Biology Sem 1

61 Mid Semester Test 1 Answers Chapter 4  Genetic Inheritance Genetic Inheritance  4.2 Deviations from the Mendelian Inheritance ● Explain briefly the types of inheritance that deviate from Mendelian inheritance: codominant alleles, incomplete dominant alleles, multiple alleles, linked genes, sex-linked genes and polygenes. Learning Outcome Codominant Alleles 1 In codominant inheritance, both alleles of a pair are fully expressed in the phenotype of an individual with heterozygous genotype. 2 The heterozygote offspring has the characteristics of both the homozygous parents. 3 An example of codominant alleles can be found in human MN blood type. 4 There are three groups in the human MN blood type. The blood groups are: (a) blood group M : genotype LMLM (b) blood group N : genotype LN LN (c) blood group MN : genotype LMLN 5 When heterozygous F1 individuals are self-crossed, the phenotypic ratio produced in the F2 generation is 1:2:1. Figure 4.18 Codominant alleles F1 genotype : Parental genotype : F1 phenotype : Gametes : Blood group Blood group M LMLM LNLN N × LMLN × LMLN F2 genotype : LMLM LMLN LMLN LNLN F2 genotypic ratio : LMLM LMLN 1 2 1 1 2 1 LNLN All MN blood group F2 phenotype : M blood group MN blood group N blood group F2 phenotypic ratio : M blood group MN blood group N blood group F1 × F1 : Parental phenotype : LM LN Gametes : LM LN LMLN LM LN C04 Matriculation Sem 1 BIO 4pp.indd 61 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

62 Mid Semester Test 1 Chapter 4 Answers  Genetic Inheritance Genetic Inheritance  Incomplete Dominant Alleles 1 Incomplete dominance is a form of intermediate inheritance. 2 It occurs when one allele for a specific trait is not completely dominant over the other allele, resulting in an intermediate phenotype. 3 The expressed physical trait is a combination of the dominant and recessive phenotypes. 4 An example of incomplete dominant alleles is the flower colour in Antirrhinum sp. (snapdragon flower). 5 The flower colour is controlled by two alleles: (a) CR is the allele that codes for red flower (b) CW is the allele that codes for white flower 6 There are three phenotypes of the Antirrhinum sp. flower colour. The phenotypes are: (a) red flower: genotype CR CR (b) white flower: genotype CWCW (c) pink flower: genotype CR CW 7 When a red-flowered snapdragon plant (CR CR ) is crossed with a white-flowered snapdragon plant (CWCW), all the offspring in the F1 generation would produce pink flowers (CR CW). 8 When heterozygous F1 individuals are self-crossed, the phenotypic ratio produced in the F2 generation is 1:2:1. Figure 4.19 Incomplete dominant allele Multiple Alleles 1 Multiple alleles is a condition when 3 or more alleles of a single gene determine the phenotype for a characteristic. 2 The alleles can occupy the same locus on a pair of homologous chromosomes, but only two of the alleles can be present in a single organism. 3 Example of a characteristic with multiple alleles is the ABO blood group in humans. 4 The ABO locus has alleles, IA, IB and i. CRCR CRCW CRCW CWCW CR CW CR CW F2 generation 1 : 2 : 1 CRCR:CRCW:CWCW F1 generation All CRCW CWCW CRCR Female gamete Male gamete C04 Matriculation Sem 1 BIO 4pp.indd 62 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

63 Mid Semester Test 1 Answers Chapter 4  Genetic Inheritance Genetic Inheritance  5 i is a recessive allele while IA and IB are both dominant allele. IA and IB are also codominant alleles. 6 The combination of any two alleles will produce four blood types: (a) blood A (c) blood AB (b) blood B (d) blood O Genotype Phenotype I AI A I Ai Type A I B I B I B i Type B I AI B Type AB ii Type O Figure 4.20 ABO blood group in humans Example 4 A couple has four children. Each child has a different blood group. Using a genetic diagram, determine the blood groups and genotypes of the parents. Answer F1 genotype : Parental genotype : F1 phenotype : Gametes : Parental phenotype : Blood AB Blood A Blood B Blood O ii Blood A I Ai I A I i AI B I Bi I Bi Blood B I A i I B i Polygenes 1 In polygenic inheritance, a characteristic is controlled by the cumulative effect of more than one gene at different loci. 2 An example of polygenic inheritance is human skin colour. 3 Skin colour is under the control of more than one gene. It depends on how many active alleles a person inherits from his or her parents. 4 Let’s say that skin colour is controlled by 3 genes, each with 2 alleles. So, there are 6 possible alleles that contribute to skin colour. 5 Each dominant allele (A, B, C) would contribute to one unit of darkness to the phenotype. 6 A person with AABBCC genotype would be very dark, while a person with aabbcc genotype would have very fair skin colour. 7 A person with AaBbCc genotype would have an intermediate skin colour. 8 Polygenic inheritance produces a bell-shaped curve, also known as normal distribution. The bell-shape curve has a mean value and extremes in either direction. 9 Environmental factors may affect the polygenic-inherited phenotype. The children have all different blood groups: A, B, AB and O. In order to obtain these phenotypes, there must be combination of all alleles. So the genotype of the parents must have all three allleles: I A, I B and i. So the parents’ blood group are blood group A with genotype I Ai and blood group B with genotype I Bi. Smart Tips! C04 Matriculation Sem 1 BIO 4pp.indd 63 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

64 Mid Semester Test 1 Chapter 4 Answers  Genetic Inheritance Genetic Inheritance  Figure 4.21 Polygenic inheritance in human skin colour Linked Genes 1 Linked genes are different genes that are linked on the same chromosome. 2 The genes remain together during the separation of homologous chromosomes at meiosis. But if crossing over occurs during prophase I of meiosis, the genes will be separated. Figure 4.22 The genes tend to be inherited together since the chromosomes are passed along as one unit 3 When crossing over occurs, new combinations of genes will be produced. Figure 4.23 Crossing over produces new combinations of genes 1/64 6/64 15/64 20/64 X AaBbCc AaBbCc 1/64 6/64 15/64 15/64 20/64 6/64 1/64 Fraction of population A B a b a b Homologous pair Meiosis I A B A B a b A B A B A b a B a b a b Homologous pair Crossing over Gametes Recombinants C04 Matriculation Sem 1 BIO 4pp.indd 64 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

65 Mid Semester Test 1 Answers Chapter 4  Genetic Inheritance Genetic Inheritance  4 If compared to Mendel’s dihybrid inheritance, the ratio produced from inheritance that involve linked gene is not the same as Mendel’s ratio. 5 In Mendel’s dihybrid test cross, the phenotypic ratio is 1:1:1:1, but for linked genes, if a test cross is done on an individual who is heterozygous for both characteristics, there will be no specific ratio. 6 The offspring produced from the linked gene test cross will consist of those that resemble the parents’ phenotypes and those with recombinant phenotypes. 7 The number of offspring with parental phenotypes are normally much larger compared to the number of offspring with recombinant phenotypes. 8 The recombinant phenotypes are due to crossing over. Figure 4.23 Linked gene test cross with crossing over Sex-linked Genes 1 In humans, every cell contains 23 pairs of chromosomes. 2 Chromosome pairs number 1 – 22 are identical in male and female. These chromosomes are known as autosomes. 3 The 23rd pair is known as the sex chromosome. The sex chromosomes are X chromosome and Y chromosome. 4 Genes that are linked on the sex chromosomes are said to be sex-linked. 5 For this topic, the sex-linked gene refer to a recessive allele located on the sex chromosome. 6 Since human females have XX chromosomes, a female will express the phenotype only if she has a homozygous genotype for that allele. For example, Xh Xh . 7 As for human males, they have XY chromosomes. So, any male that receives the recessive X-linked allele from his mother will express the trait. For example, Xh Y. Wild type Black vestigial ♀ ♂ B Vg b vg b vg b vg X b vg b vg b vg b vg b vg b vg b vg B Vg B Vg b Vg b Vg B vg B vg Black vestigial Wild type Grey vestigial Black normal 944 965 206 185 Parental phenotypes Recombinant phenotypes C04 Matriculation Sem 1 BIO 4pp.indd 65 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

