Concept explainers
(a)
Interpretation:
The most suitable matching pair of answers should be selected.
Concept introduction:
Photosynthesis is the process by which plants store energy of sunlight as chemical energy. Process happens in the chloroplast of a plant cell. Photosynthesis happens in two main stages - light reaction and dark reaction.

Answer to Problem 23P
- a) Calvin cycle − the dark reactions
Explanation of Solution
Photosynthesis happens via two main stages as light reaction and dark reaction. Light reaction happens in the presence of sunlight. Two photosystems called PS I and PS II absorbs energy of sun light. At the end of light reaction ATP, NADPH and O2 produce. Next, the dark reaction (which is also known as Calvin cycle) starts which uses ATP and NADPH.
For the dark reaction there is no need of sunlight which happens within the stroma. This is a series of
Figure 1: Calvin cycle
(b)
Interpretation:
The most suitable matching pair of answers should be selected.
Concept introduction:
Photosynthesis is the process by which plants store energy of sunlight as chemical energy. Process happens in the chloroplast of a plant cell. Photosynthesis happens in two main stages - light reaction and dark reaction.

Answer to Problem 23P
- b) Rubisco − catalyzes CO2 fixation
Explanation of Solution
Rubisco (Ribulose-1,5-bisphosphate carboxylase/oxygenase), is an enzyme used in the dark reaction of photosynthesis. It catalyzes the carbon fixation process where atmospheric carbon dioxide is converted to glucose by photosynthetic organisms..
The enzyme is capable of fixing both CO2 and O2. This enzyme is comparatively less efficient at low CO2 concentrations and tends to bind O2. This series of reactions start with O2 to finally produce CO2. The process is known as photorespiration.
(c)
Interpretation:
The most suitable matching pair of answers should be selected.
Concept introduction:
Rubisco activase enzyme is important in activating Rubisco. Rubisco (ribulose-1,5-bisphosphate carboxylase/oxygenase), is an enzyme used in the dark reaction of photosynthesis. It catalyzes the carbon fixation process where atmospheric carbon dioxide is converted to glucose.

Answer to Problem 23P
- c) Carbamate − required for the rubisco activity
Explanation of Solution
In plants and some types of green algae, an enzyme called RuBisCO activase is present. This is required to allow the rapid formation of carbamate in the active site of rubisco. RuBisCO activase is required because the ribulose 1,5-bisphosphate (RUBP) substrate binds more strongly to the active sites lacking carbamate and hence slows down the activation process.
(d)
Interpretation:
The most suitable matching pair of answers should be selected.
Concept introduction:
Starch is a

Answer to Problem 23P
- d) Starch − storage form of carbohydrates
Explanation of Solution
Starch is a polymeric carbohydrate produced by joining a large number of glucose units via glycosidic bonds.
Figure 2: Structure of starch
After plants produce glucose by photosynthesis, it is stored in the cells as starch. Starch is insoluble in plant cell. So, unlike starch glucose increases the concentration of the plant cell.
(e)
Interpretation:
The most suitable matching pair of answers should be selected.
Concept introduction:
Sucrose is a disaccharide of Glucose and fructose, having twelve carbon atoms. Disaccharide consists of two monosaccharides, having six carbon atoms each.

Answer to Problem 23P
- a) Sucrose − transport form of carbohydrates
Explanation of Solution
Sucrose consists of a glucose molecule and a fructose molecule joined through a glycosidic bond.
Figure 3: Structure of Sucrose
Plants store glucose (produced in photosynthesis) as starch in different places such as in fruits, seeds, and roots. Starch is insoluble in water and cannot be transported. When the plant needs energy, starch is converted to sucrose and transported. Even though glucose is soluble in water, plant does not convert starch to glucose. Since glucose is highly reactive compared to sucrose, it can transform into other intermediate compounds during transportation. Once it reaches the destination, sucrose is converted to glucose.
(f)
Interpretation:
The most suitable matching pair of answers should be selected.
Concept introduction:
Amylose is a polysaccharide made up of glucose units. These glucose molecules are bonded to each other through a (1→4) glycosidic bonds. Due to its tightly packed helical structure, amylose is resistant to digestive enzymes and therefore known as a resistant starch.

Answer to Problem 23P
- b) Amylose -
Explanation of Solution
As indicated in the figure, the first and fourth carbon of the glucose monomers react to form a bond between them. This bond is known as 1, 4 glycosidic bond.
Figure 4: Structure of amylose
(g)
Interpretation:
The most suitable matching pair of answers should be selected.
Concept introduction:
Amylopectin is a highly branched

Answer to Problem 23P
- c) Amylopectin − includes
Explanation of Solution
Amylopectin is a highly branched polymer. This is also known as polysaccharide. The primary unit is glucose. Like amylose, the glucose units are linked in a linear way with a (11→4) glycosidic bonds. This
Figure 5: Structure of amylopectin
(h)
Interpretation:
The most suitable matching pair of answers should be selected.
Concept introduction:
Photosynthesis is the process by which plants store energy of sunlight as chemical energy. Process happens in the chloroplast of a plant cell. Photosynthesis happens in two main stages - light reaction and dark reaction. The dark reaction consists either of the two

