/Bak (Intrinsic or Mitochondrial Pathway)

• Bax and Bak are pro-apoptotic proteins of the Bcl-2 family.

• When activated by cell stress (e.g., DNA damage, lack of survival factors), they:

  1. Oligomerize in the mitochondrial outer membrane.

  2. Form pores → cytochrome c is released into the cytosol.

  3. Cytochrome c binds Apaf-1 → forms apoptosome → activates caspase-9, leading to apoptosis.

Bcl-2 and Bcl-xL are anti-apoptotic and inhibit Bax/Bak to keep mitochondria intact.

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2. Fas and Fas Ligand (Extrinsic or Death Receptor Pathway)

• Fas (CD95): receptor on the cell surface.

• Fas ligand (FasL): often expressed on cytotoxic T cells.

• When FasL binds Fas:

  1. Fas receptors trimerize → recruit adaptor protein FADD.

  2. FADD recruits procaspase-8 → forms the DISC (death-inducing signaling complex).

  3. Caspase-8 is activated → activates executioner caspases (3, 7) → apoptosis.

This allows the immune system to eliminate infected or cancerous cells directly.

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3. Survival Factors

• Survival factors (e.g., growth factors, cytokines) bind to cell-surface receptors that activate intracellular signaling (like PI3K–Akt pathway).

• Akt phosphorylates and inactivates Bad (a pro-apoptotic Bcl-2 family member).

• When Bad is inactive, Bcl-2 and Bcl-xL inhibit Bax/Bak, preventing cytochrome c release.

➡ Without survival factors, the balance shifts toward apoptosis.

Describe the general role of p53 as a tumor suppressor. Why is one functional copy of p53 enough to increase survival rates in cancer patients?

General role

• p53 is a transcription factor activated by DNA damage, oncogene activation, or stress.

• Functions:

  1. Stops the cell cycle (by activating p21 → inhibits CDKs).

  2. Activates DNA repair genes.

  3. If damage is too severe, initiates apoptosis (by inducing Bax, Puma, Noxa).

Thus, p53 maintains genomic integrity — often called the “guardian of the genome.”

Why one functional copy helps

• p53 acts dominantly: one working allele is enough to produce enough p53 protein to activate checkpoints and apoptosis.

• Losing both copies (as in many cancers) removes this protection, allowing damaged cells to proliferate.

For each scenario, state whether p53 and mdm2 would be active or inactive/degraded. Explain your answer: Cell does not receive a mitogenic or survival signal and the DNA is damaged. Cell receives a mitogenic or survival signal and the DNA is undamaged. Cell receives a mitogenic or survival signal and the DNA is damaged.

Scenario	p53	Mdm2	Explanation
1. No mitogenic/survival signal, DNA damaged	Active	Inactive/degraded	DNA damage stabilizes p53 by preventing Mdm2 binding. p53 activates repair or apoptosis.
2. Mitogenic/survival signal, DNA undamaged	Inactive (low level)	Active	Normal growth conditions: Mdm2 ubiquitinates p53 → degradation, allowing proliferation.
3. Mitogenic/survival signal, DNA damaged	Active	Inactive/degraded	Despite growth signals, DNA damage triggers kinases (ATM/ATR) that phosphorylate p53, preventing Mdm2 binding. Cell cycle arrest/apoptosis ensues to prevent mutation propagation.