Name: …………………………………………………… Class: …….Chapter 3Exercise 1Learning Area: Heredity and VariationLearning Objective: Understanding cell division

1. The diagram below shows the process of mitosis. Fill in the blanks with the words given.

Replication align nucleus membrane divide two shorten thicken Chromatid same separate move

Stages of mitosis Simple explanation

∙ Chromosomes in the nucleus ………………. , ……………

and are more visible.

∙ ……………………of chromosomes occur.

∙ Each strand of chromosome is called a …………………….

∙ Chromosomes ......................... themselves in the centre of the cell.

∙ The ……………………………………….. disappears.

∙ Chromatids …………………… and ……………………… to the opposite ends of the cell.∙ At the same time, the cell starts to …………………………..

∙ The nucleus membrane forms again.∙ ………………………… daughter cells are formed.∙ Each progeny cell has the ………………………. number of chromosomes as the parent cell.

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Score: .…/12 x 100 = ……….. %

2. Arrange the diagram in the correct sequence:

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P Q R

W X

S

T U V

Name: ……………………………………………… Class: ……….Chapter 3Exercise 2Learning Area: Heredity and VariationLearning Objective: Understanding cell division

1. Tick (√) for the correct statement and (X) for the wrong statements about mitosis.

(i) Two daughter cells are formed in mitosis. ( )

(ii) In mitosis, cytoplasm divides once. ( )

(iii) Mitosis is important for producing new cells to replace the old or damaged cells. ( )

(iv) Mitosis cause genetic variation. ( )

(v) The number of chromosomes in the daughter cell is half the number of chromosomes in the parent cell. ( )

(vi) The parent cell and the daughter cell have the same genetic content. ( )

(vii) Crossing-over does not take place in mitosis. ( )

(viii) Mitosis occurs in somatic cells. ( )

2. Tick (√) for the correct statement and (X) for the wrong statements about meiosis.

(i) Four daughter cells are formed in meiosis. ( )

(ii) In meiosis, the number of chromosomes in each daughter cell is the same as in the parent cell. ( )

(iii) Meiosis is important to organisms that reproduce sexually because it produces gametes. ( )

(iv) Meiosis cause genetic variation. ( )

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Score: .…/27 x 100 = ……… %

3. Compare the processes of meiosis and mitosis.

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MEIOSIS MITOSIS

SIMILARITIES

(i)……..…………………………………………(ii)…….………………………………………….(iii)…….………………………………………….

DIFFERENCES

…………..Number of

divisions ………….

…………..Number of daughter

cells produced …………..

…………..

Number of chromosomes in

daughter cell compared to parent cell

………….

………….

Combination of chromosomes in

daughter cell compared to parent cell

…………..

………….Crossing

over …………..

…………..Genetic variation …………..

Name: ……………………………………………………… Class: …..Chapter 3Exercise 3Learning Area: Heredity and VariationLearning Objective: Understanding the principles and mechanism of inheritance.

1. Complete the table with the correct definition.

TERMS DEFINITIONS

……………………Describes the characteristics traits of an organism, including the physical and the physiological

…………………… Organism which has two identical genes. Example: TT or tt.

…………………… Composition of gene in an organism

……………………. Thread-like structure present in the nucleus of a cell

……………………..A gene which is weak and its effect can be masked by a dominant gene

…………………….An organism which has one dominant gene and one recessive gene for a certain characteristic. Example: Bb ( B = dominant gene for black fur , b = recessive gene for white fur )

……………………Unit carrying a characteristic trait which exists in pairs in a chromosome.

…………………….Organism which has two different members of the same gene. Example: Tt.

…………………….Exchange of genes when two chromatids cross giving rise to genetic variation.

…………………….. A gene which masks the effect of a weaker gene

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Phenotype Chromosomes Crossing over Dominant gene Gene Genotype Heterozygote Homozygote Hybrid Recessive gene

Score: .…/23 x 100 = …… %

2. A man with black hair (Bb) marries a woman with blonde hair (bb). Complete a schematic diagram to show the cross between a men who is heterozygous for black hair (Bb) and a woman with blonde hair (bb)

B : Gene carrying black hair trait (dominant) b : Gene carrying blonde hair trait (recessive)

Father Mother Parent

Meiosis

Gamete

Fertilisation

Offspring genotype

Offspring phenotype

Based on the schematic diagram, determine the phenotype ratio of their offspring

……………………………….

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Bb bb

Name: …………………………………………………. Class: ……Chapter 3Exercise 4Learning Area: Heredity and VariationLearning Objective: Understanding sex determination and the occurrence of twins in human being.1. Complete the following table.

WOMAN MANSomatic cell Ovum Somatic cell Sperm

Autosomes(non-sex

chromosomes)……………….. 22 44 ……………………

Sex chromosomes

XX ………………. ……………….. X or Y

Total 44 + XX(46 chromosomes)

…………………(23 chromosomes)

………………….(46 chromosomes)

22 + X or22 + Y

(23 chromosomes)

2. The schematic diagram below shows sex determination of an offspring. (a) Complete the schematic diagram.

Father Mother

Parents :

Meiosis

Gamete :

Fertilisation

Offspring :

Sex: …………….. ……………. …………… ……………

(b) Based on the schematic diagram above, fill in the blanks with suitable answer.