66 Mid Semester Test 1 Chapter 4 Answers  Genetic Inheritance Genetic Inheritance  8 Males normally suffer from the effects of sex-linked genetic disorders more often than females because the Y chromosome does not carry any gene or allele. 9 Examples of sex-linked inheritance are: (a) haemophilia in humans (XH is the normal allele, Xh is the recessive allele that codes for haemophilia) (b) red-green colour blindness in humans (XN is the normal allele, Xn is the recessive allele that codes for colour blind) Figure 4.24 Inheritance of haemophilia in humans Example 5 A normal female marries a normal man. But one of their children inherited a recessive allele, causing that child to suffer from a genetic disorder, that makes it difficult for him/her to differentiate between red and green colour. Draw a genetic diagram to identify the parents’ genotype and determine the genotype and probability of the child who will suffer from the disorder. Answer The genotypes of the parents are XNXn for female and XNY for male. The genotype of the child with the disorder is Xn Y and the probability to obtain a child with the disorder is ¼. Gamete Parental phenotype : Normal female XNXn XNXN Normal female Normal male Normal female Colour blind male XNY XNXn Xn Y XN XN × XNY Normal male Parental genotype : F1 phenotype : F1 genotype : Xn Y Identify the type of inheritance in this situation. The clue to identify the type of inheritance is “But one of their children inherited a recessive allele, causing that child to suffer from a genetic disorder, that makes it difficult for him/her to differentiate between red and green colour.” From this sentence, it is clear that the genetic disorder is red-green colour blindness which is an X-linked inheritance. A female who is normal could have a homozygous genotype or heterozygous genotype (also known as carrier). The mother should have a heterozygous genotype that enables the recessive allele to be inherited by her son. Smart Tips! Haemophilia is a genetic disorder that slows down the blood clotting process. Those who suffer from this disorder would experience prolonged bleeding due to injury or surgery. (Source: https:// medlineplus.gov/genetics/ condition/hemophilia/) Bio Info Carrier female Normal female XHXh XH Xh XH Y XHY XH XHXH Xh XH Y XHY Xh Y XHXh Normal female Carrier female Normal male Haemophilic male Gametes: Offspring C04 Matriculation Sem 1 BIO 4pp.indd 66 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

67 Mid Semester Test 1 Answers Chapter 4  Genetic Inheritance Genetic Inheritance  Quick Check 4.2 1 What is codominant allele? 2 State one example of incomplete dominant inheritance. 3 List the alleles of human ABO blood group. 4 What is polygenic inheritance? 4.3 Genetic Mapping ● Define genetic mapping. ● Explain genetic mapping in relation to recombinant frequency during meiosis. ● Calculate the genetic distance in map unit between genes using the given recombination data. ● Identify the position of genes along a chromosome based on recombination data. Learning Outcomes Genetic Mapping 1 Genetic mapping is the process that determines the position of genes along the length of a chromosome. 2 The distance between genes can be determined by the crossover value (COV). 3 COV refers to the percentage of linked genes that are exchanged during a crossing over. It can be obtained from linked gene test crosses. 4 1% COV is equivalent to 1 map unit. 5 The unit for distance between genes can be written as centimorgan (cM). 6 The lower the value of COV, the closer the genes would be. 7 If two genes are farther apart, the probability that crossing over will occur between them is higher. Therefore, the recombination frequency would be higher. Figure 4.25 Genetic mapping based on COV Closer genes COV Farther genes COV C04 Matriculation Sem 1 BIO 4pp.indd 67 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

68 Mid Semester Test 1 Chapter 4 Answers  Genetic Inheritance Genetic Inheritance  8 Formula to calculate the crossover value; COV = Total number of recombinant Total number of offspring × 100 Example 6 A test cross is done between a smooth seed, red flower plant with a wrinkled seed, white flower plant. Given that smooth seed (S) is dominant over wrinkled (s) seed, and red flower (R) is dominant over white (r) flower. The offspring produced from the test cross are as follows: Smooth seed, red flower = 300 Wrinkled seed, white flower = 300 Smooth seed, white flower = 100 Wrinkled seed, red flower = 100 Determine the map unit for the distance between both S and R genes on the chromosome. Answer COV = Total number of recombinant Total number of offspring × 100 = (100 + 100) (100 + 100 + 300 + 300) × 100 = 25% 25 map units S R Distance between gene S and R is 25 map units Quick Check 4.3 1 Define genetic mapping. 2 How does genetic mapping relate to recombinant frequency? 3 What type of cross is involved in determining the distance between genes? 4 State the formula to calculate COV. The offspring produced from the test cross would have phenotypes that resemble the parental phenotypes and also offspring with recombinant phenotypes. Distance between the two genes can be determined from COV. Smart Tips! C04 Matriculation Sem 1 BIO 4pp.indd 68 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

69 Mid Semester Test 1 Answers Chapter 4 1 The term ‘true-breeding’ refers to… A genetically pure lines. B organisms that have a high rate of reproduction. C organisms that are heterozygous for a given trait. D organisms that will produce identical copies of themselves upon reproduction. 2 Mating a true-breeding purple-flowered plant with a true-breeding purple-flowered plant will produce… A only plants with white flowers. B only plants with purple flowers. C plants with purple and white flowers in a 3:1 phenotypic ratio. D plants with purple and white flowers in a 1:1 phenotypic ratio. 3 The term ‘dominant’ refers to… A both alleles can be expressed in a heterozygous. B the dominant phenotype is more beneficial than the recessive phenotype. C one allele can mask the expression of another in a heterozygous genotype. D all members of the F2 generation from a F1 self-cross shows the dominant phenotype. 4 The observable characteristic of an organism for a given trait is termed… A synapsis. B genotype. C phenotype. D dominance. 5 is the location of a particular gene on a chromosome. A Trait B Locus C Allele D None of the above. 6 Mendel’s Law of Segregation states that… A alleles separate from each other during gamete formation. B true-breeding parents produce offspring of the same phenotype. C alleles from one parent mask the expression of alleles from the other parent. D hybrids will express a phenotype intermediate between the two parental phenotypes. 7 The separation of alleles of a gene takes place during… A anaphase of mitosis. B cytokinesis of mitosis. C anaphase I of meiosis. D telophase II of meiosis. 8 Mating an individual expressing a dominant phenotype, but whose genotype is unknown, with an individual expressing the corresponding recessive phenotype represents a… A test cross. B self-cross. C parental cross. D heterozygous cross. 9 Genes that tend to be inherited together are said to be… A linked. B alleles. C related. D associated. 10 Map units express the distance between… A alleles. B homologous chromosomes. C two loci on a chromosome. D chromosomes during metaphase. Summative Practice 4 Objective Questions Instruction: There are four answer options for each question. Choose the best answer. C04 Matriculation Sem 1 BIO 4pp.indd 69 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

11 The gender of a human is determined by… A the number of autosomes. B the presence of Y chromosome. C the number of sex chromosomes. D the presence of only one X chromosome. 12 refers to multiple independent pairs of genes that have similar and additive effects on the same characteristic. A Epistasis B Codominance C Complete dominance D Polygenetic inheritance 13 When heterozygous individuals with pink flower from F1 generation undergo self-cross, the F2 generation produced consists of offspring with a phenotypic ratio of 1 red flower: 2 pink flower: 1 white flower. This is an example of… A epistasis. B hybrid vigor. C polygenic inheritance. D incomplete dominance. 14 Why is colour-blindness more common in males than in females? A Because colour-blindness is an X-linked trait. B Because a male only needs to receive the recessive gene from his mother to inherit colour-blindness. C Because females would have to receive two copies of the recessive colour-blindness gene to express the trait. D All of the above. 15 Which of the following must be true for a female to inherit haemophilia? A Her mother has the disease, while her father is normal. B Her father and mother both have a heterozygous genotype. C Her mother is normal with a homozygous dominant genotype. D Her father has the disease, while her mother has a heterozygous genotype. 1 A fruit fly with wild phenotype but unknown genotype was crossed with a fruit fly with vestigial wings and grey body, which are homozygous recessive for both characteristics. Table 1 shows the phenotypes and number of offspring produced from the cross. EXAM CLONE Phenotypes Number of offspring Normal wings, black body 47 Normal wings, grey body 49 Vestigial wings, black body 46 Vestigial wings, grey body 48 Table 1 (a) State Mendel’s law for the above cross. [1 mark] (b) Explain Mendel’s law stated in 1(a). [2 marks] (c) By using the following symbols, state the genotypes of the parents. Normal wings – N Vestigial wings – n Black body – E Grey body – e [2 marks] Structured Questions Instruction: Answer all the questions. 70 Mid Semester Test 1 Answers  Genetic Inheritance Genetic Inheritance  Chapter 4 C04 Matriculation Sem 1 BIO 4pp.indd 70 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

(d) With the aid of a genetic diagram, show how the result shown in Table 1 is obtained. [5 marks] (e) State the genotypic and phenotypic ratio for the F1 generation. [2 marks] (f) Name the type of cross that produced the offspring in Table 1. [1 mark] 2 In an experiment carried on hamsters, long fur (L) is dominant against short fur (l), and grey fur (G) is dominant against brown fur (g). Ahmad’s hamster has long, grey fur but the genotype is unknown. (a) List four possible genotypes for Ahmad’s hamster. [4 marks] (b) How can Ahmad determine his hamster’s genotype? [1 mark] (c) If the genes are not linked, draw a genetic diagram showing a cross between a heterozygous hamster and a homozygous recessive hamster for both genes. [5 marks] (d) (i) Define genetic mapping. [2 marks] (ii) What information from a particular cross is required to enable mapping of two genes on a chromosome? [1 mark] 3 In Drosophila sp., the dominant G allele codes for wild-type Drosophila sp. that has a grey body. The recessive g allele codes for a yellow body. (a) A cross is done between a homozygous grey-bodied male with a yellow-bodied female, considering that the genes are not sex-linked. (i) Determine the genotype for the male and female Drosophila sp. [2 marks] (ii) Draw a genetic diagram for the above crossing. [5 marks] (b) If the gene for body colour is sex-linked, consider a cross between a grey-bodied male with a yellow-bodied female. (i) Draw a genetic diagram for the above crossing. [4 marks] (ii) What is the percentage of the F1 offspring expected to be homozygous? [1 mark] (iii) Among the male offspring, what percentage is expected to be yellow? [1 mark] 71 Mid Semester Test 1 Answers  Genetic Inheritance Genetic Inheritance  Chapter 4 C04 Matriculation Sem 1 BIO 4pp.indd 71 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

Essay Questions Instruction: Answer all the questions. 1 A tall tomato plant is controlled by a dominant allele, whereas a short tomato plant is controlled by a recessive allele. A hairy stem is controlled by a dominant allele, whereas a smooth stem is controlled by a recessive allele. Both genes are not linked. Using appropriate symbols for the allele: EXAM CLONE (a) Draw a genetic diagram showing a test cross on a tomato plant which is heterozygous for both genes. [9 marks] (b) State the genotypic and phenotypic ratios from the test cross. [2 marks] (c) Define the Law of Independent Assortment. [2 marks] 2 John has brown hair and blue eyes. He is married to a woman with black hair and brown eyes, heterozygous for both genes. Explain how their children would have black hair and blue eyes. Assume that the trait of brown eyes and black hair is dominant over blue eyes and brown hair, and the genes are not linked on the same chromosome. [13 marks] 72 Mid Semester Test 1 Answers  Genetic Inheritance Genetic Inheritance  Chapter 4 C04 Matriculation Sem 1 BIO 4pp.indd 72 19/06/2023 5:57 PM PENERBIT ILMU BAKTI SDN. BHD.

Population Genetics CHAPTER5 73 5.1 Gene Pool Concept ● Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium. Learning Outcome Population Genetics 1 Population genetics is the study of changes of allele frequencies within a population. 2 This involves studying changes in the frequencies of genetic variation in populations over space and time. 3 Table 5.1 summarises the terminologies used in population genetics. Table 5.1 Terminologies used in population genetics Terminologies Definition Population • A group of individuals of the same species that live in the same geographical area and are free to interbreed among themselves to produce fertile offspring • The individuals share the same gene pool Gene pool • Total number of genes in a population at a particular time • Consists of all alleles in all gene loci in all individuals of the population at a given time Allele frequency • Ratio of any given allele in a population, relative to all other alleles of that gene at the same locus Figure 5.1 Gene pool of a population Gene pool C05 Matriculation Sem 1 BIO 4pp.indd 73 19/06/2023 5:59 PM PENERBIT ILMU BAKTI SDN. BHD.

74 Mid Semester Test 1 Chapter 5 Answers  Population Genetics Population Genetics  Gene Pool Concept 1 In a diploid organism, each locus is represented twice in the genome. The genotype could be either homozygous (having two identical alleles) or heterozygous (having two different alleles). 2 Each of the allele type has a relative frequency in the gene pool. Example 1 A study was done on 10 000 hamsters. The study focused on the fur colour which is determined by dominant allele B and recessive allele b. Brown fur is dominant to white fur. Among the total number of hamsters studied, 6 000 are brown hamsters with homozygous dominant genotype (BB), 2 000 are brown hamsters with heterozygous genotype (Bb) and the rest are white hamsters with homozygous recessive genotype (bb). Calculate the allele frequencies for both the dominant and recessive alleles in the population. Answer Total number of BB individuals = 6 000 Total number of Bb individuals = 2 000 Total number of bb individuals = 2 000 Total number of all alleles for fur colour = 2(6 000) + 2(2 000) + 2(2 000) = 20 000 Total number of dominant allele (B) = 2(6 000) + 2 000 = 14 000 Total number of recessive allele (b) = 2(2 000) + 2 000 = 6 000 Frequency for dominant allele (B) = 14 000 20 000 = 0.7 Frequency for recessive allele (b) = 6 000 20 000 = 0.3 3 Allele frequencies in a gene pool determine the genetic change for a population. 4 Naturally, the composition of a gene pool may change over time due to certain factors. 5 Changes in allele frequency will lead to a change in the gene pool. When there is a change in the gene pool in a population, this will lead to changes in the genetic composition of a population, eventually leading to evolution. 6 If there is no change in allele frequencies from one generation to the next, the gene pool is considered to be static. 7 This means that the population is in genetic equibrilium. To answer this question, student must first identify the total number of individuals for each genotype. Each genotype possesses two alleles for a particular characteristic. So, to find out the number of alleles in the gene pool, the number of individuals with a certain genotype must be multiplied by 2. Allele frequency refers to the ratio of any given allele in a population, relative to all other alleles of that gene at the same locus. Thus, the allele frequency can be calculated by dividing the number of the respective allele with the total number of alleles. Smart Tips! C05 Matriculation Sem 1 BIO 4pp.indd 74 19/06/2023 5:59 PM PENERBIT ILMU BAKTI SDN. BHD.

Assumptions in Hardy-Weinberg Law Large population size No migration No mutation No natural selection Random mating 75 Mid Semester Test 1 Answers Chapter 5  Population Genetics Population Genetics  Quick Check 5.1 1 Define population genetics. 2 What does it mean if the population is said to be in genetic equilibrium? 3 Define gene pool. 4 In a population of 400 mice, 196 are homozygous dominant for brown fur (BB), 168 are heterozygous (Bb) and 36 are homozygous recessive (bb). Determine the allele frequencies in this population. 5.2 Hardy-Weinberg Law ● State the Hardy-Weinberg Law. ● Explain five assumptions of Hardy-Weinberg Law for genetic equilibrium. ● Calculate allele and genotype frequencies. Learning Outcomes Hardy-Weinberg Law 1 Hardy-Weinberg Law states that a population is in genetic equilibrium when the frequencies of alleles and genotypes in a population’s gene pool remain constant over the generations. 2 The population would be in genetic equilibrium as long as these five assumptions are met: (a) large population size: in a small population, the allele frequencies may change due to genetic drift. (b) random mating: the choosing of mates with certain traits only, may cause frequency of certain alleles to change. (c) no mutation: mutation may cause changes in the gene, thus will lead to change in allele frequency. (d) no migration: if individuals migrate from one place to another, this may cause gene flow that can change the gene pool composition. (e) no natural selection: in natural selection, the environment favors individuals that are well adapted to that particular environment. Once they survive, they will reproduce among themselves and pass the favoured gene to the next generation. This may also change the composition of a gene pool. 3 Hardy-Weinberg Law describes a non-evolving population which is rarely met naturally. Figure 5.2 Assumptions in Hardy-Weinberg Law C05 Matriculation Sem 1 BIO 4pp.indd 75 19/06/2023 5:59 PM PENERBIT ILMU BAKTI SDN. BHD.

76 Mid Semester Test 1 Chapter 5 Answers  Population Genetics Population Genetics  Hardy-Weinberg Equations 1 Hardy-Weinberg equation is used to estimate the allele and genotype frequencies in a population which is in genetic equilibrium. 2 Based on the five assumptions, Hardy and Weinberg derived an equation based on the theory that the total frequency of alleles is 100%. It can be expressed as decimal, 1.0 that represents 100%. 3 With that, the equation to calculate allele frequency is p + q = 1 , where p is the frequency of the dominant allele q is the frequency of the recessive allele 4 In diploid organisms, alleles occur in pairs (referring to the genotype for each individual). Therefore, the equation to calculate genotype frequency is p2 + 2pq + q2 = 1 , where p2 is the frequency of the homozygous dominant genotype 2pq is the frequency of the heterozygous genotype q2 is the frequency of the homozygous recessive genotype Example 2 The ability to taste PTC is controlled by a single gene T. In a population that is in genetic equilibrium, it was found that there are 70% tasters and 30% non-tasters. Determine the allele and genotype frequencies. Answer Frequency of non-taster individuals, q2 (tt) = 0.30 Frequency of recessive allele, q (t) = √ 0.30 = 0.55 From the equation of p + q = 1, Frequency of dominant allele, p (T) = 1 – q = 1 – 0.55 = 0.45 From the equation of p2 + 2pq + q2 = 1, Frequency of homozygous dominant genotype, p2 (TT) = (0.45)2 = 0.20 Frequency of heterozygous genotype, 2pq (Tt) = 2 × 0.45 × 0.55 = 0.50 Frequency of homozygous recessive genotype, q2 (tt) = (0.55)2 = 0.30 (The sum of the genotype frequencies must be 1) Quick Check 5.2 1 State Hardy-Weinberg law. 2 Explain how a large population size can lead to a population that is in genetic equilibrium. 3 A species of fruit fly may have normal wings and vestigial wings. The normal wings trait is dominant to vestigial wings. In a population of 5 000 fruit flies, there are 950 fruit flies with normal wings. By using Hardy-Weinberg equations, calculate all the allele frequencies and genotype frequencies. 4 State five assumptions for maintaining Hardy-Weinberg equilibrium. The first thing to find out is the frequency of the recessive phenotype which is q2 . From here, the frequency of the recessive allele, q can be determined, and eventually the frequency of the dominant allele, p and all the genotype frequencies can be determined. Smart Tips! Student tends to start the calculation by finding the frequency of dominant phenotype first. They assume that, for example p2 = 0.70. But it is actually p2 + 2pq = 0.70. So, if the allele frequency is incorrectly calculated in the beginning, the rest of the calculation would become incorrect. Common Error The Hardy-Weinberg Law was developed independently by Godfrey Hardy, an English mathematician and Wilhelm Weinberg, a German doctor. Each worked alone to come up with the founding principle of population genetics. Bio Info C05 Matriculation Sem 1 BIO 4pp.indd 76 19/06/2023 5:59 PM PENERBIT ILMU BAKTI SDN. BHD.

77 Mid Semester Test 1 Answers Chapter 5 1 All the alleles for all the loci present in a population is called the… A gene pool. B population ecology. C population variation. D evolutionary potential. 2 Because most animal species are diploid, each individual possesses… A one allele for each locus. B two alleles for each locus. C three or more alleles for each locus. D a complete set of alleles found in each chromosome. 3 A population that has unchanged allele and genotype frequencies over generations is said to be in … A genetic stability. B population stability. C genetic equilibrium. D allelic disequilibrium. 4 is one of the assumptions of the Hardy-Weinberg equilibrium. A Mutation B Natural selection C Non-random mating D Large population size 5 The symbol 2pq in the Hardy-Weinberg equation refers to the frequency of the … A recessive allele at a given locus. B recessive allele in a given population. C heterozygous genotype in a population. D homozygous recessive genotype in a population. 6 In a population with two alleles, B and b, the frequency of b is 0.50. Assume that the population is in genetic equilibrium, calculate the frequency of heterozygotes. A 0.50 B 0.25 C 0.75 D 1.00 7 In a population that is in genetic equilibrium, individuals with a particular trait have two alleles, A and a. The frequency of the recessive allele (a) is 0.2. What is the frequency of individuals with heterozygous genotype? A 0.32 B 0.78 C 0.60 D 0.80 8 A study on the ability to taste PTC was carried out on a population of students. 84% of the students were able to taste PTC. The ability to taste PTC is controlled by the dominant allele (T). Determine the frequency of the recessive allele in the student population. A 0.04 B 0.16 C 0.40 D 0.84 9 In a research done on mice, it was found that 72 out of 500 mice are albino. The allele for albino is recessive to the brown hair allele. Calculate the albino allele frequency in the population. A 0.29 B 0.38 C 0.50 D 0.75 10 Cystic fibrosis is controlled by a recessive allele, c. It is found that 1 in every 15 000 people develops cystic fibrosis. Determine the frequency of normal individuals in the population. A 0.9999 B 0.8000 C 0.8889 D 0.0001 Summative Practice 5 Objective Questions Instruction: There are four answer options for each question. Choose the best answer. C05 Matriculation Sem 1 BIO 4pp.indd 77 19/06/2023 5:59 PM PENERBIT ILMU BAKTI SDN. BHD.

11 The frequency of an allele that causes sickle-cell anemia is 0.2. Find out the frequency of carriers for this disease, if the population is in genetic equilibrium. A 0.48 B 0.32 C 0.42 D 0.24 12 If 4% of a population shows a recessive trait, determine the percentage of carriers in the next generation. A 40% B 80% C 60% D 32% 13 In a population of hamsters at a genetic equilibrium, 750 hamsters have grey fur, while the remaining 250 hamsters have white fur. Grey fur is dominant over white fur. Find out the number of hamsters with the heterozygous genotype. A 250 B 500 C 750 D 900 14 In a population of 10 000 individuals, 1 600 were found to have blue eyes. Blue eyes are controlled by a recessive allele. Find out the frequency of the dominant allele. A 0.60 B 0.40 C 0.36 D 0.16 15 Which of the following populations is not in Hardy-Weinberg equilibrium? A A small population. B A population with no mutation. C A randomly mating population. D A population with no natural selection. 1 In a population of cows, the allele for black fur (B) is dominant over the allele for white fur (b). It was found that 16% of the cows have white fur. It is assumed that the population is at genetic equilibrium. (a) Give two assumptions used in the Hardy-Weinberg equilibrium. [2 marks] (b) Calculate the frequency of black and white fur alleles. [2 marks] (c) Calculate the number of heterozygote cows in a population of 2 000 cows. [2 marks] 2 In a population of snails, brown shell is dominant over yellow shell. The total number of snails in this population is 2 000. 81% of the snails have brown shell. (Calculate up to 3 decimal places). EXAM CLONE (a) Assuming that the population is in Hardy-Weinberg equilibrium, calculate the dominant and recessive allele frequency. [2 marks] (b) How many snails have brown shell but carry the recessive allele for shell colour? [1 mark] (c) If all the yellow shell snails in the population were killed by a disease, determine the number of yellow shell snails in the next generation. [3 marks] Structured Questions Instruction: Answer all the questions. 78 Mid Semester Test 1 Answers  Population Genetics Population Genetics  Chapter 5 C05 Matriculation Sem 1 BIO 4pp.indd 78 19/06/2023 5:59 PM PENERBIT ILMU BAKTI SDN. BHD.

3 In a population of beetles, black beetle (B) is dominant over brown beetle (b). The population consists of 600 beetles. Table 1 below shows the number of beetles for each phenotype. EXAM CLONE PhenotypesGenotypes Number of individual Phenotypes Genotypes Number of individuals Black beetles BB 350 Black beetles Bb 130 Brown beetles bb 120 Table 1 (a) What is the gene pool size for the population of beetles? [1 mark] (b) Determine the frequency of dominant allele, B. [1 mark] (c) Determine the frequency of recessive allele, b. [1 mark] (d) If the beetles were left to breed randomly and the population remains in equilibrium, how many individuals are expected to be heterozygous in the next generation of 2 000 beetles? [3 marks] Essay Questions Instruction: Answer all the questions. 1 A cross between pea plants produces 9 744 tall pea plants and 256 dwarf pea plants. Calculate all the expected genotype frequencies of the pea plant population. EXAM CLONE (Calculation must use up to 4 decimal places). [6 marks] 2 In a particular country, a study done on a population of humans showed that 0.085% of the human population within that country suffers from cystic fibrosis. Cystic fibrosis is controlled by a pair of alleles, C and c. Individuals that suffer from cystic fibrosis have the homozygous recessive genotype. Using the Hardy-Weinberg equation, calculate the percentage of the two genotypes of individuals that are not suffering from the disease. (Calculation must use up to 3 decimal places). [6 marks] 79 Mid Semester Test 1 Answers  Population Genetics Population Genetics  Chapter 5 C05 Matriculation Sem 1 BIO 4pp.indd 79 19/06/2023 5:59 PM PENERBIT ILMU BAKTI SDN. BHD.

1 Molecules of Life Quick Check 1.1 1 Covalent bond 2 Hydrogen bond 3 High specific heat capacity/High latent heat of vaporisation/Cohesion of water molecules. [Any two answers] 4 Universal solvent/High specific heat capacity/High latent heat of vaporisation/Cohesion of water molecules/Maximum density at 4 °C. [Any two answers] Quick Check 1.2 1 Sweet, soluble in water, can be crystallised. 2 (i) Monomers for disaccharides and polysaccharides. (ii) Energy source in plants and animals. 3 Glycogen 4 Condensation; hydrolysis Quick Check 1.3 1 Carbon, hydrogen, oxygen. 2 Three 3 (i) A phospholipid has a hydrophilic head and (two) hydrophobic tails of fatty acids while a triglyceride has only (three) hydrophobic tails of fatty acids. (ii) A phospholipid contains two fatty acids while a triglyceride contains three fatty acids. 4 Hydroxyl group and carboxylic group. Quick Check 1.4 1 Based on the side chain or R group. 2 Chemical reaction: Condensation Bond: Peptide 3 Fibrous protein is insoluble in water while globular protein is water-soluble. 4 Amino acid Quick Check 1.5 1 DNA (deoxyribonucleic acid) and RNA (ribonucleic acid) 2 Cytosine, adenine and guanine. 3 Adenine pairs with thymine, guanine pairs with cytosine 4 Hydrogen bond Summative Practice 1 Objective Questions 1 C 2 B 3 A 4 B 5 D 6 B 7 D 8 C 9 B 10 D 11 D 12 A 13 B 14 B 15 D Structured Questions 1 (a) Molecule P: Glycerol Molecule Q: Fatty acid (b) Chemical reaction: Condensation End products: Triglyceride and water (c) Answers (d) Lipids have a higher ratio of hydrogen atoms to carbon atoms compared to carbohydrates. (e) Insulator/Emulsifier/Transport of fat-soluble vitamins/Shock absorber [Any one answer] 2 (a) Amino acid (b) Chemical reaction: Condensation New compound: Dipeptide Bond: Peptide bond Byproduct: Water (c) Functional groups Characteristics Amino group Basic Carboxyl group Acidic (d) 3 (a) P: Nitrogenous base/Cytosine Q: Pentose/Deoxyribose sugar R: Phosphate group (b) 5’ – GTCA – 3’/GTCA (c) • DNA has double strands while RNA has a single strand. • The pentose sugar for DNA is deoxyribose while the pentose sugar for RNA is ribose. • The nitrogenous bases for DNA are adenine, thymine, cytosine and guanine while the nitrogenous bases for RNA are adenine, uracil, cytosine and guanine. R2 C C O OH H H N R1 C C O OH H H H N H + H H H O O O O O O H H H H H HO O O O HO HO H H OH R" R" R" 3 water molecules + 3H2 O R" R" R" OH OH C C C C C C C C C C C C condensation ester linkage 1 glycerol 3 fatty acids triglyceride 169 Mid Semester Test 1 Answers C10 Matriculation Sem 1 BIO (ANS) 5pp.indd 169 20/06/2023 3:45 PM PENERBIT ILMU BAKTI SDN. BHD.

• The DNA molecule is large while the RNA molecule is small. [Any two answers] (d) (i) Organism Adenine Cytosine Guanine Thymine A 25 21 21 25 B 40 30 30 40 (ii) Based on the pairing of purine bases with pyrimidine bases: / pairing of adenine with thymine and cytosine with guanine. Therefore, the number of adenine is equal to thymine and the number of cytosine is equal to guanine. Essay Questions 1 (a) Similarities: (i) Both are organic compounds. (ii) Both are polymers. (b) Differences: (i) Globular proteins are folded into a spherical shape while fibrous proteins are long and parallel polypeptide chains. (ii) Globular proteins are tertiary or quaternary structures while fibrous proteins are secondary structures. (iii) Globular proteins are soluble in water while fibrous proteins are insoluble in water. (iv) Globular proteins function in the metabolic activity while fibrous proteins are for structural support. (v) Globular proteins are relatively unstable while fibrous proteins are stable. (vi) Globular proteins are colloidal while fibrous proteins are non-colloidal. (vii) The structure of globular proteins is maintained by hydrogen bond, ionic bond, Van der Waals, disulfide bridge and hydrocarbon bond while the structure of fibrous proteins is maintained by hydrogen bond. (c) An example of globular protein is haemoglobin/myoglobin/ antibodies while an example of fibrous protein is collagen/ keratin/fibrin. 2 • The monomer for DNA is nucleotide. • Each nucleotide consists of a five-carbon atom pentose sugar, phosphate group and nitrogenous base. • The nitrogenous bases in DNA are adenine, thymine, guanine and cytosine. • Adenine pairs with thymine using two hydrogen bonds. • Guanine pairs with cytosine using three hydrogen bonds. • The nucleotides are linked to each other by phosphodiester bonds between sugar and phosphate groups forming a polynucleotide strand. • DNA consists of two polynucleotide strands/chains. • The (two) strands run antiparallel/ in opposite directions to each other. • One strand is oriented in the 5’ to 3’ direction while the other strand is oriented in the 3’ to 5’ direction. • The two polynucleotide strands are held together by hydrogen bonds between the complementary nitrogenous base pairs. • The polynucleotide strands twist around each other forming a double helix. • Each full turn of helix has 10 base pairs. 2 Cell Structures and Functions Quick Check 2.1 1 Cells of algae/protozoa/fungi/plants/ animals [Any two answers] 2 (a) Both cells are enclosed by the plasma membrane. (b) Both cells contain DNA as genetic material. 3 The cell wall of prokaryotic cells contains peptidoglycan while the cell wall of eukaryotic cells does not contain peptidoglycan. 4 The flagella of bacteria do not contain a 9 + 2 microtubules arrangement like the flagella of protozoa. Quick Check 2.2 1 Both of the cells are bounded by a plasma membrane. Both of the cells have organelles such as nucleus, mitochondria, endoplasmic reticulum, Golgi body and ribosomes. 2 Centrioles, lysosomes. 3 Integral protein and peripheral protein. 4 Fluid: Phospholipids and proteins can move laterally. Mosaic: Proteins are embedded in the phospholipid bilayer like mosaic tiles embedded in mortar. Quick Check 2.3 1 Epithelium tissue, nerve tissue, muscle tissue and connective tissue. 2 Neuron has a long fibre (axon) so that it can carry messages up and down the body over long distances. 3 Xylem and phloem. 4 Blood vessels. They transport products from digested food to cells throughout the body. Quick Check 2.4 1 Simple diffusion, facilitated diffusion, osmosis. 2 Sodium-potassium pump 3 Bulk transport 4 Aquaporins 170 Mid Semester Test 1 Answers  Answers C10 Matriculation Sem 1 BIO (ANS) 5pp.indd 170 20/06/2023 3:45 PM PENERBIT ILMU BAKTI SDN. BHD.

Summative Practice 2 Objective Questions 1 B 2 D 3 C 4 A 5 B 6 B 7 C 8 D 9 D 10 A 11 B 12 C 13 D 14 B 15 A Structured Questions 1 (a) Organelle X: Chloroplast Organelle Y: Golgi body Organelle Z: Mitochondrion (b) A: Thylakoid B: Cisterna/ Cisternae C: Matrix of mitochondrion (c) A: Involved in photosynthesis. B: Modify/ Package/ Sort proteins/ lipids/carbohydrates received from the (smooth and rough) endoplasmic reticulum. C: Site for Krebs cycle// Site for oxidation of fatty acids. (d) X and Z// Chloroplast and mitochondrion 2 (a) (i) Blood/ Blood tissue (ii) A: Red blood cell// Erythrocyte B: Platelet C: Plasma D: White blood cell/ Leukocyte (iii) Has a biconcave shape to increase the surface area for effective gas exchange// Has no nucleus to provide more space for haemoglobin storage// Has an elastic membrane that change shape easily for ease of movement in blood capillary. (iv) Xylem and phloem (b) Cell wall limits the movement of human cells/growth of human cells. 3 (a) X: Sclerenchyma Y: Parenchyma Z: Collenchyma (b) X: Give strength and support for plants/ Sclereids protect seeds/ Fibers like hemp are used commercially. Y: Store various organic products like starch through photosynthesis/ For gases exchange/ Turgid cells give support for herbaceous plants. Z: Give support to young and herbaceous plants/ Give flexibility to plants without restraining growth. (c) • Tissue Z consists of living cells/ while tissue X consists of nonliving cells/ cells that are dead at maturity • Tissue Z has primary and secondary cell walls, while tissue X has primary cell wall only. • Tissue Z has cell wall thickened with lignin, while tissue X has cell wall thickened with cellulose. • Tissue Z has cell wall that are evenly thickened, while tissue X has cell wall that are unevenly thickened. • Tissue Z are rigid structural tissues, while tissue X are flexible structural tissues. [Any two answers] Essay Questions 1 • Plasma membrane consists of the phospholipid bilayer. • Various protein molecules are embedded in it or attached to it. • The hydrophilic heads of phospholipids point outwards and are attracted to water. • The hydrophobic tails of phospholipids face inwards. • The peripheral proteins are embedded on the outer or inner surface of the membrane. • The intrinsic proteins are embedded partially or fully in the membrane. • The intrinsic proteins move laterally in the membrane. • Glycoproteins are carbohydrates attached to proteins. • Glycolipids are carbohydrates attached to lipids. • Cholesterol molecules are found between phospholipid molecules. • Cholesterol molecules maintain the structure of the plasma membrane. [Any ten answers] 2 (a) Diffusion • Involves the movement of small solutes, down the concentration gradient// without using energy. (b) Facilitated diffusion • Involves the movement of small polar molecules/ ions, down the concentration gradient// without using energy. • Movement is aided by carrier/ transport proteins. (c) Active transport • Involves the movement of large polar/ionic molecules, against the concentration gradient. • Movement is aided by energy and carrier/transport proteins. (d) Endocytosis • Influx of materials into the cell through the invagination of the plasma membrane. • Requires energy. (e) Exocytosis • Releases of substances from cells • Requires energy. (f) Osmosis • Involves the movement of water molecules, down the water potential gradient. • Movement does not require energy. 3 Cell Division Quick Check 3.1 1 No, not all cells divide. An example is neurons that cannot divide because they do not have centrosomes to produce centrioles. Centrioles initiate the formation of spindle fibres needed for cell division. 171 Mid Semester Test 1 Answers Answers  C10 Matriculation Sem 1 BIO (ANS) 5pp.indd 171 20/06/2023 3:45 PM PENERBIT ILMU BAKTI SDN. BHD.

2 Karyokinesis and cytokinesis 3 Karyokinesis is the division of the nucleus. 4 Mitosis Quick Check 3.2 1 Interphase and mitotic phase 2 A: G1 phase/first gap B: S phase/Synthesis gap C: G2 phase/second gap 3 Mitosis and cytokinesis 4 DNA replicates and histone proteins are synthesised. Quick Check 3.3 1 Mitosis 2 Root tip/Shoot tip/Cambium 3 Q, P, S, R Quick Check 3.4 1 The pairing of homologous chromosomes during prophase 1 in meiosis. 2 Germ cells 3 19 Summative Practice 3 Objective Questions 1 A 2 C 3 C 4 D 5 B 6 A 7 A 8 D 9 C 10 B 11 D 12 D 13 B 14 A 15 B Structured Questions 1 (a) Cell A: Mitosis Cell B: Meiosis I/Meiosis (b) Cell A: Anaphase Cell B: Anaphase I (c) Cell A: Sister chromatids separate and move to the opposite poles/ Centromeres of the chromosomes split. Cell B: Homologous chromosomes split and move to the opposite poles. (d) Cell A: 4 Cell B: 2 (e) Cell A: Microsporocyte/ Microspore mother cell/ Megasporocyte/Megaspore mother cell Cell B: Root tip/Shoot tip/ Cambium 2 (a) (i) Mitosis (ii) The daughter cells have the same number of chromosomes as the parent cell. (b) C D B A (c) Chromosome/Chromatid (d) 46 (e) (i) Metaphase (ii) Metaphase plate/Equator of the cell (f) 2 (g) Undergoes mutation 3 (a) P: Centromere Q: Chiasma R: Bivalent/Tetrad/Homologous chromosomes (b) Type of cell division: Meiosis Stage of cell division: Prophase I (c) • Chromosomes are condensed and become visible. • Homologous chromosomes pair up to form a bivalent. • Crossing over occurs between two non-sister chromatids at the chiasma. • Genetic materials are exchanged. • Then the segments at the chiasma break off and rejoin. [Any three answers] (d) (i) Genetic variation (ii) Genetic recombination Essay Questions 1 Mechanism of cytokinesis in animal cells: • The membrane is pulled inwards by the cytoskeleton. • A shallow groove is formed in the surface of cells near the metaphase plate. • A cleavage furrow occurs. • A contractile ring forms and when the ring contracts, the diameter of the ring is reduced. • Two cells are formed. [Any five answers] Mechanism of cytokinesis in plant cells: • Plant cells form vesicles inside the cell. • The vesicles enlarge and fuse to form a cell plate. • Cell plate is formed across the equatorial of the cell forming two membranes that grow laterally and unite with existing membranes to form two plant cells. • A new cell wall forms between the membranes. [Any five answers] 2 Similarities: • Both do not have crossing over. • Both have no pairing of homologous chromosomes//no synapsis • Both have individual chromosomes with two sister chromatids aligned at the metaphase plate. • Both undergo division of the centromere during anaphase. • Both have sister chromatids pulled to opposite poles during anaphase. • Both have cytokinesis that take place at the end of the cell division process. [Any four answers] Differences: • DNA replication occurs in mitosis, whereas no DNA replication occurs in meiosis II. • Chromosomes become condensed and visible in mitosis, whereas chromosomes are already condensed in meiosis II. • Mitosis occurs in somatic cells whereas meiosis occurs in reproductive cells. • Daughter cells of mitosis have the same number of chromosomes as the parent cell, whereas daughter cells of meiosis are genetically different from each other. • Daughter cells are genetically identical to parental cells in mitosis, whereas daughter cells in meiosis are genetically different from each other. 172 Mid Semester Test 1 Answers  Answers C10 Matriculation Sem 1 BIO (ANS) 5pp.indd 172 20/06/2023 3:45 PM PENERBIT ILMU BAKTI SDN. BHD.

• Two daughter cells are produced in mitosis, whereas four daughter cells are produced in meiosis. [Any six answers] Summative Assessment Test (UPS) 1 1 A 2 A 3 D 4 B 5 A 6 A 7 B 8 D 9 A 10 B 11 D 12 C 13 D 14 D 15 C 16 B 17 C 4 Genetic Inheritance Quick Check 4.1 1 Gene refers to a unit of hereditary information consisting of a specific nucleotide base sequence in DNA. Allele refers to the alternative version of a gene that determines distinguishable phenotypes. Alleles can either be dominant or recessive. 2 The characteristics of Mendel’s pea plant could be easily observed because its different traits can be distinguished easily, for example by their flower colours, purple and white. 3 Mendel’s first law is the Law of Segregation, which states that the pair of alleles segregate during meiosis and only one allele of each pair can be present in a single gamete. 4 Mendel’s second law is the Law of Independent Assortment, which states that each member of a pair of alleles may combine randomly with either one of the other pair of alleles during gamete formation. Quick Check 4.2 1 A codominant allele refers to both alleles of a gene that are fully expressed in the phenotype of an individual with a heterozygous genotype. The heterozygote offspring has the characteristics of both of the homozygous parents. 2 Flower colour of Antirrhinum sp. (snapdragon flower). 3 I A, IB , i 4 Polygenic inheritance refers to a characteristic that is controlled by the cumulative effect of more than one gene at different loci. Quick Check 4.3 1 Genetic mapping is the process that determines the position of genes along the length of a chromosome. 2 If two genes are farther apart, the probability that crosssing over will occur between them is higher. Therefore, the recombination frequency would be higher. 3 Test cross 4 COV = Total number of recombinants Total number of offspring × 100 Summative Practice 4 Objective Questions 1 A 2 B 3 C 4 C 5 B 6 A 7 C 8 A 9 A 10 C 11 B 12 D 13 D 14 D 15 D Structured Questions 1 (a) Law of Independent Assortment // Mendel’s Second Law (b) • Each member of a pair of alleles combines randomly with the other member of another pair of alleles. • During gamete formation (c) NnEe and nnee (d) Parental Normal wings, Vestigial wings, phenotype : Black body Grey body Parental NnEe × nnee genotype : Gamete : F1 genotype : NnEe Nnee nnEe nnee F1 phenotype : Normal Normal Vestigial Vestigial wings wings wings wings Black body Grey body Black body Grey body (e) Genotypic ratio: 1NnEe : 1Nnee : 1nnEe : 1nnee Phenotypic ratio: 1 Normal : 1 Normal : 1 Vestigial : 1 Vestigial wings, wings, wings, wings, Black body Grey body Black body Grey body (f) Test cross 2 (a) LLGG LlGg LLGg LlGG (b) By carrying out a test cross // By crossing it with a hamster with a recessive phenotype / short, grey fur NE Ne nE ne ne 173 Mid Semester Test 1 Answers Answers  C10 Matriculation Sem 1 BIO (ANS) 5pp.indd 173 20/06/2023 3:45 PM PENERBIT ILMU BAKTI SDN. BHD.

(c) Parental Long and grey Short and brown phenotype : fur fur Parental LlGg × llgg genotype : Gamete : F1 genotype : LlGg Llgg llGg llgg F1 phenotype : Long, Long, Short, Short, grey brown grey brown fur fur fur fur (d) (i) Genetic mapping is a process that determines the position of genes along the length of a chromosome. (ii) Percentage / total number of recombinant // COV 3 (a) (i) Male : GG Female : gg (ii) Parental phenotype : Grey body Yellow body Parental GG × gg genotype : Gamete : F1 genotype : Gg F1 phenotype : All grey-bodied (b) (i) Parental phenotype : Grey body Yellow body Parental XGY × XgXg genotype : Gamete : F1 genotype : XGXg XgY F1 Grey-bodied Yellow-bodied phenotype : female male (ii) 0% (iii) 100% Essay Questions 1 (a) T – dominant tall allele t – recessive short allele H – dominant hairy allele h – recessive smooth allele Parental Tall and hairy Short and smooth phenotype : stem stem Parental TtHh × tthh genotype : Gamete : F1 genotype : TtHh Tthh ttHh tthh F1 phenotype : Tall, Tall, Short, Short, hairy smooth hairy smooth (b) Genotypic ratio: 1TtHh : 1Tthh : 1ttHh : 1tthh Phenotypic ratio: 1 Tall, hairy : 1 Tall, smooth : 1 Short, hairy : 1 Short, smooth (c) The Law of Independent Assortment states that different pairs of alleles segregate independently of each other during gamete formation. 2 • This is a dihybrid inheritance. Assume that H is the dominant allele that codes for black hair, while B is the dominant allele for brown eyes. John’s genotype would be homozygous recessive, hhbb. Whereas his wife’s genotype is heterozygous, HhBb. • John will produce only one type of gamete, which is hb. While his wife will produce four types of gametes; HB, Hb, hB and hb. • John’s gamete, hb fertilizes with one of his wife’s gamete, Hb. This results in the production of a diploid zygote with the genotype of Hhbb. The phenotype of this zygote would be black hair and blue eyes as it receives the dominant allele for hair colour from the mother and recessive alleles for eye colour from both parents. The percentage of John’s children who would have black hair and blue eyes is 25%. 5 Population Genetics Quick Check 5.1 1 Population genetics is the study of genetic variability within a population and the forces that change the allele frequencies. LG Lg lG lg lg TH Th tH th XG Y Xg G g th 174 Mid Semester Test 1 Answers C10 Matriculation Sem 1 BIO (ANS) 5pp.indd 174 20/06/2023 3:45 PM PENERBIT ILMU BAKTI SDN. BHD.

2 The population is said to be in genetic equilibrium when the allele frequencies remain constant from one generation to another. 3 Gene pool refers to the total number of genes in a population at a particular time. It consists of all alleles at all gene loci in all individuals of the population at a given time. 4 Total number of alleles in the gene pool = 2 × 400 = 800 196 BB mice contribute 196 × 2 = 392 B alleles 168 Bb mice contribute 168 B alleles Total number of B alleles = 392 + 168 = 560 alleles 36 bb mice contribute 36 × 2 = 72 b alleles 168 Bb mice contribute 168 b alleles Total number of b alleles = 72 + 168 = 240 alleles Frequency of B allele = 560 800 = 0.7 Frequency of b allele = 240 800 = 0.3 Quick Check 5.2 1 Hardy-Weinberg law states that the frequency of alleles and genotypes in a population’s gene pool remain constant over the generations. 2 Genetic drift does not happen in a large population. Thus, it will not cause any change in the allele frequency. But if the population size is small, genetic drift may easily change the allele frequency. 3 Fruit flies with vestigial wings = 5 000 – 950 = 4 050 Frequency of homozygous recessive genotype, q2 = 4 050 5 000 = 0.81 Frequency of recessive allele, q = √0.81 = 0.9 From the equation of p + q = 1, Frequency of dominant allele, p = 1 – q = 1 – 0.9 = 0.1 From the equation of p2 + 2pq + q2 = 1, Frequency of homozygous dominant genotype, p2 = (0.1)2 = 0.01 Frequency of heterozygous genotype, 2pq = 2 × 0.1 × 0.9 = 0.18 Frequency of homozygous recessive genotype, q2 = (0.9)2 = 0.81 4 The five assumptions are large population size, random mating, no mutation, no migration and no natural selection. Summative Practice 5 Objective Questions 1 A 2 B 3 C 4 D 5 C 6 A 7 A 8 C 9 B 10 A 11 B 12 D 13 B 14 A 15 A Structured Questions 1 (a) • Large population size • Random mating • No mutation • No migration • No natural selection [Any two answers] (b) Frequency of the homozygous recessive genotype, q2 = 0.16 Frequency of recessive allele, q(b) = √0.16 = 0.4 Frequency of dominant allele, p (B) = 1 – 0.4 = 0.6 (c) Frequency of heterozygous genotype, 2pq (Bb) = 2 × 0.6 × 0.4 = 0.48 Number of heterozygote cows = 0.48 × 2 000 = 960 cows 2 (a) Frequency of homozygous recessive genotype, q2 = 0.19 Frequency of recessive allele, q = √0.19 = 0.436 Frequency of dominant allele, p = 1 – 0.436 = 0.564 (b) Frequency of heterozygous genotype, 2pq = 2 × 0.564 × 0.436 = 0.492 Number of heterozygous snails = 2pq × 2 000 = 0.492 × 2 000 = 984 snails (c) Number of yellow snails = 0.19 × 2 000 = 380 New population = 2 000 – 380 = 1 620 Total number of all alleles in the new population = 2 × 1 620 = 3 240 New recessive allele frequency, q = 984 3 240 = 0.304 Number of yellow shell snails = q2 × 1 620 = 0.304 × 1 620 = 492 snails 3 (a) Total number of individuals = 350 + 130 + 120 = 600 Total number of alleles = 2 × 600 = 1 200 (b) Frequency of dominant allele, p(B) = 2(350) + 130 1 200 = 0.69 (c) Frequency of recessive allele, q(b) = 2(120) + 130 1 200 = 0.31 (d) Frequency of heterozygous genotype, 2pq (Bb) = 2 × 0.69 × 0.31 = 0.43 175 Mid Semester Test 1 Answers Answers  C10 Matriculation Sem 1 BIO (ANS) 5pp.indd 175 20/06/2023 3:45 PM PENERBIT ILMU BAKTI SDN. BHD.

