CBSE Sample Papers for Class 11 Biology Set 5 with Solutions

Students can access the CBSE Sample Papers for Class 11 Biology with Solutions and marking scheme Set 5 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 11 Biology Set 5 with Solutions

Time : 3 Hours
Maximum Marks : 70

General Instructions:

1. All questions are compulsory.
2. The question paper has five sections and 33 questions. All questions are compulsory.
3. Section-A has 16 questions ofl mark each; Section-B has 5 questions of 2 marks each; Section- C has 7 questions of 3 marks each; Section- Dhas 2 case-based questions of 4 marks each; and Section-E has 3 questions of 5 marks each.
4. There is no overall choice. However, internal choices have been provided in some questions. A student has to attempt only one of the alternatives in such questions.
5. Wherever necessary, neat and properly labeled diagrams should be drawn.

Section – A

Question 1.
The reaction that is responsible for the primary fixation of C02 is catalysed by: 1
(A) RuBP carboxylase
(B) PEP carboxylase
(C) Both (A) and (B)
(D) PCA synthase
Answer:
Option (C) is correct.
Explanation: In the C₃ cycle, the enzyme RuBP Carboxylase is responsible for fixing atmospheric carbon dioxide. It facilitates the interaction between CO₂ and RuBP the major acceptor of CO₂. It is found in both C₃ and C₄ plants.

PEP carboxylase: PEP carboxylase is a catalyst for the C₄ pathway, which fixes atmospheric carbon dioxide with phosphoenolpyruvate (PEP). Only C₄ plants contain it; C₃ plants do not.

Question 2.
Electron Transport System (ETS) is located in mitochondrial: 1
(A) Outer membrane.
(B) Inter membrane space.
(C) Inner membrane.
(D) Matrix.
Answer:
Option (C) is correct.
Explanation: A group of coenzymes and cytochromes that participate in the movement of electrons from one carrier to another and to their final acceptor is known as an electron transport system (ETS) or electron transport chain (ETC). In the inner mitochondrial membrane, the various ETS components are grouped in five different types of complexes, each in a particular sequence.

Question 3.
Match the following and mark the correct option. 1

Column I	Column II
(a) Sternum	(i) Synovial fluid
(b) Glenoid Cavity	(ii) Vertebrae
(c) Freely movable joint	(iii) Pectoral girdle
(d) Cartilaginous joint	(iv) Flat bones

(A) (a) – (ii), (b) – (i), (c) – (iii), (d) – (iv)
(B) (a) – (iv), (b) – (iii), (c) – (i), (d) – (ii)
(C) (a) – (ii), (b) – (i), (c) – (iv), (d) – (iii)
(D) (a) – (iv), (b) – (i), (c) – (ii), (d) – (iv)
Answer:
Option (B) is correct.
Explanation: The thoracic basket contains the sternum, a flat bone which connects the ribs together.
The upper arm’s pectoral girdle contains the glenoid cavity.
The synovial fluid on freely moving joints allows for greater movement directions.
The cartilaginous joint is present between the vertebrae.

Question 4.
Amino acids have both an amino group and a carboxyl group in their structure. Which one of the following is an
amino acid? 1
(A) Formic acid
(B) Methane
(C) Phenol
(D) Glycine
Answer:
Option (D) is correct.
Explanation: The only amino acid with both an amino and a carboxyl group is glycine. While glycerol is a fatty acid and glycolic acid is a carboxylic acid substituted with a hydroxyl group, formic acid is the simplest carboxylic acid.

Question 5.
Coconut water contains: 1
(A) ABA
(B) Auxin
(C) Cytokinin
(D) Gibberellin
Answer:
Option (C) is correct.
Explanation: Cytokinin is abundant in coconut milk.

Question 6.
Plasticity in plant growth means that: 1
(A) Plant roots are extensible.
(B) Plant growth is dependent on the environment.
(C) Stems can extend.
(D) None of the above
Answer:
Option (B) is correct.
Explanation: The capacity of a plant to change its growth, development, and metabolic rate in order to adapt to a certain environment is known as plasticity. It enables the plant to begin cell division from any tissue, regenerate missing organs, and go through many developmental routes for its survival.

Question 7.
The giant redwood tree (Sequoia sempervirens) is a/ an: 1
(A) Angiosperm.
(B) Free fern.
(C) Pteridophyte.
(D) Gymnosperm.
Answer:
Option (D) is correct.
Explanation: A gymnosperm is the giant redwood tree. These coniferous gymnosperms have naked seeds and are always present with their green needle-shaped leaves. They are gymnosperms since they lack flowers and typically grow in mountainous regions where precipitation is more prominent.

Question 8.
The figure given below shows a schematic plan of blood circulation in humansjyith labels (i) to (iv). Identify the correct label with its functions? 1

(A) (i) Pulmonary vein – takes impure blood from the body part.
(B) (ii) Pulmonary artery – takes blood from lung to heart.
(c) (iii) Aorta – takes blood from heart to body parts.
(d) (iv) Vena cava – takes blood from body parts to right auricle
Answer:
Option (D) is correct.
Explanation: (i) Pulmonary vein – collects the oxygenated blood and carries it from the lungs back to the heart.
(ii) Pulmonary artery – carry blood from the right ventricle of the heart to the lungs.
(iii) Aorta – carries oxygen-rich blood from the left ventricle of the heart to other parts of the body.
(iv) Vena cava – carries deoxygenated blood from other areas of the body to the heart.

Question 9.
The term synergistic action of hormones refers to: 1
(a) When two hormones act together but bring about opposite effects.
(b) When two hormones act together and contribute to the same function.
(c) When one hormone affects more than one function.
(d) When many hormones bring about any one function.
Answer:
Option (B) is correct.
Explanation: It has been discovered that two or more phytohormones operating synergistically or antagonistically influence the growth, differentiation, and developmental processes of plants. For example, auxins and cytokinin’s functioning synergistically increase cell division. Auxins, gibberellins, and cytokinin’s control cell proliferation. Auxins and cytokinin’s interact during morphogenesis, and both substances must be present at a specific concentration.

Question 10.
Given below is the representation of a certain event at a particular stage of a type of cell division. Identify the stage?

(A) Prophase of mitosis.
(B) Both prophase and metaphase of mitosis.
(C) Prophase I during meiosis.
(D) Prophase II during meiosis.
Answer:
Option (C) is correct.
Explanation: The given stage of cell division is the crossing over during pachytene of prophase I of meiosis – I.

Question 11.
Energy required for ATP synthesis in PS II comes from:
(A) Proton gradient
(B) Electron gradient
(C) Reduction of glucose
(D) Oxidation of glucose
Answer:
Option (A) is correct.
Explanation: The protons produced as a result of this reaction are confined in the thylakoid lumen because it occurs on the inner side of the membrane. These protons are transferred across the membrane by electrons passing via photosystems. This gradient is significant because it causes the release of energy as it breaks down.

Question 12.
Which chlorophyll is termed as universal photosynthetic pigment ?
Wavelength of light (nm) and color

(A) Chlorophyll ‘b’
(B) Chlorophyll ‘a’
(C) Carotenoids
(D) Chloroplast
Answer:
Option (B) is correct.
Explanation: Chlorophyll ‘a’ is termed as universal photosynthetic pigment.

Question No. 13 to 16 consist of two statements – Assertion (A) and Reason (R). Answer the questions selecting the appropriate option given below.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

Question 13.
Assertion (A): Linnaeus is regarded as the father of taxonomy. 1
Reason (R): He developed the binomial system of nomenclature and the system of classification.
Answer:
Option (A) is correct.
Explanation: Linnaeus has been considered as the “father of taxonomy” since, he developed the binomial system of nomenclature and system of classification.

Question 14.
Assertion (A): Haemoglobin has a quaternary structure. 1
Reason (R): Haemoglobin has four chains, two α-chains and two β-chains.
Answer:
Option (A) is correct.
Explanation: Large proteins such as haemoglobin have quaternary structure. It has four chains, two α-chains and two β-chains.

Question 15.
Assertion (A): C₄ photosynthetic pathway is more efficient than the C₃ pathway. 1
Reason (R): Photorespiration is suppressed in C₄ plants.
Answer:
Option (A) is correct.
Explanation: C₄ pathway ensures that calvin cycle is to be operated only in the bundle sheath cell. This is an adaptation to photorespiratory loss. Hence, C₄ pathways are adapted to photorespiratory loss and are more efficient than C₃ pathway.

Question 16.
Assertion (A): The last two pairs of ribs are called floating ribs. 1
Reason (R): The anterior end of floating ribs is not attached to either the sternum or the cartilage of another rib.
Answer:
Option (A) is correct.
Explanation: The last two pairs of ribs are called floating ribs because their anterior ends are not attached to either the sternum or the cartilage of another rib. The first seven pairs of ribs are called true ribs.

Section B

Question 17.
State differences between aestivation and hibernation. 2
Answer:
Aestivation- It is an organism’s dormant state during the summer. This assists the organism in surviving the summer’s extreme heat.

Hibernation- An organism’s dormant state during the winter is known as hibernation or winter sleep. This assists the organism in surviving the bitter cold of winter. At this stage, the organism’s metabolic and physiological activities are minimal.

Question 18.
The diagram given shows the process of crossing over. 2

(a) Define crossing over.
(b) Name the enzyme on which the process of crossing over is dependent.
Answer:
(a) Crossing over is the exchange of genetic material between the non-sister chromatids of homologous chromosomes during the pachytene of prophase I.
(b) Recombinase.

Question 19.
In the figure given below, the black line (upper) indicates the action spectrum for photosynthesis and lighter line (lower) indicates the absorption of the spectrum of chlorophyll a.
What does the action spectrum indicate? Explain with an example. 2

Answer:
Action spectrum indicates the rate of photosynthesis. This is measured by release of oxygen. We can plot an action spectrum by plotting the amount of oxygen released on Y-axis and wavelength on X-axis. This can be shown by plotting absorption of different wavelengths by any pigment. e.g. chlorophyll a as shown in this graph.

Question 20.
The image below shows the emersion effect. On the basis of your observations answer the following questions. 2

(a) What do you understand by Emerson effect?
(b) How it was observed?
Answer:
(a) “The fall in photosynthetic yield beyond the red region of the spectrum is called Emerson’s first effect or red drop.”

(b) Emerson et. al. (1957) exposed unicellular alga Chlorella to monochromatic light and measured the photosynthetic yield in terms of quantum yield. They observed that the quantum yield of photosynthesis decreased very sharply towards far-red region of the spectrum.

Question 21.
Complete the following table:

Groups of Algae	Chlorophyceae	Phaeophyceae	Rhodophycea
Stored food	Starch	(a) ……………….	(b) ………………
Cell wall	(c) ……………..	Cellulose and algin	(d) ……………..
Major pigments	(e) …………….	(f) ……………….	Chl-a, d and phycoerythrin

OR
Write a short note on protonema.
Answer:
(a) Mannitol
(b) Floridian starch
(c) Cellulose
(d) Cellulose and pectin
(e) Chlorophyll a, b
(f) Chl- a, c, Fucoxanthin
OR
Protonema is a thread-like chain of cells that forms the haploid gametophytic phase in the life cycle of mosses and liverworts. When moss grows from the spore, it grows as a protonema, which develops into a leafy gametophore.

Section – C

Question 22.
(a) Identify the structure given below. 3

(b) Write a short note on PPLO.
Answer:
(a) The structure given below is of Mycoplasma.

(b) PPLO (Pleuro Pneumonia Like organism). It is also called as Mycoplasma. It is the smallest organism. It lack cell wall surrounding the cell membrane. Due to the absence of cell wall, it is resistant to certain antibiotics like penicillin. It survives without oxygen.

Question 23.
What is the economic importance of diatoms? 3
Answer:
(i) Diatoms are an important source of food to aquatic animals.
(ii) Diatomite is porous and chemically inert, therefore used in filtration of sugar, alcohols and antibiotics.
(iii) It is employed as a cleansing agent in toothpastes and metal polishes.
(iv) It is also employed as insulation material in refrigerators, boilers and furnaces. (Any three)

Question 24.
(a) What are leucoplasts?
(b) Name and write the function of different kinds of leucoplast in plant cell. 3
Answer:
(a) Leucoplasts are the colourless plastids and do not contain any pigment. They usually perform the function of storage of reserve food material.
(b) They are classified into the following three types :

1. Amyloplasts: Amyloplasts are generally found in tubers, endosperms and cotyledons. They store starch as reserve food material.
2. Aleuroplasts: Aleuroplasts are found in the seeds of Ricinus, maize, etc. They store proteins as a reserve food material.
3. Elaioplasts: Elaioplasts are found in most monocotyledonous plants. They store oil as reserve food material.

Question 25.
Identify the type of leaf given below. How it is different from one another. 3

Answer:
The leaf labelled as (i) is simple leaf.
In simple leaf, lamina is not divided into distinct lobes or leaflets i.e., it has a single lamina. Axillary bud is present in the axil of a simple leaf. Simple leaves are in acropetal succession on stem. Base of the leaf may have stipules. Simple leaves appear in one or more plane.

(ii) The leaf labelled as (ii) is compound Leaf.
In this leaf, lamina is incised into two or more distinct leaflets. The leaflets are borne either at the tip of petioles or on the sides of rachis. Leaflets do not bear any axillary bud. Axillary bud is present in the axil of the whole leaf. Leaflets of a compound leaf are not in acropetal succession on rachis. Leaflets in a compound leaf lie in one plane only.

Question 26.
What is the importance of active transport? 3
Answer:
Active transport plays an important role in :

1. Absorption of most nutrients from the intestine.
2. Rapid and selective absorption of nutrients by cells.
3. Maintaining a membrane potential.
4. Maintaining water and ionic balance between cells and extracellular fluid. (Any three)

Question 27.
(a) What is meant by chloride shift?
(b) Where does it occur in human body?
(c) What is its utility? 3
OR
(a) What is partial pressure? How does it help in gaseous exchange during respiration?
(b) Where does exchange of gases occur?
Answer:
(a) Chloride shift is a process which occurs in the body of the humans during the circulation of oxygen and carbon dioxide through the blood. The carbon dioxide is taken up the RBCs and the enzyme carbonic anhydrase converts the CO₂ to H₂CO₃, which breaks to give bicarbonate ion and hydrogen ion. There is an exchange of bicarbonate and the chloride ions through the membrane of the red blood cells. The chloride ions move inside and the bicarbonate is moving outside the red blood cells in the plasma.

(b) It occurs from plasma to RBCs (erythrocytes) in human body.

(c) Utility of chloride shift: To maintain ionic balance and electrochemical neutrality.
OR
(a) (i) The pressure exerted by an individual gas in a mixture of gases is called partial pressure. It is directly proportional to its concentration in a mixture. It is expressed as pO₂, pCO₂, etc.

(ii) A gas diffuses across a membrane from the side where its partial pressure is higher to the side where it is lower. In lungs, pO₂ is higher in alveolar air than in blood capillaries, so O₂ diffuses from lungs into the blood. Sometimes, pCO₂ is higher in blood than in alveoli and hence it diffuses out.

(b) Exchange of gases occurs in lung alveoli and tissue cells.

Question 28.
What are nodal tissues? Explain their role in heartbeat. 3
Answer:
Nodal tissues are those types of cardiac muscular fibers which help in sending electrical signals from SA nodes (sino-atrial nodes) to the conduction system of the heart.

Conduction system of the heart consists of SAN, AVN, Bundle of His, Bundle branches and Purkinje tissues.
The heartbeat results from a wave of electrical potential called the cardiac impulse, which originates from nodal tissue like sino-atrial node (SAN), auriculo-ventricular node (AVN), bundle of His and Purkinje fibres.

Functions:

1. SAN is located in the right auricle at the opening of the vena cava. It is the part of the conducting system of the heart. The stimulation arises initially in the sino-atrial node.
2. The AVN is located in the right auricle in the inter-auricular septum. It gives rise to a bundle of His, a muscular bridge that conducts stimulation from the auricles to the ventricles.
3. The bundle of His divides into two branches. The terminal branches of the conducting system are represented by a network of Purkinje fibres which convey stimuli to the myocardium.

Section – D

Question 29.
Centrosome is a small naked organelle found in the cytoplasm of animal cell near the outer surface of the nucleus. It consists of two bundles of microtubules called centrioles that lie at right angles to each other. Centrioles are short cylinders and possess a whorl of 9 peripheral fibrils. The fibrils are absent in the centre. Each fibril is made of 3 sub-fibres. There is a proteinaceous hub in the central part of a centriole. 4

(a) Describe the ultrastructure of centrioles and basal bodies.
(b) (i) The central hub is connected to the triplets through …………… .
(ii) Out of the two centrioles in the sperm, the distal centriole forms ……………….. .
OR
Write down the functions of centriole and basal bodies.
Answer:
(a) Centrioles are cylindrical structures. They appear as small granules associated with a meiotic spindle during cell division. They occur in all lower plants, protozoa and higher animals.

Ultrastructure:

1. Under the electron microscope, each centriole is seen to be formed of nine sets of evenly spaced peripheral fibrils of tubulin protein.
2. Each of these sets is a triplet composed of three microtubules. Each microtubule has a diameter of about 250 Å. The triplets are found in the matrix.
3. Sometimes, delicate strands appear to connect sets of the triplets to each other. Also, from the central core of the cylinder, delicate strands connect sets of the triplets to each other giving a cartwheel appearance.
4. Basal bodies are structures similar to the centrioles. They produce cilia and flagella.

(b) (i) Radial spokes are delicate strands which appear to connect sets of triplets.
(ii) Neck of sperm contains distal centrioles, which step by step form axial filament.
OR
Functions of centrioles are:

1. Centrioles are the self-duplicating bodies. They contain DNA and RNA.
2. The centrioles which give rise to cilia and flagella are known as basal bodies.
3. Out of the two centrioles in the sperm, the distal centriole forms the axial filament. (Any two)

Functions of basal bodies are:

1. Cilia and flagella arise from the basal bodies.
2. Basal bodies are converted into centrioles.

Question 30.
Observe the graph given below and answer the following questions. 4

(a) Define ECG.
(b) Name the instrument used to measure the heartbeat.
(c) Explain the different segments in ECG.
OR
A cardiologist observes an enlarged QR wave in the ECG of a patient. What does this indicate? What is the importanc’of ECG?
Answer:
(a) ECG is a graphic record of the electric current produced by the excitation of the cardiac muscles.

(b) The instrument used to record the change is an electrocardiogram or ECG machine.

(c) Different segments in ECG:

1. 1. A normal electrocardiogram (ECG) is composed of a P-wave, a QRS wave (complex) and a T-wave.
   2. The P wave is a small upward wave that indicates the depolarisation of the atria. It is caused by the activation of SA node.
   3. The QRS wave (complex) begins after a fraction of a second of the P wave. It represents ventricular depolarisation (ventricular contraction). It is caused by the impulses of the contraction from AV node through the bundle of His and Purkinje fibres and marks the beginning of the systole.
   4. The T-wave is a dome-shaped which indicates ventricular repolarisation (ventricular relaxation). The potential generated by the recovery of the ventricle from the depolarisation state is called the repolarisation wave.

OR
The enlarged QR wave indicates myocardial infarction,

The importance of ECG is that it gives accurate information about the heart. Therefore, ECG is of great diagnostic value in cardiac diseases.

Section – E

Question 31.
(a) What are chemoautotrophic bacteria? How they obtained energy? 5
(b) State differences between autotrophic bacteria and heterotophic bacteria on the basis of characteristics given in the table.

Characteristics	Autotrophic bacteria	Heterotrophic bacteria
Carbon source
Energy source
Metabolic pathways
Nutrient requirements
Examples

OR
What is heterospory? Briefly comment on its significance. Give two examples.
Answer:
(a) Chemoautotrophic bacteria are able to manufacture their organic food from inorganic raw materials with
the help of energy derived from exergonic chemical reactions involving the oxidation of an inorganic substance present in the external medium.

Characteristics	Autotrophic bacteria	Heterotrophic bacteria
Carbon source	Obtain carbon from inorganic sources (carbon dioxide).	Obtain carbon from organic sources (preformed organic compounds).
Energy source	Obtain energy from various sources (light, inorganic compounds).	Obtain energy from organic compounds through respiration or fermentation.
Metabolic pathways	Can perform photosynthesis or chemosynthesis.	Depend on organic compounds for energy and carbon.
Nutrient requirements	Require inorganic nutrients for growth.	Require organic nutrients for growth.
Examples	Cyanobacteria, green sulfur bacteria	E. coli, Salmonella, Streptococcus

OR
Heterospory refers to the generation of two types of spores by a single plant. These spores are different in size and structure; one is called a microspore, while the other is called a megaspore. The microspore is smaller and gives rise to the male gamete after germination.

On the other hand, megaspores are larger and give rise to the female gamete after germination. The megaspore is enclosed within megasporangia, ensuring the zygote’s proper development. A zygote is formed after the process of fertilisation of the microgamete and the megagamete.

Heterospory was first observed in plants like Selaginella and Salvinia.

Question 32.
(a) How diabetes insipidus is different from diabetes mellitus? 5
(b) (i) What is ADH?
(ii) State its functions. (c)
What is diuretic ? Give example.
OR
(a) (i) Why is haemoglobin called conjugated protein ?
(ii) What happens to the molecule at high and low partial pressure of oxygen ?
(b) Write short note on:
(i) Carbaminohaemoglobin.
(ii) Oxyhaemoglobin.
Answer:
(a) Deficiency of ADH leads to diabetes insipidus. It is a condition marked by the output of huge amounts of urine and intense thirst. The name itself (diabetes =overflow; insipidus = tasteless) distinguishes it from diabetes mellitus (mel = honey), in which insulin deficiency causes large amounts of blood sugar to be lost in the urine.

(b) (i) Antidiuretic hormone (ADH) is one of the hormones that efficiently monitors and regulates the functioning of the kidneys.

(ii) Antidiuretic hormone released from the posterior pituitary, prevents wide swings in water balance, helping to avoid dehydration or water overload. ADH facilitates reabsorption of water by the distal parts of the kidney tubules and thereby prevents diuresis.

(c) A substance which increase the volume of urine excreted is called diuretic. E.g., Tea, Coffee, etc.
OR
(a) (i) Haemoglobin is called conjugated protein because it consists of a basic protein globulin and a non-protein called haem.

(iii) The haemoglobin, when exposed to the high partial pressure of oxygen, combines with it to form oxyhaemoglobin, which carries four molecules of oxygen loosely bound to the four Fe²⁺ ions. When this oxyhaemoglobin reaches the tissues where there is low oxygen pressure, oxyhaemoglobin dissociates into oxygen and deoxyhaemoglobin.

(b) (i) Carbon dioxide when enters the erythrocytes combines with globin, part of deoxyhaemoglobin thereby forming carbaminohaemoglobin.
(ii) Oxygen diffuses into erythrocytes and combines with the iron ions of haemoglobin and forms oxyhaemoglobin.

Question 33.
Deciduous plants shed their leaves during hot summer or in autumn. This process of shedding of leaves is called abscission. Apart from physiological changes what anatomical mechanism is involved in the abscission of leaves?5
OR
Briefly describe the alimentary canal and digestive glands of the frog.
Answer:
Abscission is the process of shedding of leaves during hot summer or in autumn by deciduous plants. The abscission layer is made up of two layers of cambium-like cells. The cells are thin- walled and without deposition of lignin or suberin. At the time of abscission, the middle lamella may dissolve between the cells of two middle layers but the primary wall remains intact. The middle lamella as well as the primary walls of the adjacent cells is dissolved. Ultimately the whole cells of middle layer found in the abscission layer get dissolved completely. Thus, the leaf gets separated from the stem and falls down from the stem by wind and rain.
OR
Alimentary canal of frog: The alimentary canal of the frog is short. It starts with a mouth that is terminal in position. It opens into the buccopharyngeal cavity which contains the maxillary and vomerine teeth and carries the openings of the eustachian tube, vocal sac, gullet and glottis. The gullet opens in a narrow and short tubed oesophagus which
continues in the large stomach. Stomach walls are highly muscular which helps in converting the food into chyme. The stomach is followed by a coiled small intestine. The small intestine bears a number of finger-like folds called villi and microvilli which increase the surface area for absorption of digested food. Intestine continues into a wider section, opening into the cloaca. The urinary bladder opens into the cloaca. The urinary opens into the cloacal chamber through the ureter.

The gastric glands of the stomach bring about the digestion of protein. The liver secretes bile that is temporarily stored in the gall bladder. Bile helps in the digestion of food by changing its pH from acidic to alkaline and by emulsifying the fat. The liver does not secrete any enzyme. The pancreas secretes pancreatic juice that causes the digestion of protein starch and fats with the help of trypsin, amylase and lipase.