(i) The probability ( in percentage ) of a son or a daughter being born at each birth is ……………………

(ii) The sex of a zygote depends on whether the ovum is fertilized by a sperm carrying an X-chromosome or a sperm carrying a Y-chromosome. If the ovum is fertilized by a sperm carrying a Y-chromosome, a …………….. zygote will be produced.

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Score: .…/33 x 100 = …… %

44 + XY 44 + XX

……… …….. ………. ……….

………... ………. ………. ………..

3. The comparison between identical twins with non-identical twins:

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……………….Number of ovum

involved

Genetic make-up

………………..Number of sperms

involved

………………….Fertilised ovumsplits into two

………………….

Sharing of placenta…………………..

Sex…………………..

………………..

…………………

…………………

…………………

…………………..

………………….

Differences

IDENTICAL TWINSNON-IDENTICAL TWINS

Similarities:Number of babies born:

……………

Name: ………………………………………………… Class: ……..Chapter 3Exercise 5Learning Area: Heredity and VariationLearning Objective: Understanding mutation.

1. Match the disease with the correct description

DISEASES DEFINITION

Albinism……..

[A] Lack of a protein in the blood that is necessary for blood-clotting

Down’sSyndrome

……...

[B] The mutant genes cause the red blood cells to be sickle shaped. The patient is pale because the red blood cells carry little oxygen due to lack of haemoglobin and are easily broken.

Colour-blindness ………

[C] A female who has only one X chromosome. The female has no ovaries and menstrual cycle.

Turner’sSyndrome

………

[D] Hereditary lack of pigmentation in an organism. The animals or human beings have no colour in their skin, hair or eyes.

Sickle-cellanaemia

……..

[E] Possession of an extra chromosome 21. The child has short neck, short nose, downward-sloping eyes and protruding tongue.

Klinefelter’sSyndrome

……..

[F] The existence of a recessive gene that is defective and causes an inability to see and differentiate between red and green colour

Haemophilia………

[G] Presence of three sex chromosomes (XXY) instead of the usual two. Males are sterile and have small testes

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Score: .…/7 x 100 = …… %

Name: ………………………………………………… Class: ……Chapter 3Exercise 6Learning Area: Heredity and VariationLearning Objective: Evaluating the effects of genetic research on human life.

1. Complete the table with the correct answer.

Technique Explanation

………………

The techniques involved in altering the characteristics of an organism by inserting genes from another organism into its DNA.

………………The crossing of two parents, each with some good traits, to produce offspring with the good traits of both parents.

……………….A coordinated international project, begun in 1988 to map the entire human genome so that the genes could be isolated and sequenced. It involved the production of a DNA library.

………………..The application of genetic engineering techniques to alter or replace defective genes.

…………………The process of producing offspring asexually from a single parent.

…………………The growth of the tissues of living organisms outside the body in a suitable culture medium.

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Gene therapy Genetic engineering Cloning Selective breeding Tissue culture Human Genome Project

Score: .…/6 x 100 = …… %

Name: …………………………………………………… Class: ….. Chapter 3Exercise 7Learning Area: Heredity and VariationLearning Objective: Analysing variation among living things.

1. Fill in the blanks with correct word(s).

2. A study had been carried out to gather the body weight (in kg) of 40 students in Form 4 Omega . The results are shown in Table 2.1.

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32.1 41.7 38.1 40.8 42.8 44.1 40.5 42.7

40.4 35.7 44.5 42.4 37.3 47.7 38.3 49.3

46.7 43.2 48.6 43.7 42.1 33.8 48.2 41.4

42.3 53.0 31.1 35.3 43.7 51.3 41.7 47.336.8 47.3 41.5 49.3 37.9 43.5 37.7 39.3

Table 2.1

Score: .…/21 x 100 = …… %

Definition :

Natural ………………… among individual of the ………………species.CONTINUOUS

DISCONTINUOUS

Definition:

Variation that shows a whole

range of intermediate

between two ………………...Example:1. Body weight2. Height3. Intelligence

Influenced by environmental

factors such as ……………. ,

……………. and …………..

Definition:

Variation that shows

…………………… differences.

Example:1. Gender /sex2. Blood group3. Type of ear lobe

Influenced by ………………. factor.

VARIATION

(a) Based on the data shown above, complete Table 2.2.

Body weight / kg Number of students

30.0-34.9 ……………

35.0-39.9 …………..

40.0-44.9 …………..

45.0-49.9 ……………

50.0-54.9 ……………

Table 2.2

(b) Based on the results in Table 2.2, draw a histogram to show the number of students against the weight.

(c) (i) State type of variation for the trait above.

………………………………………………………………………………………………….

(ii) Explain your answer.

………………………………………………………………………………………………….

(d) Give another one example of the same type of variation mentioned in (c).

………………………………………………………………………………………………………..

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2

4

8

10

30.0 –34.9

35.0 –39.9

12

14

16

18

20

40.0 –44.9

45.0 –49.9

50.0 –54.9

Weight / kg

Numberof student

6

3. A study had been carried out to gather the variation in blood group type in a population of Kampar Town. Table 3 shows the data collected.

Blood group type Number of individuals

A 40

B 60

AB 30

O 80

Table 3 (a) Based on the results in Table 3, draw a bar chart to show the number of individuals against the groups of blood group.

(b) (i) State type of variation for trait above.

………………………………………………………………………………………………….

(ii) Explain your answer.

…………………………………………………………………………………………………..

(c) Give another one example of the same type of variation mentioned in (b).

……………………………………………………………

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20

40

60

80

Type of blood group

Number of individual

A B AB O