Answer to Problem 23P
- d) C3 plants − 3 − Phosphoglycerate is formed after carbon fixation
Explanation of Solution
The dark reaction of photosynthesis consists of two metabolic pathways, that is, C3 and C4 cycle. All the green plant uses one of these two pathways. The nomenclature is based on the number of carbon atoms present in the first metabolite produced after CO2 fixation.
The first metabolite produced in C3 cycle composed of three carbons while a four carbon metabolite is produced in C4 cycle.
The light reaction is common to both C3 and C4 cycle. But the type of enzymes and intermediate produced during the dark reaction are different.
RUBP (Ribulose Bisphosphate) carboxylase is the enzyme of C3 cycle which fixes CO2, present in the atmosphere. After Co2 fixation, a five-carbon molecule, RUBP (Ribulose bisphosphate) reacts with CO2 to form a six-carbon complex. The six-carbon molecule break immediately and form two 3- phosphoglycerate. 3- phosphoglycerate consists of three carbon atoms.
So, this is the first stable product of CO2 fixation. This is the reason why the mechanism is known as C3 cycle.
Figure 6: Calvin cycle of C3photosynthesis representing the carbon fixation.
(i)
Interpretation:
The most suitable matching pair of answers should be selected.
Concept introduction:
Photosynthesis is the process by which plants store energy of sunlight as chemical energy. Process happens in the chloroplast of a plant cell. Photosynthesis happens in two main stages - light reaction and dark reaction. The dark reaction consists either of the two metabolic pathways, that is, C3 and C4 cycle.

Answer to Problem 23P
- e) C4 plants − carbon fixation results in oxaloacetate formation
Explanation of Solution
The dark reaction of photosynthesis consists of two metabolic pathways, that is, C3 and C4 cycle. All the green plant uses one of these two pathways. The nomenclature is based on the number of carbon atoms present in the first metabolite produced after CO2 fixation.
As explained earlier, the first metabolite produced in C3 cycle composed of three carbons while a four-carbon metabolite is produced in C4 cycle.
Out of the two steps of photosynthesis (light and dark reaction), light reaction is common. But the type of enzymes and intermediate produced during the dark reaction are different.
PEP (Phosphoenolpyruvate) carboxylase is the primary enzyme of C4 mechanism which catalyzes the fixation of CO2 in chloroplast. For this, a three-carbon molecule (PEP) reacts with CO2 to form a four-carbon molecule which is known as oxaloacetate. Oxaloacetate is the first stable product of C4 cycle.
This is the reason why the mechanism is called C4 mechanism. The reactions take place in mesophyll cells.
(j)
Interpretation:
The most suitable matching pair of answers should be selected.
Concept introduction:
Stomata are pores, found in the epidermis of leaves, stems, and other organs that facilitate gaseous exchange.