Number of heterozygous beetles in the next generation = 2pq × 2 000 = 0.43 × 2 000 = 860 beetles Essay Questions 1 Number of dominant phenotype (tall) = 9 744 Number of recessive phenotype (dwarf) = 256 Total number of pea plants = 10 000 Frequency of homozygous recessive genotype, q2 = 256 10 000 = 0.0256 Frequency of recessive allele, q = √ 0.0256 = 0.16 Frequency of dominant allele, p = 1 – 0.16 = 0.84 Frequency of homozygous dominant genotype, p2 = (0.84)2 = 0.7056 Frequency of heterozygous genotype, 2pq = 2 × 0.84 × 0.16 = 0.2688 2 Frequency of homozygous recessive genotype, q2 (cc) = 0.00085 Frequency of recessive allele, q (c) = √ 0.00085 = 0.029 Frequency of dominant allele, p (C) = 1 – 0.029 = 0.971 Frequency of homozygous dominant genotype, p2 (CC) = (0.971)2 = 0.943 Percentage of individuals with homozygous dominant genotype = 0.943 × 100 = 94.3% Frequency of heterozygous genotype, 2pq (Cc) = 2 × 0.971 × 0.029 = 0.056 Percentage of individuals with heterozygous genotype = 0.056 × 100 = 5.6% The percentage of those not suffering from cystic fibrosis = 94.3% + 5.6% = 99.9% 6 Expression of Biological Information Quick Check 6.1 1 Central dogma states that the flow of genetic information is one-way from DNA to RNA to protein. 2 Francis Crick Quick Check 6.2 1 Primers are required in DNA replication as a starter/pre-existing chain to initiate the synthesis of the new DNA strand as DNA polymerase III can only add DNA nucleotide to the 3’ end of a pre-existing chain (DNA polymerase cannot inititate the synthesis of polynucleotides). 2 The leading strand is synthesised when DNA polymerase III continuously adds DNA nucleotides to the new complementary strands as the replication fork progresses. 3 Catalyses the synthesis of new DNA strand by adding free DNA nucleotides to the free 3’end of RNA primer or the growing DNA strand. 4 DNA ligase catalyses the joining of the Okazaki fragments to form the lagging strands through phosphodiester bonds. Quick Check 6.3 1 DNA acts as the template for the synthesis of mRNA during transcription. 2 Aminoacyl-tRNA synthetase. 3 The complementary base sequence on mRNA is 5’ UGCUUAAGGUAUCCG 3’. 4 Translation cannot occur in the absence of the small subunit of ribosome because the initiation stage of translation cannot take place// because the initiation complex cannot be formed. Quick Check 6.4 1 Operon can be defined as the control of a cluster of genes as a single unit. 2 In the absence of lactose, the operon is said to be switched off. The lac repressor protein is active and binds to the lac operator, preventing RNA polymerase from binding to the lac promoter. Therefore, the transcription of the three structural genes cannot occur and no mRNA is produced. Thus, no enzymes are synthesized. Summative Practice 6 Objective Questions 1 C 2 C 3 D 4 D 5 A 6 D 7 C 8 D 9 A 10 B 11 D 12 C 13 C 14 B 15 C Structured Questions 1 (a) Nucleus (b) Semi-conservative model. James Watson and Francis Crick (c) Q: Single-strand binding protein. Its function is to bind and stabilise the single-stranded DNA from reforming until it can be used as a template for DNA replication. (d) The new DNA strand cannot be synthesised// DNA replication cannot occur. (e) Strand A is synthesised in the 5’ to 3’ direction continuously towards the replication fork. (f) DNA polymerase III// DNA polymerase I// DNA ligase (g) DNA replication occurs during the S phase of the cell cycle, in between the G1 and G2 phases. This means that the DNA replicates completely before the mitotic phase begins. 176 Mid Semester Test 1 Answers  Answers C10 Matriculation Sem 1 BIO (ANS) 5pp.indd 176 20/06/2023 3:45 PM PENERBIT ILMU BAKTI SDN. BHD.

2 (a) 3’ UAC 5’ (b) Y: mRNA // messenger RNA Function : Acts as a template for the synthesis of polypeptide chain during translation. (c) (i) RNA polymerase (ii) RNA polymerase cannot bind to the promoter. Transcription cannot occur. mRNA cannot be synthesised. (d) Process A uses DNA as the template while process B uses mRNA as the template. Process A produces an mRNA strand while process B produces a polypeptide chain/protein. Process A uses RNA polymerase to make copy of RNA from DNA while process B uses ribosomes of synthesise proteins from RNA transcripts. 3 (a) Allolactose (b) To hydrolyse lactose into glucose and galactose. (c) Molecule R : lac repressor protein Binds to the lac operator (d) • RNA polymerase cannot bind to the lac promoter • Thus, transcription of the structural genes/lacZ cannot occur. • No enzymes produced / β-galactosidase enzyme cannot be produced. • The lactose operon is switched off. [Any three answers] (e) • A small amount of lactose will convert to its isomer, allolactose (which acts as an inducer). • The allolactose will bind to the lac repressor protein. • The lac repressor protein changes its conformation, thus can no longer bind to the lac operator. • RNA polymerase binds to the lac promoter. • Transcription of the structural gene/ lacZ can occur • β-galactosidase enzyme is synthesised. Essay Questions 1 The elongation process has three stages which are codon recognition, peptide bond formation and translocation. During codon recognition, the upcoming aminoacyltRNA binds to a codon in A site by complementary base-pairing between the tRNA’s anticodon and the codon on mRNA. During formation of peptide bond, enzyme peptidyl transferase catalyses the formation of peptide bond between the amino end of the amino acid in the A site and the free carboxyl group of the growing polypeptide chain in the P site. The growing polypeptide chain (peptidyl-tRNA) occupying the P site becomes attached by a peptide bond to the amino acid linked to the tRNA at the A site. During translocation, the ribosome translocates the tRNA in the A site to the P site by moving down the mRNA by one codon. The mRNA moves along with its bound tRNAs, bringing the next codon to be translated at the A site and ready to bind the new aminoacyl-tRNA. The empty/uncharged tRNA in the P site is moved to the E site where it is released. The process is repeated until a complete polypeptide chain is formed. 2 The components of the lac operon in E. coli are lac promoter, lac operator, lacZ, lacY and lacA. Lac promoter acts as the binding site for RNA polymerase while the lac operator acts as the binding site for the lac repressor protein. lacZ codes for β-galactosidase, lacY codes for permease and lacA codes for transacetylase. Summative Assessment Test (UPS) 2 1 B 2 C 3 C 4 B 5 D 6 B 7 B 8 C 9 B 10 D 11 B 12 D 13 D 14 B 15 D 16 C 17 D 7 Mutation Quick Check 7.1 1 Mutation is defined as the sudden permanent change in the amount, arrangement or structure of the DNA of an organism. 2 Mutation is classified into two which are spontaneous mutation and induced mutation. 3 Induced mutation occurs when an organism is exposed to mutagens physically or chemically. 4 Mutagen is an agent that can induce changes in the DNA and causes mutation. An example of physical mutagen is ultraviolet (UV) light/Xrays/gamma rays while an example of chemical mutagen is colchicine/ mustard gas/ethidium bromide/ formaldehyde. Quick Check 7.2 1 Three types of gene mutation are base substitution, base insertion and base deletion. 2 Sickle cell anaemia is an example of missense mutation due to base substitution. The base thymine (T) in the DNA template 3’-CTC-5’ is replaced with base adenine (A), resulting in a mutant template DNA strand, 3’-CAC-5’. This causes the codon to change from 5’-GAG-3’ to 5’-GUG-3’. Thus, the codon that is supposed to code for glutamic acid (Glu) is changed to valine (Val). Therefore, abnormal/ sickle-cell haemoglobins (Hb-S) are produced. 177 Mid Semester Test 1 Answers Answers  C10 Matriculation Sem 1 BIO (ANS) 5pp.indd 177 20/06/2023 3:45 PM PENERBIT ILMU BAKTI SDN. BHD.