Biology Sem 1

PENERBIT ILMU BAKTI SDN. BHD.

Answering Guidelines for Matriculation Biology Examination vi Chapter 1 Molecules of Life 1.1 Water 1 1.2 Carbohydrates 3 1.3 Lipids 5 1.4 Proteins 7 1.5 DNA and RNA Molecules 9 Summative Practice 1 12 Chapter 2 Cell Structures and Functions 2.1 Prokaryotic and Eukaryotic Cells 16 2.2 Structures and Functions: Plasma Membrane and Organelles 18 2.3 Animal and Plant Tissues 23 2.4 Cell Transport 28 Summative Practice 2 30 Chapter 3 Cell Division 3.1 The Concept of Cell Division 33 3.2 The Cell Cycle 34 3.3 Mitosis 36 3.4 Meiosis 38 Summative Practice 3 42 Summative Assessment Test (UPS) 1 46 Contents iii C00 Matriculation Sem 1 BIO (CTN/GUIDE) 3pp.indd 3 20/06/2023 3:17 PM PENERBIT ILMU BAKTI SDN. BHD.

 Number System Chapter 4 Genetic Inheritance 4.1 Mendelian Genetics: Monohybrid and Dihybrid 48 4.2 Deviations from the Mendelian Inheritance 61 4.3 Genetic Mapping 67 Summative Practice 4 69 Chapter 5 Population Genetics 5.1 Gene Pool Concept 73 5.2 Hardy-Weinberg Law 75 Summative Practice 5 77 Chapter 6 Expression of Biological Information 6.1 DNA and Genetic Information 80 6.2 DNA Replication 81 6.3 Protein Synthesis: Transcription and Translation 88 6.4 Gene Regulation and Expression – lac Operon 97 Summative Practice 6 100 Summative Assessment Test (UPS) 2 104 Chapter 7 Mutation 7.1 Overview the Classification and Types of Mutation 106 7.2 Gene Mutation 107 7.3 Chromosomal Mutation 112 Summative Practice 7 121 iv C00 Matriculation Sem 1 BIO (CTN/GUIDE) 3pp.indd 4 20/06/2023 3:17 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter 8 Recombinant DNA Technology 8.1 Recombinant DNA Technology 125 8.2 Methods in Gene Cloning 129 8.3 Application of Recombinant DNA Technology 132 Summative Practice 8 135 Summative Assessment Test (UPS) 3 139 Chapter 9 Reproduction and Development 9.1 Sexual Reproduction in Flowering Plants 141 9.2 Human Reproductive System 146 9.3 Fertilisation and Foetal Development 153 9.4 Role of Hormones during Pregnancy and Parturition 156 9.5 Growth Patterns in Human and Plants 158 Summative Practice 9 161 Semester Examination 165 Answers 169 v C00 Matriculation Sem 1 BIO (CTN/GUIDE) 3pp.indd 5 20/06/2023 3:17 PM PENERBIT ILMU BAKTI SDN. BHD.

 Number System Answering Guidelines for Matriculation Biology Examination A Structured Questions 1 Read through the statement given in each question. 2 Study the diagram in each question, if given. 3 Be mindful of the marks allocated for each question. The marks showing the number of points should be written in the answers. For instance: 2 points are equal to 2 marks. 4 If you are only required to give two points, write only two points. Never provide more answers because certain questions apply the concept “wrong cancels right,” meaning that one wrong answer can diminish one correct answer. 5 There is no need to write answers in sentences for questions starting with “State”, “Name” and “Label”. 6 If a question requires you to differentiate, write one difference in one full sentence using the conjunctions “while”, “whereas” or “but”. 7 If a question requires you to label parts in a diagram, carefully write the labels with accurate spelling, keeping in mind that the first letter must be a capital letter for a proper noun and if necessary, the singular or plural noun based on the number of arrows. B Essay Questions 1 Candidates can identify essay questions by the action words and the minimum total marks allocated, which are usually four or more. 2 Read the questions carefully. 3 Distinguish the relevant information from the extraneous information in the questions. 4 Underline the keywords in the question. Keywords are important as guidelines for writing essays. 5 Sketching a simple diagram or figure can also help you focus your thoughts. 6 Make sure all sentences are complete. 7 All sentences must be short, concise and not drag on. 8 All sentences must consist of points. A sentence can have at least one point or more. Your essays should be concise. 9 Only draw diagrams if the question specifically instructs you to do so. 10 For incorrectly spelled science terminology, no points are awarded. Please be mindful. 11 Interpreting the action verbs in essay questions: Action verb Interpretation Differentiate / Contrast Write the differences only between the two items in sentence form. You are strictly prohibited to write the differences in a table. Compare Write the similarities as well as the differences between the two items. Normally, only one or two marks are allocated for the similarities. Start the similarities with the word “both”. Describe / Discuss / Explain You must elaborate in detail to cater to these keywords. You are suggested to write more than the marks allocated because no mark will be deducted. Define The definition must be complete and accurate to get a mark. So, you must memorise all the definitions given in the lecture. Normally the mark allocated is either 2 or zero. There is no intermediate mark. Illustrate Draw appropriately labelled diagrams to explain. 12 Your handwriting must be clear, neat and in the appropriate size. vi C00 Matriculation Sem 1 BIO (CTN/GUIDE) 3pp.indd 6 20/06/2023 3:17 PM PENERBIT ILMU BAKTI SDN. BHD.

Molecules of Life CHAPTER1 1 1.1 Water ● State the structure and properties of water molecules. ● Relate the properties of water and its importance. Learning Outcomes Structure and Properties of Water Molecule 1 A water molecule consists of an oxygen atom and two hydrogen atoms. 2 The two hydrogen atoms are bound to the oxygen atom by covalent bonds. 3 The angle between the covalent bonds is 104.5°, forming a wide V-shaped water molecule. 4 One water molecule can bind to four other water molecules through hydrogen bonds. A hydrogen bond occurs when the partially positive hydrogen atoms attract the partially negative oxygen atoms of adjacent water molecules. 5 The structure of water molecules and related bonds is shown in Figure 1.1. Figure 1.1 Structure of water molecules H H H H H H H H H H O O O O O 104.5o 104.5o 104.5o 104.5o Hydrogen bond Covalent bond C01 Matriculation Sem 1 BIO 4pp.indd 1 19/06/2023 5:11 PM PENERBIT ILMU BAKTI SDN. BHD.

2 Mid Semester Test 1 Chapter 1 Answers  Molecules of Life Molecules of Life  An oxygen atom is more electronegative compared to a hydrogen atom. Therefore, shared electrons are pulled more towards oxygen. This situation produces two regions of partial negative charge on oxygen atoms and a partial positive charge on each hydrogen atom. Bio Info 6 Properties of water: (a) Universal solvent due to its polarity Water dissolves polar substances such as sugar, ionic substances such as Na2SO4 and organic molecules such as carboxylic acid. As such, it is a universal solvent. (b) High specific heat capacity A high amount of heat energy is required to break hydrogen bonds to change the temperature of 1 g of water by 1 °C. As such, water can stabilise the temperature of body cells. (c) High latent heat of vapourisation A high amount of heat energy is required to turn 1 g of water from its liquid state to water vapour. The evaporation of sweat on the skin gives a cooling effect as body heat is removed. (d) Cohesion of water molecules Water molecules tend to stick to each other due to hydrogen bonds. This produces the cohesion of water. This property enables aquatic insects to walk on the surface of water (surface tension) and helps water move up plants from the ground to the tip of the plant (cohesion and adhesion of water). (e) Maximum density at 4 °C As water reaches maximum density at 4 ºC, during winter or in the Arctic, ice floats on top of the lake and insulates the water below. As such, aquatic organisms can continue to live below the ice layer. Example 1 Draw two molecules of water. Answer Quick Check 1.1 1 Name the bond found between the oxygen and hydrogen atoms in a water molecule. 2 Name the bond found between water molecules. 3 State two significance of hydrogen bonding in water molecules. 4 State two properties of water. Draw two water molecules separately. Then connect an oxygen atom from one water molecule to the hydrogen atom of the adjacent water molecule with dotted lines to represent the hydrogen bond. Smart Tips! H H H H O O 104.5o 104.5o Hydrogen bond Covalent bond H H O δ- δδ+ δ+ C01 Matriculation Sem 1 BIO 4pp.indd 2 19/06/2023 5:11 PM PENERBIT ILMU BAKTI SDN. BHD.

3 Mid Semester Test 1 Answers Chapter 1  Molecules of Life Molecules of Life  1.2 Carbohydrates ● State the classes of carbohydrates such as monosaccharides, disaccharides and polysaccharides. ● Illustrate the formation and breakdown of maltose. ● Compare the structures and functions of starch, glycogen and cellulose. Learning Outcomes Classes of Carbohydrates 1 Carbohydrates consist of carbon, hydrogen and oxygen atoms in the ratio 1:2:1. 2 There are three classes of carbohydrates: (a) Monosaccharides Simple sugars and monomers for disaccharides and polysaccharides. The main energy source in plants and animals. They are sweet, soluble in water, can be crystallised and are reducing sugars. Examples of monosaccharides: glucose, fructose and galactose are shown in Figure 1.2. Figure 1.2 Monosaccharides (b) Disaccharides Consist of two monosaccharides that are linked together by a glycosidic linkage. They are sweet, soluble in water, can be crystallised and are non-reducing sugars. Examples: Sucrose (glucose + fructose), lactose (glucose + galactose) and maltose (glucose + glucose) (Figure 1.3). Figure 1.3 Disaccharides (c) Polysaccharides Consist of hundreds of monosaccharides that are linked together by glycosidic linkages. They are complex, not sweet, insoluble in water, cannot be crystallised and are non-reducing sugars. Examples: Starch, glycogen and cellulose. CH2 OH O H HO OH OH OH H H H CH2 OH O H HO OH OH OH H H H CH2 OH O H OH OH H HO H CH2 OH Glucose Fructose Galactose CH2 OH O HO OH H H H CH2 OH O O H HO OH OH H H H H CH2 OH O CH2 OH O H HO OH OH H H H H CH2 OH O H OH OH H H H H OH O CH2 OH O H HO OH OH H H H H CH2 OH O H OH OH OH H H H H Sucrose Lactose Maltose C01 Matriculation Sem 1 BIO 4pp.indd 3 19/06/2023 5:11 PM PENERBIT ILMU BAKTI SDN. BHD.

4 Mid Semester Test 1 Chapter 1 Answers  Molecules of Life Molecules of Life  3 Maltose is formed through the condensation reaction of two molecules of α-glucose. Maltose is broken down into two molecules of α-glucose through hydrolysis as shown in Figure 1.4. Figure 1.4 Formation and breakdown of a maltose 4 Comparison between the structures and functions of starch, glycogen and cellulose are shown in Table 1.1. Table 1.1 Comparison between starch, glycogen and cellulose Polysaccharide Starch Glycogen Cellulose Monomer α-glucose α-glucose β-glucose Bonds • α-1,4 glycosidic bond (amylose) • α-1,4 glycosidic bond and α-1,6 glycosidic bond (amylopectin) • α-1,4 glycosidic bond • α-1,6 glycosidic bond β-1,4 glycosidic bond Branch • Unbranched (amylose) or • Branched (amylopectin) Extensively branched Unbranched Functions Stores energy in plants Stores energy in animals and fungi The structural component of plant cell wall 5 The structure of amylopectin as a branched starch is shown in Figure 1.5. Figure 1.5 Amylopectin Cellulose is the most abundant organic compound on Earth and is produced by plants. Plants produce almost 1014 kg or 100 billion tonnes of cellulose per year. Bio Info −O CH2 OH O H OH OH H H H H CH2 OH O H OH OH H H H H O O CH2 OH O H OH OH H H H H CH2 OH O H OH OH H H H H CH2 O H OH OH H H H H O O O CH2 OH O H OH OH H H H H O CH2 OH O H OH OH H H H H O α-1,6 glycosidic linkage α-1,4 glycosidic linkage CH2 OH O H HO OH OH H H H H OH CH2 OH O H H O OH OH H H H H OH + O CH2 OH O H HO OH OH H H H H CH2 OH O H OH OH OH H H H H Condensation Hydrolysis α-1,4 glycosidic linkage α-glucose α-glucose Maltose 1 4 + H2 O C01 Matriculation Sem 1 BIO 4pp.indd 4 19/06/2023 5:11 PM PENERBIT ILMU BAKTI SDN. BHD.

5 Mid Semester Test 1 Answers Chapter 1  Molecules of Life Molecules of Life  Example 2 State three structural differences between cellulose and glycogen. Answer Cellulose is an unbranched molecule while glycogen is an extensively branched molecule. The monomer for cellulose is β-glucose while the monomer for glycogen is α-glucose. Cellulose has β-1,4 glycosidic bonds while glycogen has α-1,4 glycosidic bonds and α-1,6 glycosidic bonds. Quick Check 1.2 1 State the three common properties of monosaccharides and disaccharides. 2 Name two benefits of glucose. 3 Name the polysaccharide stored in the human body. 4 _____________ is the process of forming molecules from monomers while _____________ is the process of breaking down molecules into monomers. 1.3 Lipids ● State the types of lipids: triglycerides, phospholipids and steroids. ● Describe the structure of fatty acids and glycerol. ● Explain the formation and breakdown of triglycerides. Learning Outcomes Lipids 1 Lipids are large organic compounds. 2 Lipids are insoluble in water but soluble in non-polar organic solvents such as chloroform. 3 The three types of lipids are triglycerides, phospholipids and steroids as shown in Figure 1.6. Figure 1.6 Types of lipids H H H C O O C R1 R2 R3 C C O O O O C C H H O OH H H H Hydrophilic head Phosphate group Hydrophobic tail Glycerol 1 CH2 −O− −CH2 −CH2 −CH2 −CH2 −CH2 −CH2 −CH2 −CH2 −CH2 −CH2 −CH2 −CH3 −CH2 −CH2 −CH2 −CH2 −CH2 −CH O C O C 2 CH−O− CH CH2 CH2 CH2 CH2 CH2 - O O R−O−P−O−1 CH2 Fatty acids Triglycerides Phospholipids Steroids C01 Matriculation Sem 1 BIO 4pp.indd 5 19/06/2023 5:11 PM PENERBIT ILMU BAKTI SDN. BHD.

6 Mid Semester Test 1 Chapter 1 Answers  Molecules of Life Molecules of Life  4 A triglyceride consists of one glycerol and three fatty acids. A fatty acid consists of a long linear hydrocarbon chain (symbol: R) and a carboxyl group (-COOH). Glycerol is a type of alcohol that contains three carbon atoms and three hydroxyl (OH-) groups. 5 A triglyceride is formed by the condensation of one glycerol and three molecules of fatty acids followed by the release of three water molecules. Hydrolysis is the reverse of the condensation process. The formation and breakdown of triglycerides are shown in Figure 1.7. Figure 1.7 Formation and breakdown of triglycerides Example 3 Draw and label the structure of a triglyceride molecule. Answer Quick Check 1.3 1 Both carbohydrates and lipids are made up of the same elements. State the elements. 2 How many molecules of water are lost in the formation of one triglyceride molecule? 3 State two differences between the structures of phospholipid and triglyceride. 4 Which groups are involved in the formation of ester linkages in triglycerides? Most animal fats are saturated fats. Fats in plants and fish are unsaturated. A diet rich in saturated fats is the main factor that causes cardiovascular diseases, as plaque develops within blood vessel walls. Bio Info H H H H OH OH OH C C C H HO C O R’ HO C O R’’ HO C O R’’’ + 1 Glycerol 3 Fatty acids Triglyceride Condensation Hydrolysis H H H H O O O C C C H C O R’ C O R’’ C O R’’’ + 3H2 O 3 Water molecules Ester linkage H H H H O O O C C C H C O R1 C O R2 C O R3 Ester linkage C01 Matriculation Sem 1 BIO 4pp.indd 6 19/06/2023 5:11 PM PENERBIT ILMU BAKTI SDN. BHD.

7 Mid Semester Test 1 Answers Chapter 1  Molecules of Life Molecules of Life  1.4 Proteins ● Describe the basic structure of amino acids. ● State how amino acids are grouped. ● Describe primary, secondary, tertiary and quaternary levels of proteins and the types of bonds. ● Describe the effect of pH and temperature on the structure of a protein. ● Explain the formation and breakdown of the dipeptide. ● Classify proteins according to structure and composition. Learning Outcomes Amino Acid and Proteins 1 The basic structure of amino acids is shown in Figure 1.8. Figure 1.8 Structure of an amino acid 2 There are 20 types of amino acids that are grouped into four groups based on the characteristics of the side chain or R group: (a) Non-polar amino acids: The side chain is hydrophobic and non-polar. (b) Polar amino acids: The side chain is hydrophilic and polar. (c) Acidic amino acids: The side chain has a negative charge. (d) Basic amino acids: The side chain has a positive charge. 3 Proteins are formed from amino acids which are joined together by peptide bonds (polypeptides). 4 Proteins can be divided into four levels: (a) Primary structure. A linear polypeptide chain joined by peptide bonds. Example: Glucagon. (b) Secondary structure. The linear polypeptide chain coils to form an alpha helix or folds to form a beta-pleated sheet. Examples: Keratin and silk protein. (c) Tertiary structure. The linear polypeptide chain coils into a globular shape which is maintained by disulfide bridges, ionic bonds, hydrogen bonds, hydrophobic interactions and van der Waals interactions. Example: Myoglobin. (d) Quaternary structure. Two or more polypeptide chains joined together by disulfide bridges, ionic bonds, hydrogen bonds, hydrophobic interactions and van der Waals interactions. Examples: Haemoglobin and collagen. Side chain Amino group Carboxyl group R C C O OH H H H N C01 Matriculation Sem 1 BIO 4pp.indd 7 19/06/2023 5:11 PM PENERBIT ILMU BAKTI SDN. BHD.

8 Mid Semester Test 1 Chapter 1 Answers  Molecules of Life Molecules of Life  Figure 1.9 Levels of proteins 5 Proteins can become denatured at extreme pH and extremely high temperatures. Denatured proteins lose their original conformation or structure due to the breakdown of disulfide bridges, ionic bonds, hydrogen bonds, hydrophobic interactions and van der Waals interactions. 6 Dipeptide is formed by two amino acids through a condensation reaction and broken down by hydrolysis. Figure 1.10 shows the reactions. Figure 1.10 Formation and breakdown of dipeptide 7 Proteins are classified into three classes based on their structure and composition: (a) Fibrous proteins are long polypeptide chains, with secondary and stable structures, are insoluble in water and used for mechanical support. Examples: Collagen, keratin and fibrin. (b) Globular proteins are polypeptides folded into spherical shapes, have tertiary or quaternary structure, are unstable and soluble in water and are important for chemical processes. Examples: Haemoglobin, myoglobin and antibodies. (c) Conjugated proteins are proteins with non-protein materials as prosthetic groups. Examples: Glycoprotein, lipoprotein and flavoprotein. Figure 1.11 Classes of proteins H H N C R C O H OH H H N C R C O H OH H N N C O H H H C R H R C C O OH + + H2 O Amino acid Amino acid Condensation Hydrolysis Peptide bond Dipeptide Water Polypeptide chain α-Chain β-Chain Iron Heme Non-protein part Fibrous protein: Collagen Globular protein: Haemoglobin Conjugated protein: Flavoprotein Primary structure Secondary structure Tertiary structure Quaternary structure Amino acids α-helices β-sheets β-sheets α-helices C01 Matriculation Sem 1 BIO 4pp.indd 8 19/06/2023 5:11 PM PENERBIT ILMU BAKTI SDN. BHD.

9 Mid Semester Test 1 Answers Chapter 1  Molecules of Life Molecules of Life  Example 4 Draw the structure resulting from the combination of two molecules of amino acids. Answer Quick Check 1.4 1 What distinguishes amino acids from one another? 2 Name the chemical reaction involved and the type of bond formed during the formation of a dipeptide. 3 State the difference between fibrous protein and globular protein in terms of water solubility. 4 Name the monomer of haemoglobin. 1.5 DNA and RNA Molecules ● State the structure of nucleotides as the basic composition of nucleic acids (DNA and RNA). ● Illustrate the structure of DNA based on the Watson and Crick Model. ● Explain the structure of DNA and RNA. ● State the types of RNA. Learning Outcomes Nucleotides, DNA and RNA 1 The basic unit of deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) is nucleotides. 2 A nucleotide consists of a phosphate group, a five-carbon sugar (a pentose) and a nitrogenous base. Figure 1.12 shows the structure of a nucleotide. Figure 1.12 Structure of a nucleotide An X-ray crystallography is used to determine the three-dimensional structure of nucleic acids and proteins such as antibodies. Bio Info H N N CC R R O O OH C C H H H H Peptide bond Dipeptide O O O O P CH2 − − H H OH OH H O H N N N NH2 N Nitrogenous base Pentose Phosphate group C01 Matriculation Sem 1 BIO 4pp.indd 9 19/06/2023 5:11 PM PENERBIT ILMU BAKTI SDN. BHD.

10 Mid Semester Test 1 Chapter 1 Answers  Molecules of Life Molecules of Life  3 Nucleotides are linked to each other by phosphodiester bonds to form a polynucleotide strand or a sugar-phosphate backbone. 4 There are two groups of nitrogenous bases: pyrimidines and purines. Pyrimidines are cytosine (C), thymine (T) and uracil (U). Purines are guanine (G) and adenine (A). 5 Based on the Watson and Crick Model (1953), DNA consists of two polynucleotide strands that run anti-parallel and are held together by hydrogen bonds formed between opposite nitrogenous bases. These two strands form a double helix with ten base pairs in each full turn and contain sugar-phosphate backbones. Figure 1.13 shows the structure of DNA. Figure 1.13 Structure of DNA based on the Watson and Crick Model 6 The structure of DNA differs from RNA in certain characteristics, as shown in Table 1.2. Table 1.2 Differences between DNA and RNA Characteristics DNA RNA Number of strands Double Single Pentose sugar Deoxyribose Ribose Nitrogenous bases Adenine, thymine, cytosine, guanine Adenine, uracil, cytosine, guanine Molecule size Relatively large Relatively small Stability More stable Less stable Ratio of bases The ratio of adenine to thymine and guanine to cytosine is one. The ratio of adenine to uracil and guanine to cytosine varies. 7 There are three types of RNA: messenger RNA (mRNA), ribosomal RNA (rRNA) and transfer RNA (tRNA). G C G C C G A A C G A T A T TA AT A G C A T 3.4 nm 0.34 nm 1 nm Sugar-phosphate backbone O T A O O O −O O H2 C OH P O −O O H2 C OH P O −O O H2 C OH P O −O O H2 C OH OH P OH O O O G C O O A T C G CH2 O O− O O P CH2 O O− O O P CH2 O O− O O P CH2 O O− O HO P Hydrogen bond Phosphodiester linkage Nucleotide unit C01 Matriculation Sem 1 BIO 4pp.indd 10 19/06/2023 5:11 PM PENERBIT ILMU BAKTI SDN. BHD.

11 Mid Semester Test 1 Answers Chapter 1  Molecules of Life Molecules of Life  Example 5 Draw a nucleotide. Answer Example 6 Draw a section of DNA. Answer Quick Check 1.5 1 State the types of nucleic acids. 2 State the common nitrogenous bases for DNA and RNA. 3 State the pairing of bases in DNA. 4 Name the bond that links together the polynucleotide strands. At least 4 nucleotides must be drawn for each polynucleotide strand. The directions of the polynucleotide strands must be opposite each other. Must label the 3’ and 5’ ends according to the number of carbons. Make sure the line that joins the 4th carbon of the pentose and the phosphate group is bent to indicate a 5th carbon. All the important components must be labelled. Smart Tips! Nitrogenous bases G C T A C G A O O O O O O O O T Hydrogen bond Phosphodiester linkage Phosphate group Pentose 3’ end 5’ end 3’ end 5’ end O O O O P CH2 − − H H OH OH H O H N N N NH2 N Nitrogenous base Pentose Phosphate group C01 Matriculation Sem 1 BIO 4pp.indd 11 19/06/2023 5:11 PM PENERBIT ILMU BAKTI SDN. BHD.

12 Mid Semester Test 1 Answers Chapter 1 Molecules of Life  1 The partial negative charge at the oxygen atom of one water molecule is attracted to the partial positive charge of the hydrogen atom of another water molecule by … A ionic bond. B covalent bond. C hydrogen bond. D Van der Waals attraction. 2 At what temperature is water at its densest? A 0 °C B 4 °C C –4 °C D –10 °C 3 Glycogen is a and is made up of . A polymer; glucose B protein; amino acids C polypeptide; monomers D carbohydrate; fatty acids 4 Lactose, a type of sugar in milk is a … A lipid. B disaccharide. C polysaccharide. D monosaccharide. 5 Which of the following is not a characteristic of monosaccharides? A Sweet-tasting B Water-soluble C Forms white crystal D Not soluble in water 6 Which of the following is true about both starch and cellulose? A Both can be digested by humans B Both have glucose as monomers C Both are stored in plants for energy D Both are geometric isomers of each other 7 Which of the following is true about both maltose and amylose? A Both are disaccharides B Both are polysaccharides C Both contain β-glucose as monomers D Both contain α-glucose as monomers 8 Which of the following is true about cellulose? A A polymer composed of sucrose as monomers B It is a storage polysaccharide for energy in plant cells C It is a major structural component in the cell wall of plants D It is a major structural component in the plasma membrane of animals 9 Why are humans able to digest starch but not cellulose? A Humans have enzymes to digest the β-glycosidic linkage of starch but not the α-glycosidic linkage of cellulose B Humans have enzymes to digest the α-glycosidic linkage of starch but not the β-glycosidic linkage of cellulose C The monomer of starch is glucose while the monomer of cellulose is galactose D Humans have starch-digesting bacteria in their digestive tract 10 Which of the following statements best describes how a triglyceride is formed? A Hydrolysis of three molecules of glycerol and one molecule of fatty acid B Hydrolysis of one molecule of glycerol and three molecules of fatty acids C Condensation of three molecules of glycerol and one molecule of fatty acid D Condensation of one molecule of glycerol and three molecules of fatty acids Summative Practice 1 Objective Questions Instruction: There are four answer options for each question. Choose the best answer. C01 Matriculation Sem 1 BIO 4pp.indd 12 19/06/2023 5:11 PM PENERBIT ILMU BAKTI SDN. BHD.

11 Which of the following is true about lipids? A Lipids are not soluble in water B Lipids are important components in the plasma membrane of cells C Lipids can release twice the amount of energy compared to the same weight of polysaccharides D All of the above 12 The different properties of the 20 different amino acids can be attributed to the … A side chain. B amino group. C carboxyl group. D hydrocarbon chain. 13 The formation of a dipeptide requires the … A formation of a hydrogen bond. B removal of a water molecule. C addition of a water molecule. D formation of an ionic bond. 14 The number of peptide bonds present in a polypeptide that contains 100 amino acids is … A 0. C 100. B 99. D 101. 15 The phosphodiester linkage in DNA or RNA is formed between … A two pentose sugars. B two nitrogenous bases. C a pentose sugar and a nitrogenous base. D a pentose sugar and a phosphate group. 1 Figure 1 shows the structural formula of molecules P and Q. EXAM CLONE Molecule P Molecule Q Figure 1 (a) Name molecules P and Q. [2 marks] (b) Molecules P and Q bind to form a new compound. Name the type of chemical reaction and the end products produced after the reaction. [2 marks] (c) Show how molecules P and Q are combined to form the compound stated in 1(b). Circle the atoms that are removed during the reaction. [4 marks] (d) Carbohydrates and lipids function as energy stores. Why do lipids store more energy per gram compared to carbohydrates? [1 mark] (e) State one other function of lipids in living organisms. [1 mark] Structured Questions Instruction: Answer all the questions. 13 Mid Semester Test 1 Answers Chapter 1 Molecules of Life  H H H H OH OH OH C C C H O R HO C C01 Matriculation Sem 1 BIO 4pp.indd 13 19/06/2023 5:11 PM PENERBIT ILMU BAKTI SDN. BHD.

2 Figure 2 shows monomers M and N. EXAM CLONE M N Figure 2 (a) State the common name for monomers M and N. [1 mark] (b) Name the chemical reaction, the new compound produced, the bond formed and the byproduct produced when monomers M and N are combined. [4 marks] (c) Identify two functional groups in monomers M and N and their characteristics to complete Table 1. Functional groups Characteristics Table 1 [4 marks] (d) In Figure 2, circle the atoms that are removed when the bond between monomers M and N is formed. [1 mark] 3 Figure 3 shows a section of the DNA molecule. EXAM CLONE Figure 3 14 Mid Semester Test 1 Answers  Molecules of Life Molecules of Life  Chapter 1 R1 C C O OH H H H N R2 C C O OH H H H + N G T A G T O O O O O O O O 3’ end 5’ end 3’ end 3’ end P Q R C01 Matriculation Sem 1 BIO 4pp.indd 14 19/06/2023 5:11 PM PENERBIT ILMU BAKTI SDN. BHD.

(a) Name P, Q and R. [3 marks] (b) State the base sequence for 5’ to 3’ strand. [1 mark] (c) State two structural differences between a DNA molecule and a RNA molecule. [2 marks] (d) (i) Table 2 shows the percentage composition of bases in the DNA of two organisms. Complete the table below. Organism Adenine Cytosine Guanine Thymine A 25 21 B 30 40 Table 2 [2 marks] (ii) Explain your answer in 3(d)(i). [2 marks] Essay Questions Instruction: Answer all the questions. 1 Compare globular protein and fibrous protein. State one example for each. EXAM CLONE [10 marks] 2 Describe the structure of DNA based on the Watson and Crick model. EXAM CLONE [12 marks] 15 Mid Semester Test 1 Answers  Molecules of Life Molecules of Life  Chapter 1 C01 Matriculation Sem 1 BIO 4pp.indd 15 19/06/2023 5:11 PM PENERBIT ILMU BAKTI SDN. BHD.

CHAPTER2 Cell Structure and Functions Plasma membrane Ribosome Large and circular DNA Nucleoid Cell wall Capsule Pili Flagellum 16 2.1 Prokaryotic and Eukaryotic Cells ● State the three principles of cell theory. ● Explain the structures of prokaryotic and eukaryotic cells. ● Illustrate and compare the structures of prokaryotic and eukaryotic cells. Learning Outcomes Prokaryotic and Eukaryotic Cells 1 The three principles of cell theory: (a) All organisms are made up of one or more cells. (b) New cells are produced from the pre-existing cells. (c) Cells are the structural and functional unit of all living organisms. 2 There are two types of cells; eukaryotic cell and prokaryotic cell. 3 An example of prokaryotic cell is bacterium (plural: bacteria), as shown in Figure 2.1. Figure 2.1 Bacterium C02 Matriculation Sem 1 BIO 4pp.indd 16 19/06/2023 5:38 PM PENERBIT ILMU BAKTI SDN. BHD.

17 Mid Semester Test 1 Answers Chapter 2  Cell Structure and Functions Cell Structure and Functions  4 Eukaryotic cells are cells of algae, protozoa, fungi, plants and animals. 5 Both the structures of prokaryotic and eukaryotic cells have similarities: (a) enclosed by the plasma membrane. (b) contain DNA as its genetic material. 6 However, some of the structures of prokaryotic and eukaryotic cells also have differences: Structures Prokaryotic cells Eukaryotic cells Nucleus Have no true nucleus Have true nucleus Membraneenclosed organelles Have no membrane enclosed organelles Have membrane enclosed organelles DNA Circular and not associated with histone protein Linear and associated with histone protein Ribosome Smaller, 70S Larger, 80S Cell wall Consist of peptidoglycan Do not consist of peptidoglycan Flagella Have simple flagella without a 9 + 2 microtubule arrangement Have flagella with a 9 + 2 microtubule arrangement Example 1 State the two main differences between prokaryotic and eukaryotic cells. Answer Prokaryotic cells have no true nucleus, while eukaryotic cells have true nucleus. Prokaryotic cells have no membrane-enclosed organelles, while eukaryotic cells have membrane-enclosed organelles. Quick Check 2.1 1 State two examples of eukaryotic cells. 2 State two similarities between prokaryotic cells and eukaryotic cells. 3 How is the cell wall of prokaryotic cells different from the cell wall of eukaryotic cells? 4 Why are the flagella of bacteria so simple compared to the flagella of protozoa? Eukaryotic cells are generally bigger than prokaryotic cells. The diameter for typical eukaryotic cells is 10 – 100 μm. The diameter for typical prokaryotic cells is 0.1 – 5 μm. Bio Info Many students think that eukaryotes are all multicellular, but that is not true. Prokaryotes are always unicellular organisms while eukaryotes can be either unicellular or multicellular. For instance, most protists are single-celled eukaryotes. Common Error C02 Matriculation Sem 1 BIO 4pp.indd 17 19/06/2023 5:38 PM PENERBIT ILMU BAKTI SDN. BHD.

18 Mid Semester Test 1 Chapter 2 Answers  Cell Structure and Functions Cell Structure and Functions  2.2 Structures and Functions: Plasma Membrane and Organelles ● State the structures and functions of organelles in animal and plant cells: nucleus, rough endoplasmic reticulum, smooth endoplasmic reticulum, Golgi body, lysosome, ribosome, mitochondria, chloroplast and centriole. ● Explain the structures and functions of the endomembrane systems which includes the nuclear envelope, rough endoplasmic reticulum, smooth endoplasmic reticulum, Golgi body, vesicles, vacuoles and plasma membrane. ● Show the structure of the plasma membrane based on the Fluid Mosaic Model. ● Explain the structure of the plasma membrane and the functions of each of its components. Learning Outcomes Structures of Animal and Plant Cell 1 The typical animal cell and plant cell are shown in Figure 2.2 and Figure 2.3 respectively. Figure 2.2 Typical animal cell Cytoskeleton is a network of fibres extending throughout the cytoplasm of eukaryotic cells. Bio Info Microvilus Rough endoplasmic reticulum with ribosomes Smooth endoplasmic reticulum Plasma membrane Centrosome Lysosome Nucleolus Nuclear envelope Nuclear pore Golgi body Mitochondrion Nucleus C02 Matriculation Sem 1 BIO 4pp.indd 18 19/06/2023 5:38 PM PENERBIT ILMU BAKTI SDN. BHD.

19 Mid Semester Test 1 Answers Chapter 2  Cell Structure and Functions Cell Structure and Functions  Figure 2.3 Typical plant cell 2 Both types of cells are bound by a plasma membrane and contain membrane-bound organelles including nucleus, mitochondrion (plural: mitochondria), endoplasmic reticulum (smooth and rough), Golgi body and ribosomes. 3 However, animal cells and plant cells have certain differences. Table 2.1 Differences between animal and plant cells Structures/Organelles Animal cells Plant cells Cell wall No cell wall Have a cellulose cell wall Plasmodesmata No plasmodesmata Present in the cell wall Chloroplast No chloroplasts Have chloroplasts Vacuoles No vacuoles, if present they are small and temporary Have large and permanent vacuoles Tonoplast No tonoplast Tonoplast surrounds vacuole Nucleus Often at central Usually at cell periphery Centrioles Have centrioles No centrioles Lysosomes Have lysosomes No lysosomes Cytoplasm Throughout the cell Usually at the cell periphery body Cell wall Plasma membrane Nuclear envelope Nucleolus Rough endoplasmic reticulum with ribosomes Golgi body Central vacuole Chloroplast Mitochondrion Smooth endoplasmic reticulum Nucleus C02 Matriculation Sem 1 BIO 4pp.indd 19 19/06/2023 5:38 PM PENERBIT ILMU BAKTI SDN. BHD.

20 Mid Semester Test 1 Chapter 2 Answers  Cell Structure and Functions Cell Structure and Functions  Endomembrane System 1 The endomembrane system is composed of a nuclear envelope, rough endoplasmic reticulum, smooth endoplasmic reticulum, vesicles, Golgi body, vacuoles and plasma membrane. 2 Table 2.2 shows the functions of each component in the endomembrane system. 3 This system performs a variety of functions in the cell, including protein transport for metabolism, lipid transport and toxin detoxification. Table 2.2 Structure and functions of organelles Structure Functions of organelles (a) Nucleus (i) Nucleus contains the genetic information of the cells in the form of DNA. • Function: Controls the activities, cell division and the life span of the cell. (ii) Nuclear envelope consists of an inner membrane and an outer membrane with a space in between. It is also perforated by pores. • Function: Regulates the movement of molecules into and out of the nucleus. (b) Rough endoplasmic reticulum • Appears rough because the outer membrane is studded with ribosomes. • Functions: Packs and transports proteins such as insulin synthesised by ribosomes attached to it. (c) Smooth endoplasmic reticulum • Appears smooth because the outer membrane is not studded with ribosomes. • Functions: Site for lipid synthesis, metabolism of carbohydrates, detoxification of drugs and poisons as well as storage of calcium ions. (d) Golgi body Functions: • Produces lysosomes. • Produces secretory vesicles. • Site for modifying, packaging, sorting and transporting of proteins and lipids received from the endoplasmic reticulum. Cisternae Ribosomes Cisternal space Rough endoplasmic reticulum Nuclear envelope Nuclear pores Nucleolus Chromatin Nucleoplasm Endoplasmic reticulum Ribosomes Cis face Trans face Newly forming vesicle Cisternae Incoming transport vesicle Lumen Secretory vesicle Lumen Tubules C02 Matriculation Sem 1 BIO 4pp.indd 20 19/06/2023 5:38 PM PENERBIT ILMU BAKTI SDN. BHD.

21 Mid Semester Test 1 Answers Chapter 2  Cell Structure and Functions Cell Structure and Functions  Structure Functions of organelles (e) Lysosome • Contains hydrolytic enzymes and lysosomal membranes that are made up by rough endoplasmic reticulum and then further processed in the Golgi body. • Function: Hydrolytic enzymes digest macromolecules by phagocytosis or autophagy. (f) Ribosome • Made up of ribosomal RNA and proteins. • Consists of a small subunit and a large subunit. • Not an organelle because it is not bound by a membrane. • Found free in the cytosol or bound to the endoplasmic reticulum. • Function: Site for the synthesis of proteins. (g) Mitochondrion • Rod-shaped • Enclosed by a double-membrane system: outer membrane and inner membrane separated by an intermembrane space. • The inner membrane forms numerous folds (cristae), which extend into the interior (or matrix) of the organelle. • A semiautonomous organelle because it is able to replicate freely using its own DNA. • Function: Site for cellular respiration. (h) Chloroplast • Biconvex disc-shaped. • Enclosed by a double-membrane system: an outer membrane and an inner membrane. • A third internal membrane called the thylakoid membrane, which is extensively folded and appears as stacks of flattened disks contains chlorophyll. • A semiautonomous organelle because it is able to replicate freely using its own DNA. • Function: Site for photosynthesis. (i) Centrosome • Contains a pair of centrioles, arranged at a 90º to each other. • Centrioles are hollow cylinders. • Function: Centre for growth of microtubules. Large subunit Small subunit Membrane Lumen Digestive enzyme Proteins Enzyme-substrate complex Proximal end Distal end Microtubule triplet Inner membrane Outer membrane Matrix Intermembrane space Cristae DNA Ribosome Thylakoid Granum Stroma Outer membrane Inner membrane C02 Matriculation Sem 1 BIO 4pp.indd 21 19/06/2023 5:38 PM PENERBIT ILMU BAKTI SDN. BHD.

22 Mid Semester Test 1 Chapter 2 Answers  Cell Structure and Functions Cell Structure and Functions  Plasma Membrane and Fluid Mosaic Model 1 In 1972, S. Jonathan Singer and Garth Nicolson proposed the Fluid Mosaic Model which explains the plasma membrane, as shown in Figure 2.4. 2 Fluid: Phospholipids and proteins can move laterally. Mosaic: Proteins are embedded in the phospholipid bilayer like mosaic tiles embedded in mortar. 3 The plasma membrane is composed of two layers of phospholipids with embedded globular proteins. The hydrophilic heads of phospholipids face outwards while the hydrophobic tails of phospholipids face inwards. Figure 2.4 Fluid Mosaic Model 4 Figure 2.5 shows the structure of a phospholipid that consists of a hydrophilic head and hydrophobic tails. Figure 2.5 A phospholipid Glycolipid Cholesterol Phospholipid Integral protein Peripheral protein Channel protein Alpha-helix protein Globular protein Glycoprotein Carbohydrate PO O Phosphate O O O O CH2 CH2 C CO O CH R Glycerol Hydrophillic head Saturated fatty acid Unsaturated fatty acid Hydrophobic tails A phospholipid is an amphipathic molecule which means that it has both a hydrophilic (“water-loving”) region and a hydrophobic (“water-fearing”) region. Bio Info glyco = carbohydrate Smart Tips! C02 Matriculation Sem 1 BIO 4pp.indd 22 19/06/2023 5:38 PM PENERBIT ILMU BAKTI SDN. BHD.

23 Mid Semester Test 1 Answers Chapter 2  Cell Structure and Functions Cell Structure and Functions  Example 2 Label the structure below. Answer A: Plasma membrane B: Centrosome C: Lysosome D: Golgi body E: Nucleolus F: Nuclear envelope G: Nuclear pore H: Mitochondrion I: Rough endoplasmic reticulum J: Microvilus Quick Check 2.2 1 State two similarities between plant cells and animal cells. 2 What are the organelles found in animal cells but not in plant cells? 3 State two types of proteins found in the plasma membrane. 4 Why is the plasma membrane structure proposed by Singer and Nicholson called the Fluid Mosaic Model? 2.3 Animal and Plant Tissues ● Describe animal tissues and plant tissues. ● Explain animal cells and tissues (epithelial cells, nerve cells, muscle cells and connective tissues). ● Explain plant cells and tissues (apical meristem, ground and vascular). Learning Outcomes Animal Tissues 1 Cells with similar functions are grouped into certain tissues. Different tissues work together as an organ. Different organs work together as a system. Different systems work together as an organism. 2 Animal tissues consist of four types of tissues; epithelial, nerve, muscle and connective. 3 Epithelial tissue (a) Structure: Squamous, cuboidal or columnar. Packed together tightly. One free layer exposed to air or fluid and another surface is attached to the basement membrane. A B C E F D H I J G C02 Matriculation Sem 1 BIO 4pp.indd 23 19/06/2023 5:38 PM PENERBIT ILMU BAKTI SDN. BHD.

24 Mid Semester Test 1 Chapter 2 Answers  Cell Structure and Functions Cell Structure and Functions  (b) Distributions: Mouth cavity, alveolus, kidney tubules, blood vessels and lymphatic vessels. (c) Functions: Allows diffusion of nutrients and gases and regulates the movement of substances. Figure 2.6 Types of epithelial tissues 4 Nerve cell or neurone (a) Structure: Neurones are the basic unit of the nervous system. A neurone consists of a cell body, dendrites and an axon. (b) There are three types of neurones: sensory neurone, motor neurone and interneurone. (c) Distribution: Most found in the brain and throughout the body. (d) Function: To transmit nerve impulses. Figure 2.7 Nerve cells 5 Muscle tissue (a) Cardiac muscle: Lines the walls of the heart for pumping blood through the heart. (b) Smooth muscle: Lines the digestive tract, urinary bladder, uterus and blood vessels for involuntary actions. (c) Skeletal muscle: Attached to the skeleton for movement of the skeleton. Figure 2.8 Muscle tissues Simple columnar epithelial Simple cuboidal epithelial cells Simple squamous epithelial cells Pseudostratified columnar epithelial cells Stratified cuboidal epithelial cells Stratified squamous epithelial cells Cardiac muscle Smooth muscle Skeletal muscle Dendrites Direction of conduction Axon Motor neurone cell body Interneurone Cell body Dendrites Axon Cell body Sensory neurone 1 3 2 C02 Matriculation Sem 1 BIO 4pp.indd 24 19/06/2023 5:38 PM PENERBIT ILMU BAKTI SDN. BHD.

25 Mid Semester Test 1 Answers Chapter 2  Cell Structure and Functions Cell Structure and Functions  6 Connective tissue (a) Compact bone: Protects internal organs, for example, the cranium protects the brain and the rib cage protects the heart and lungs. (b) Hyaline cartilage: Covers the ends of bones and reduces friction between joints during movement. (c) Blood: Consists of erythrocytes, leucocytes, platelets and plasma. (i) Erythrocytes carry oxygen and carbon dioxide (ii) Leucocytes are needed for body defence (iii) Platelets are needed for blood clotting (iv) Plasma carries nutrients, hormones, metabolic wastes, respiratory gases, vitamins and antibodies. Figure 2.9 Connective tissues Plant Tissues 1 Plant tissues consist of meristem tissues and permanent tissues. 2 Meristem tissues: (a) Apical meristem: (i) Distributed at the shoot apex and root tips for primary growth. (ii) Increases the height of the plant. Leucocyte Lamella Blood Hyaline cartilage Compact bone Platelet Red blood cell (Erythrocyte) Plasma Lacuna Central (Haversian) canal Canaliculi C02 Matriculation Sem 1 BIO 4pp.indd 25 19/06/2023 5:38 PM PENERBIT ILMU BAKTI SDN. BHD.

26 Mid Semester Test 1 Chapter 2 Answers  Cell Structure and Functions Cell Structure and Functions  (b) Lateral meristem: (i) Distributed at the older stems and roots of woody plants for secondary growth. (ii) Increases the diameter of stems and roots. 3 Permanent tissues: (a) Ground tissues (i) Parenchyma tissue • Consists of living cells. The cells are isodiametric in shape, loosely packed together and have large intercellular spaces. • Functions: – Store various organic products like starch through photosynthesis – For gases exchange – Turgid cells give support for herbaceous plants. (ii) Collenchyma tissue • Consists of living cells. The cells are polygonal in shape, have no secondary wall, have primary cell walls that are thickened at the corner, are closely packed together and have no intercellular spaces. • Functions: – Give support to young and herbaceous plants – Give flexibility to plants without restraining growth. (iii) Sclerenchyma tissue • The cells are dead at maturity, have primary and secondary walls, are tightly packed together and have no intercellular spaces. • Functions: – Give strength and support for plants – Sclereids protect seeds – Fibers like hemp are used commercially. Figure 2.10 Ground tissues Parenchyma Collenchyma Cross section Cross section Cross section Longitudinal Longitudinal Longitudinal Sclerenchyma Intercellular space Thick secondary cell wall Thin primary cell wall Lumen Thin primary cell wall Nucleus Cytoplasm Starch grain Vacuole Thick primary cell wall C02 Matriculation Sem 1 BIO 4pp.indd 26 19/06/2023 5:38 PM PENERBIT ILMU BAKTI SDN. BHD.

27 Mid Semester Test 1 Answers Chapter 2  Cell Structure and Functions Cell Structure and Functions  (b) Vascular tissues (i) Xylem Xylem consists of tracheids and vessel elements. This tissue helps transport water and mineral ions from roots. (ii) Phloem Phloem consists of sieve tube elements and companion cells. This tissue transports sucrose and other organic substances from leaves to other parts of the plant through the translocation process. Figure 2.11 Vascular tissues: Xylem and phloem Example 3 Name this tissue. Explain the structural characteristics. Answer Parenchyma tissue. The cells are isodiametric in shape, loosely packed together, and have a thin primary cell wall as well as large intercellular spaces. Quick Check 2.3 1 State the types of animal tissues. 2 State one structural adaptation of the nerve tissue to enable it to carry out its function. 3 State the vascular tissues used for transportation in plants. 4 Which tissue in animals has the same function as the phloem tissue in plants? Give a reason to support your opinion. Epidermis Cortex Pith Cambium Xylem Phloem Sieve tube element Companion cell Vessel element Xylem Cambium Phloem C02 Matriculation Sem 1 BIO 4pp.indd 27 19/06/2023 5:38 PM PENERBIT ILMU BAKTI SDN. BHD.

28 Mid Semester Test 1 Chapter 2 Answers  Cell Structure and Functions Cell Structure and Functions  2.4 Cell Transport ● Provide an overview of the transport mechanisms across the plasma membrane. ● Describe the following transport mechanisms: – Passive transport: simple diffusion, facilitated diffusion and osmosis. – Active transport: sodium-potassium pump. – Bulk transport: endocytosis and exocytosis. Learning Outcomes Overview 1 Plasma membrane is a selectively permeable membrane. It regulates the movement of substances into and out of the cell. 2 The significance of the movement of substances is that it supplies oxygen for respiration, regulate the pH, secrete useful substances for cell activities and excrete toxic waste substances. Passive Transport 1 Passive transport does not need energy. There are three types of passive transport: (a) Simple diffusion is the movement of substances down their concentration gradient across a biological membrane. Example: uptake of oxygen by cells. (b) Facilitated diffusion is the passive diffusion of polar molecules and ions such as potassium ions with the help of transport proteins in the plasma membrane. The transport proteins are channel proteins and carrier proteins. (c) Osmosis is the movement of water molecules from a region of higher water potential to a region of lower water potential through a selectively permeable membrane using aquaporins. Active Transport 1 Active transport involves the movement of solutes against their concentration gradients through transport proteins with the usage of energy. The transport proteins are called carrier proteins. Example: sodium-potassium pump which exchanges Na+ for K+ across the plasma membrane of animal cells. Figure 2.12 Sodium-potassium pump A p p p P A p p Na+ Na Na + + K K+ + K+ K+ P P C02 Matriculation Sem 1 BIO 4pp.indd 28 19/06/2023 5:38 PM PENERBIT ILMU BAKTI SDN. BHD.

29 Mid Semester Test 1 Answers Chapter 2  Cell Structure and Functions Cell Structure and Functions  Bulk Transport 1 Bulk transport across the plasma membrane occurs by exocytosis and endocytosis. Exocytosis is the fusion of vesicles with the plasma membrane for the secretion of certain molecules out of cells. Endocytosis is when cells take in molecules and particulate matter by forming new vesicles from the plasma membrane. Figure 2.13 Bulk transport: Endocytosis and exocytosis Example 4 State two differences between active transport and passive transport. Answer Active transport requires energy/ATP, whereas passive transport does not require energy/ATP. The movement of molecules in active transport occurs against the concentration gradient, whereas the movement of molecules in passive transport occurs down the concentration gradient. Quick Check 2.4 1 State three types of passive transport. 2 State one type of active transport across the animal cell membrane. 3 State the transport that forms vesicles. 4 State the transport proteins that facilitate osmosis. Endocytosis Exocytosis Vesicle Plasma membrane Plasma membrane Solutes Cytoplasm Cytoplasm Outside of cell C02 Matriculation Sem 1 BIO 4pp.indd 29 19/06/2023 5:38 PM PENERBIT ILMU BAKTI SDN. BHD.

30 Mid Semester Test 1 Answers Chapter 2 1 Which of the following organelles is found in both eukaryotic and prokaryotic cells? A Nucleus B Ribosome C Chloroplast D Golgi body 2 Why are the cell walls of prokaryotes strong and rigid? A Contain lignin B Contain pectin C Contain suberin D Contain peptidoglycan 3 Which of the following are semiautonomous organelles? A Nucleus and chloroplast B Mitochondria and Golgi body C Mitochondria and chloroplast D Endoplasmic reticulum and Golgi body 4 Which of the following organelles is not found in plant cells? A Centrioles B Cytoskeleton C Mitochondria D Endoplasmic reticulum 5 Which of the following vesicles are formed by Golgi body? A Vacuoles B Lysosomes C Phagosome D Food vacuole 6 Proteins are produced by but then modified by A nucleus; Golgi body B ribosomes; Golgi body C mitochondria; Golgi body D endoplasmic reticulum; Golgi body 7 Which of the following cells is involved in the transmission of impulse? A Sperm C Neurone B Muscle D Red blood cell 8 Which part of the human body is lined with stratified epithelium cells? A Heart C Lymph B Blood D Esophagus 9 Which of the following tissues is rigid, has no intercellular space and has pits? A Meristem B Parenchyma C Collenchyma D Sclerenchyma 10 Which of the following pass through the plasma membrane most easily? A Small polar molecules B Large polar molecules C Small non-polar molecules D Large non-polar molecules 11 How do integral proteins differ from the other membrane proteins? A Attached to the lipids in the plasma membrane B Fully or partially penetrate the plasma membrane C Attached to the carbohydrates on the plasma membrane D Embedded on the outer surface of the plasma membrane 12 In a plasma membrane, the fatty acid tails face each other in the phospholipid bilayer while the heads face the cytoplasm and extracellular fluid. A hypertonic; hypotonic B hypotonic; hypertonic C hydrophobic; hydrophilic D hydrophilic; hydrophobic Summative Practice 2 Objective Questions Instruction: There are four answer options for each question. Choose the best answer. C02 Matriculation Sem 1 BIO 4pp.indd 30 19/06/2023 5:38 PM PENERBIT ILMU BAKTI SDN. BHD.

13 Why is the plasma membrane called as Fluid Mosaic Model by Singer and Nicolson? A It is because of the solubility of water in the plasma membrane B It is because of the transport of substances across the plasma membrane C It is because of the diffusion of lipid-soluble substances through the phospholipid bilayer D It is because of the movement of lipids and integral proteins within the phospholipid bilayer 14 Which of the following types of transport is classified as active transport? A Osmosis C Simple diffusion B Phagocytosis D Facilitated diffusion 15 Water molecules move from a hypotonic solution to a hypertonic solution. What type of transport is this? A Osmosis C Simple diffusion B Active transport D Facilitated diffusion 1 Figure 1 shows three types of organelles. EXAM CLONE Organelle X Organelle Y Organelle Z Figure 1 (a) Name organelles X, Y and Z. [3 marks] (b) Name structures A, B and C. [3 marks] (c) State one function each for structures A, B and C. [3 marks] (d) Which organelles are enclosed by double membranes? [1 mark] 2 (a) Figure 2 shows a type of human connective tissue as seen under a microscope. EXAM CLONE Figure 2 Structured Questions Instruction: Answer all the questions. A B C D 31 Mid Semester Test 1 Answers  Cell Structure and Functions Cell Structure and Functions  Chapter 2 A B C C02 Matriculation Sem 1 BIO 4pp.indd 31 19/06/2023 5:38 PM PENERBIT ILMU BAKTI SDN. BHD.

(i) Name this tissue. [1 mark] (ii) Identify components A, B, C and D. [4 marks] (iii) State one adaptation and one function of component A. [2 marks] (iv) One main function of this tissue is transportation. Name two plant tissues that perform the same function. [2 marks] (b) Why donʼt human cells have cell walls? [1 mark] 3 Figure 3 shows three supporting tissues X, Y and Z which are present in flowering plants. X Y Z Figure 3 (a) Name tissues X, Y and Z. [3 marks] (b) State one main function for each of tissues X, Y and Z. [3 marks] (c) State two differences between tissue X and tissue Z. [4 marks] Essay Questions Instruction: Answer all the questions. 1 Describe the structure of the plasma membrane based on the Fluid Mosaic Model. EXAM CLONE [10 marks] 2 Explain how substances move in and out of the cellʼs plasma membrane. EXAM CLONE [12 marks] 32 Mid Semester Test 1 Answers  Cell Structure and Functions Cell Structure and Functions  Chapter 2 C02 Matriculation Sem 1 BIO 4pp.indd 32 19/06/2023 5:38 PM PENERBIT ILMU BAKTI SDN. BHD.

Genetic Inheritance CHAPTER4 48 4.1 Mendelian Genetics: Monohybrid and Dihybrid ● Define the terminologies used in genetic inheritance. ● State the characteristics of Mendel’s pea plants. ● State Mendel’s first law (Law of Segregation). ● Construct a genetic diagram on a Mendelian monohybrid cross and include the genotypic ratio (1:2:1) and phenotypic ratio (3:1) of F2 generation. ● Construct a genetic diagram on a Mendelian monohybrid test cross and include the genotypic ratio (1:1) and phenotypic ratio (1:1). ● State Mendel’s second law (Law of Independent Assortment). ● Construct a genetic diagram on a Mendelian dihybrid cross and include only phenotypic ratio (9:3:3:1) of F2 generation. ● Construct a genetic diagram on a Mendelian dihybrid test cross and include phenotypic ratio (1:1:1:1) of F2 generation. Learning Outcomes Genetic Inheritance 1 Genetic inheritance is a fundamental idea in genetics that describes how traits are handed down from one generation to the next. 2 Genetic inheritance occurs due to genetic material, in the form of DNA, being passed from parents to their offspring. 3 Table 4.1 summarises the terminologies used in genetic inheritance. Table 4.1 Terminologies used in genetic inheritance Terminology Definition Allele • Alternative versions of a gene • Determines distinguishable phenotypes • Appears as dominant or recessive C04 Matriculation Sem 1 BIO 4pp.indd 48 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

49 Mid Semester Test 1 Answers Chapter 4  Genetic Inheritance Genetic Inheritance  Terminology Definition Dominant allele • The allele that is fully expressed in the phenotype of a heterozygote • Usually represented as capital letters Figure 4.1 Dominant and recessive alleles Recessive allele • The allele that is not fully expressed in the phenotype when two different alleles are present (heterozygous genotype) • Usually represented as lowercase letters Gene • A basic unit of inheritance for a given characteristic or trait • It also refers to a unit of hereditary information consisting of a specific nucleotide base sequence in DNA Figure 4.2 Genes on chromosome Locus • The position of a gene on a chromosome • The alleles of a gene occupy the same locus on homologous chromosomes Figure 4.3 Locus of gene on homologous chromosomes P a B P a b Gene loci Dominant allele Recessive allele Genotype: PP Homozygous for the dominant allele Homozygous for the recessive allele Heterozygous aa Bb Gene 1 Gene 2 Chromosome DNA Allele for purple flowers Allele for white flowers Locus for flower-colour gene Homologous pair of chromosomes C04 Matriculation Sem 1 BIO 4pp.indd 49 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

50 Mid Semester Test 1 Chapter 4 Answers  Genetic Inheritance Genetic Inheritance  Terminology Definition Genotype • The genetic makeup of an organism • It also refers to the combination of alleles it possesses Homozygous • Having two identical alleles for a certain gene • Example: AA (homozygous dominant) or aa (homozygous recessive) Figure 4.4 Homozygous genotype Homozygote • An organism that has two identical alleles for a certain gene Heterozygous • Have two different alleles for a certain gene • Example: Aa Figure 4.5 Heterozygous genotype Heterozygote • An organism that has two different alleles for a certain gene Phenotype • The observable characteristics or physical appearance of an organism • Phenotype is determined by its genotype (genetic makeup) True-breeding / Pure-breeding • Organisms with homozygous genotype that produce offspring of the same genotype over many generations Figure 4.6 True-breeding individuals Homologous pair of chromosomes Allele for pink flowers Allele for Locus for white flowers flower-colour gene P a P a Genotype: PP Homozygous dominant Homozygous recessive aa × YY YY YY yy × yy yy C04 Matriculation Sem 1 BIO 4pp.indd 50 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

51 Mid Semester Test 1 Answers Chapter 4  Genetic Inheritance Genetic Inheritance  Terminology Definition Self-cross • Cross that involves mating between individuals of the same generation • Example: F1 × F1 Figure 4.7 Self-cross Test cross • Cross that involves mating between individuals of a dominant phenotype but unknown genotype with individuals that display a recessive phenotype • Test cross is done to determine the unknown genotype of the dominant phenotype Figure 4.8 Test cross P GENERATION (True-breeding parents) Purple flowers White flowers Fertilisation among F1 plants (F1 × F1 ) All plants have purple flowers F1 GENERATION F2 GENERATION of plants have purple flowers 3 4 of plants have white flowers 1 4 × Dominant phenotype, unknown genotype: PP or Pp? Recessive phenotype, known genotype: pp × If all the offspring are purple, then the unknown genotype would be PP If half of the offspring are purple and half of the offspring are white, then the unknown genotype would be Pp Pp Pp Pp Pp P P p p Pp pp Pp pp P p p p C04 Matriculation Sem 1 BIO 4pp.indd 51 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

52 Mid Semester Test 1 Chapter 4 Answers  Genetic Inheritance Genetic Inheritance  Characteristics of Mendel’s Pea Plant 1 Gregor Mendel was an Austrian monk who studied the inheritance of characteristics in garden peas (Pisum sativum) which he grew on the vegetable farm in his monastery. 2 Four reasons for the selection of the pea plant for Mendel’s experiments: (a) The plants could be easily grown in large numbers. (b) The plants have a short life cycle. (c) The fertilisation process could be controlled; the reproductive structures are covered. (d) Their characteristics could be easily observed because the different traits are distinctive. Figure 4.9 Characteristics of Mendel’s pea plants Flower colour Purple White Axial Terminal Yellow Green Tall Dwarf Round Wrinkled Inflated Constricted Green Yellow Flower position Seed colour Pod colour Stem length Seed shape Pod shape C04 Matriculation Sem 1 BIO 4pp.indd 52 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

53 Mid Semester Test 1 Answers Chapter 4  Genetic Inheritance Genetic Inheritance  Mendel’s Monohybrid Cross 1 Mendel’s monohybrid cross focused on one characteristic only, for example, stem length. 2 From Mendel’s experiment, he observed that: (a) there was no intermediate traits that were produced from the cross between two pure-breeding parents. (b) in the F1 generation, only one trait can be observed. (c) in the F2 generation, the traits of the parents were observed, but the ratio was 3(dominant phenotype) : 1(recessive phenotype). (d) the parents must be pure-breeding. Example of pure-breeding parents are tall plants (TT) that cross with dwarf plants (tt). (e) the F1 generation produced displays a dominant phenotype (tall) with a heterozygous genotype (Tt). (f) the F1 generation is produced from a combination of gametes from the parents; T allele from the tall parent and t allele from the dwarf parent. (g) specific factors responsible for controlling the characteristics of the plants are known as genes or a pair of alleles. (h) only one of the pair was passed down to the gamete during meiosis. Figure 4.10 Mendel’s monohybrid cross Mendel’s Monohybrid Self-cross of Pea Plants in Reference to Flower Colour 1 When Mendel crossed true-breeding purple-flowered and white-flowered plants, all the offsprings of the F1 generation were heterozygotes (Pp) with a dominant phenotype (purple flowers). 2 When individuals from the F1 generation were self-crossed, the offsprings produced in the F2 generation consisted of both purple (dominant phenotype) and white (recessive phenotype) in the ratio of 3:1. 3 The genotypic ratio produced in the F2 generation was 1(PP) : 2(Pp) : 1(pp). Parental generation: P Filial 1 generation: F1 Filial 2 generation: F2 Phenotypic ratio × 3 : 1 All are tall plants F1 × F1 Self-cross C04 Matriculation Sem 1 BIO 4pp.indd 53 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

54 Mid Semester Test 1 Chapter 4 Answers  Genetic Inheritance Genetic Inheritance  Figure 4.11 Phenotypic and genotypic ratio in Mendel’s monohybrid self-cross Law of Segregation 1 Based on the monohybrid cross, Mendel proposed his first law, known as the Law of Segregation. 2 The characteristics of an organism are determined by a pair of alleles. 3 Law of Segregation states that the pair of alleles segregates during meiosis and only one allele of each pair can be present in a single gamete. TT T T tt t t × Figure 4.12 Segregation of alleles into different gametes Purple flowers All dominant Parental phenotype: Parental genotype: Gamete: Gamete: Purple Purple Purple White F1 genotype: F1 phenotype: Phenotypic ratio: 3 purple (dominant) : 1 white (recessive) Genotypic ratio: 1 PP : 2 Pp : 1 pp F2 genotype: F2 phenotype: F1 × F1 : (self-cross) × Purple flowers PP White flowers pp P P p p Pp Pp Pp Pp × Purple flowers Pp Purple flowers Pp P p P p PP Pp Pp pp C04 Matriculation Sem 1 BIO 4pp.indd 54 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

55 Mid Semester Test 1 Answers Chapter 4  Genetic Inheritance Genetic Inheritance  4 Segregation in genetics is defined as the separation of pairs of homologous chromosomes during meiosis which leads to the separation of the allele pairs. Figure 4.13 Separation of homologous chromosomes during meiosis Monohybrid Test Cross of Pisum sativum 1 A test cross is done to determine the unknown genotype of an individual that displays a dominant phenotype. 2 The genotype of the organism with a dominant phenotype could be homozygous or heterozygous. 3 A test cross is done by crossing an organism with a dominant phenotype with a recessive phenotype. 4 The genotype of the organism with a recessive phenotype must always be homozygous for the recessive allele. 5 When Mendel crossed true-breeding parents of tall and dwarf plants, all the offsprings of the F1 generation were heterozygotes (Tt) with a dominant phenotype (tall plants). 6 When individuals from the F1 generation were test crossed, the offsprings produced in the F2 generation consisted of both tall (dominant phenotype) and dwarf (recessive phenotype) in the phenotypic ratio of 1:1. y Y y Y y Y y y Y Y Prophase I Metaphase I Anaphase I Telophase I MEIOSIS I MEIOSIS II Prophase II Heterozygous (Yy) diploid cell from a plant with yellow seeds Metaphase II Anaphase II Telophase II y Y Possible haploid gametes C04 Matriculation Sem 1 BIO 4pp.indd 55 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

56 Mid Semester Test 1 Chapter 4 Answers  Genetic Inheritance Genetic Inheritance  To answer this question, the student must first identify whether the cross involves one or two characteristics. In this case, this is a monohybrid cross because it involves only one characteristic. Then identify the alleles that control each dominant and recessive phenotype. The assumption made that the genes for body colour are not linked, implies that this is Mendelian inheritance. Smart Tips! 7 The genotypic ratio produced in the F2 generation was 1(Tt) : 1(tt). Figure 4.14 Mendel’s monohybrid test cross Example 1 In Drosophila sp., wild-type grey body is controlled by dominant allele G, while yellow body is controlled by recessive allele g. Assuming that the genes for body colour are not linked, construct a genetic diagram of a cross between a homozygous grey-bodied male with a yellow-bodied female Drosophila sp. Answer Parental phenotype : Parental genotype : Gamete : F1 genotype : F1 phenotype : GG gg × Grey body Yellow body Gg Grey body G g Parental genotype : Gamete : Gamete : F1 genotype : F1 phenotype : F2 genotype : F2 phenotype : F 1(tall) 1(dwarf) : 2 phenotypic : F1 (test cross) : TT × t t tall dwarf Tt tall T t Tt × t t tall dwarf Tt tall t t dwarf T t t A test cross on a tall pea plant whose genotype is Tt will produce tall and dwarf offspring in the phenotypic ratio of 1 : 1 C04 Matriculation Sem 1 BIO 4pp.indd 56 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

57 Mid Semester Test 1 Answers Chapter 4  Genetic Inheritance Genetic Inheritance  Mendel’s Dihybrid Cross 1 A dihybrid cross is a cross between individuals that involves two characteristics that are controlled by two distinct genes, for example, shape and colour of seeds. 2 Mendel crossed plants from two pure-breeding strains, one with round and yellow seeds (RRYY), the other plant had wrinkled and green seeds (rryy). 3 All the offspring in the F1 generation produced from the cross were individuals with dominant phenotypes for both characteristics: round with yellow seeds. The genotype for the F1 generation was heterozygous for both characteristics (RrYy). 4 When the F1 generation was self-crossed, the F2 generation produced a phenotypic ratio of 9 (Round, Yellow seeds) : 3 (Round, Green seeds) : 3 (Wrinkled, Yellow seeds) : 1 (Wrinkled, Green seeds). Figure 4.15 Mendel’s dihybrid self-cross A cross between homozygous parents (pure breed) All are round and yellow P : G : F1 : RRYY RY ry rryy Round / Yellow Wrinkled / Green × RrYy Self-cross between F1 (F1 × F1 : heterozygotes) F1 × F1 : RrYy × Round / Yellow RrYy Round / Yellow G : RY Ry rY ry RY Ry rY ry F2 Punnett square: ♀ gametes ♂ gametes RY RRYy RRyy RRYY RRYy RrYY RrYy RrYY RrYy RrYy rrYY rrYy rrYy rryy RrYy Rryy RY Ry Ry rY rY ry ry Rryy Round, yellow (RRYY, RrYY, RRYy, RrYy) Round, green (RRyy, Rryy) Wrinkled, yellow (rrYY, rrYy) Wrinkled, green (rryy) : 9 : 3 : 3 : 1 Dihybrid ratio C04 Matriculation Sem 1 BIO 4pp.indd 57 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

58 Mid Semester Test 1 Chapter 4 Answers  Genetic Inheritance Genetic Inheritance  5 Mendel observed that two phenotypes of the F2 generation resembled both parents, while the other two had combined traits of both parents. 6 He concluded that the two pair of alleles that controlled the two characteristics segregate independently of each other. 7 This led him to formulate his second law, the Law of Independent Assortment. Law of Independent Assortment 1 Mendel’s Law of Independent Assortment states that each member of a pair of alleles may combine randomly with either one of another pair of alleles during gamete formation. 2 This law can be related to the independent arrangement and separation of homologous chromosomes during metaphase I and anaphase I, respectively. 3 As a result, four possible arrangements of alleles can be found in each of the male and female gametes. 4 Further, each gamete from one parent is randomly fertilised with any of the four gametes from the other parent. Figure 4.16 Law of Independent Assortment MEIOSIS I Prophase I Heterozygous (YyRr) diploid cell from a plant with round yellow seeds r y Y R y Y R r y Y r R y R Y r R y Y r y R y R Y r Y r y r y r Y R Y R Metaphase I Anaphase I Telophase I OR MEIOSIS II Prophase II Metaphase II Possible haploid gametes Anaphase II Telophase II C04 Matriculation Sem 1 BIO 4pp.indd 58 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

59 Mid Semester Test 1 Answers Chapter 4  Genetic Inheritance Genetic Inheritance  Dihybrid Test Cross of Pisum sativum 1 Mendel’s dihybrid test cross involved two characteristics. 2 When Mendel crossed true-breeding parents of round and yellow seeds (RRYY) with wrinkled and green seeds (rryy), all the offsprings of the F1 generation were heterozygotes (RrYy) with the dominant phenotype (round and yellow seeds). 3 When individuals from the F1 generation were test crossed, the offsprings produced in the F2 generation produced a genotypic and phenotypic ratio of 1:1:1:1. Figure 4.17 Mendel’s dihybrid test cross Example 2 Differentiate between monohybrid and dihybrid cross. Answer A monohybrid cross is defined as the cross happening in the F1 generation offspring of parents differing in one trait only. A dihybrid cross is a cross happening in the F1 generation offspring of parents differing in two traits. All are round and yellow Gamete : F1 genotype : F1 : (Test cross) F2 : RY ry RRYY rryy Round / Yellow Wrinkled / Green × RrYy RrYy Rryy rrYy rryy rryy Round / Yellow Wrinkled / Green × RrYy G : Genotypic : ratio :1 :1 :1 1 Phenotypic : ratio Parental : genotype Round, yellow seed Round, green seed Wrinkle, yellow seed Round, green seed :1 :1 :1 1 RY Ry rY ry ry C04 Matriculation Sem 1 BIO 4pp.indd 59 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

60 Mid Semester Test 1 Chapter 4 Answers  Genetic Inheritance Genetic Inheritance  Example 3 Pure-breeding pea plants with round, yellow seeds were crossed with pure-breeding plants with wrinkled, green seeds. All the offspring in the F1 generation had round, yellow seeds. Using a genetic diagram, explain the phenotypic ratio you would expect if the F1 generation were self-crossed. Answer F1 genotype : Parental genotype : Parental phenotype : F1 phenotype : F1 × F1 : Gametes : Gametes : F2 Punnet square: F2 phenotypic ratio: 9 (round, yellow seeds) : 3 (round, green seeds) : 3 (wrinkled, yellow seeds) : 1 (wrinkled, green seeds) RY Ry rY ry RY Ry rY ry Round, yellow seeds Wrinkled, green seeds RRYY × rryy RrYy × RrYy RY ry RrYy Round, yellow seeds Gametes RY Ry rY ry RY RRYY RRYy RrYY RrYy Ry RRYy RRyy RrYy Rryy rY RrYY RrYy rrYY rrYy ry RrYy Rryy rrYy rryy Quick Check 4.1 1 Define gene and allele. 2 State why the characteristics of Mendel’s pea plant can easily be observed. 3 State Mendel’s first law. 4 State Mendel’s second law. Find out the keywords and identify the type of inheritance. In the question, it is stated that the parents are purebreeding plants and have two characteristics (colour and shape of seed). So, this is a dihybrid inheritance. To determine which phenotype is dominant and which one is recessive, look at the phenotype of the F1 generation because from the concept that has been learned, when pure-breeding individuals are crossed together, all the offspring produced will have the dominant phenotype. Smart Tips! C04 Matriculation Sem 1 BIO 4pp.indd 60 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.