Answer to Problem 23P
- f) Stomata − allow exchange of gases
Explanation of Solution
Stomata are the pores which are mostly found in the epidermis of leaves and stems. It facilitates gaseous exchange. This stomatal pore is bordered by the guard cells which regulates the stomatal opening and closing.
Atmospheric air consists of carbon dioxide and oxygen enters the plant through these openings. These gases are crucial for photosynthesis and respiration, respectively. Also, water vapor releases to the atmosphere through the stomata. The process is known as transpiration.
So, stomata are the structures, present in the plant leaves crucial for all types of gaseous exchanges.
Figure 8: Microscopy view of stomata
Want to see more full solutions like this?
Chapter 20 Solutions
BIOCHEMISTRY (LOOSELEAF)-W/ACCESS
- 2. Which one is the major organic product obtained from the following reaction sequence? HO A OH 1. NaOEt, EtOH 1. LiAlH4 EtO OEt 2. H3O+ 2. H3O+ OH B OH OH C -OH HO -OH OH D E .CO₂Etarrow_forwardwhat is a protein that contains a b-sheet and how does the secondary structure contributes to the overall function of the protein.arrow_forwarddraw and annotate a b-sheet and lable the hydrogen bonding. what is an example that contains the b-sheet and how the secondary structure contributes to the overall function of your example protein.arrow_forward
- Four distinct classes of interactions (inter and intramolecular forces) contribute to a protein's tertiary and quaternary structures. Name the interaction then describe the amino acids that can form this type of interaction. Draw and annotate a diagram of the interaction between two amino acids.arrow_forwardExamine the metabolic pathway. The enzymes that catalyze each step are identified as "e" with a numeric subscript. e₁ e3 e4 A B с 1° B' 02 e5 e6 e7 E F Which enzymes catalyze irreversible reactions? ப e ez ☐ ez e4 ☐ ப es 26 5 e7 Which of the enzymes is likely to be the allosteric enzyme that controls the synthesis of G? €2 ез e4 es 26 5 e7arrow_forwardAn allosteric enzyme that follows the concerted model has an allosteric coefficient (T/R) of 300 in the absence of substrate. Suppose that a mutation reversed the ratio. Select the effects this mutation will have on the relationship between the rate of the reaction (V) and substrate concentration, [S]. ㅁㅁㅁ The enzyme would likely follow Michaelis-Menten kinetics. The plot of V versus [S] would be sigmoidal. The enzyme would mostly be in the T form. The plot of V versus [S] would be hyperbolic. The enzyme would be more active.arrow_forward
- Penicillin is hydrolyzed and thereby rendered inactive by penicillinase (also known as ẞ-lactamase), an enzyme present in some penicillin-resistant bacteria. The mass of this enzyme in Staphylococcus aureus is 29.6 kDa. The amount of penicillin hydrolyzed in 1 minute in a 10.0 mL. solution containing 1.00 x 10 g of purified penicillinase was measured as a function of the concentration of penicillin. Assume that the concentration of penicillin does not change appreciably during the assay. Plots of V versus [S] and 1/V versus 1/[S] for these data are shown. Vo (* 10 M minute"¹) 7.0 6.0 5.0 4.0 3.0 20 1.0 0.0 о 10 20 30 1/Vo (* 10 M1 minute) 20 103 90 BO 70 50 [S] (* 100 M) 40 50 60 y=762x+1.46 × 10" [Penicillin] (M) Amount hydrolyzed (uM) 1 0.11 3 0.25 5 0.34 10 0.45 30 0.58 50 0.61arrow_forwardConsider the four graphs shown. In each graph, the solid blue curve represents the unmodified allosteric enzyme and the dashed green curve represents the enzyme in the presence of the effector. Identify which graphs correctly illustrate the effect of a negative modifier (allosteric inhibitor) and a positive modifier (allosteric activator) on the velocity curve of an allosteric enzyme. Place the correct graph in the set of axes for each type of modifier. Negative modifier Reaction velocity - Positive modifier Substrate concentration - Reaction velocity →→→→ Substrate concentration Answer Bankarrow_forwardConsider the reaction: phosphoglucoisomerase Glucose 6-phosphate: glucose 1-phosphate After reactant and product were mixed and allowed to reach at 25 °C, the concentration of each compound at equilibrium was measured: [Glucose 1-phosphate] = 0.01 M [Glucose 6-phosphate] = 0.19 M Calculate Keq and AG°'. Код .0526 Incorrect Answer 7.30 AG°' kJ mol-1 Incorrect Answerarrow_forward
- Classify each phrase as describing kinases, phosphatases, neither, or both. Kinases Phosphatases Neither Both Answer Bank transfer phosphoryl groups to acidic amino acids in eukaryotes may use ATP as a phosphoryl group donor remove phosphoryl groups from proteins catalyze reactions that are the reverse of dephosphorylation reactions regulate the activity of other proteins catalyze phosphorylation reactions PKA as an example turn off signaling pathways triggered by kinasesarrow_forwardConsider the reaction. kp S P kg What effects are produced by an enzyme on the general reaction? AG for the reaction increases. The rate constant for the reverse reaction (kr) increases. The reaction equilibrium is shifted toward the products. The concentration of the reactants is increased. The activation energy for the reaction is lowered. The formation of the transition state is promoted.arrow_forwardThe graph displays the activities of wild-type and several mutated forms of subtilisin on a logarithmic scale. The mutations are identified as: • The first letter is the one-letter abbreviation for the amino acid being altered. • The number identifies the position of the residue in the primary structure. ⚫ The second letter is the one-letter abbreviation for the amino acid replacing the original one. • Uncat. refers to the estimated rate for the uncatalyzed reaction. Log₁(S-1) Wild type S221A H64A -5 D32A S221A H64A D32A -10 Uncat. How would the activity of a reaction catalyzed by a version of subtilisin with all three residues in the catalytic triad mutated compare to the activity of the uncatalyzed reaction? It would have more activity, because the reaction catalyzed by the triple mutant is approximately three-fold faster than the uncatalyzed reaction. It would have less activity, because the reaction catalyzed by the triple mutant is approximately 1000-fold slower than the…arrow_forward
- Biology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage LearningBiochemistryBiochemistryISBN:9781305577206Author:Reginald H. Garrett, Charles M. GrishamPublisher:Cengage LearningBiology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning
- Anatomy & PhysiologyBiologyISBN:9781938168130Author:Kelly A. Young, James A. Wise, Peter DeSaix, Dean H. Kruse, Brandon Poe, Eddie Johnson, Jody E. Johnson, Oksana Korol, J. Gordon Betts, Mark WomblePublisher:OpenStax CollegeBiology: The Unity and Diversity of Life (MindTap...BiologyISBN:9781305073951Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa StarrPublisher:Cengage LearningBiology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStax





