ch 14 inheritance teacher

PANITIA BIOLOGI NEGERI KELANTAN 2007

CHAPTER 14: INHERITANCE

Multiple Choice Questions

For Question 1 to 15, each question is followed by four alternative answers A, B, C or D.

Choose one correct answer for each question and blacken the corresponding space in your

objective answer sheet.

14.1 Concept of Inheritance Based on Mendel’s Experiment

1. Diagram 14.1 shows a pair of homologous chromosomes. The alphabets represent the

gene in the chromosomes.

Q and q represent

A. linked genes

B alleles

C. genotypes

D. phenotype

2. If T represents the allele for tallness and t the allele for dwarfness, then an individual

with Tt is

A. homozygous for tallness

B heterozygous for tallness

C. homozygous for dwarfness

D. heterozygous for dwarfness

3. Two pure-bred plants with the genotypes BBRR and bbrr respectively are crossed. F1

offspring were allowed to self-pollinate. How many types of phenotypes are produced if

B and R are dominant?

A. 2

B 4

C. 8

D. 16

BIOLOGY QUESTION BANK14-1

P q R s T U

p Q R S t u

DIAGRAM 14.1


4. Diagram 14.2 shows the inheritance of type of eye colour in humans. Gene for blue

eyes is recessive.

First generation

Second generation

Third generation

Key :

Male – brown eyes

Male – blue eyes

Female – blue eyes

Female – brown eyes

Which of the following conditions are true of P and Q?

P Q

A heterozygous Homozygous dominant

B heterozygous Homozygous recessive

C Heterozygous dominant heterozygous

D Homozygous recessive heterozygous

5. The allele for black hair in human is dominant to the allele for brown hair. A man with

black hair is heterozygous while his wife has brown hair. The probably of getting a child

with brown hair is

A. 1

B ¾

C. ½

D. ¼


P

Q

DIAGRAM 14.2


14.2 Inheritance

6. A married couple has different blood group. Ahmad has a blood group A, whereas his

wife has a blood group B.

Determine the possible blood group of their children.

A. It could only be blood group AB

B. It could only be blood group A or B

C. It could only be blood group A, B or O

D. It could only be blood group O

7. Which of the following represented the alleles of the human blood group?

A. ABO

B. Xa Xb Xo

C. Aa B bOo

D. IA IB IO

8. Erythroblastosis fetalis is the problem of new born baby that related to

A. blood transfusion incompatibility

B. Rh factor

C. malnutrition

D. mother’s drug abuse.

9. Diagram 14.3 shows the human karyotype.

The abnormal number of the chromosome 21 is a result of non-disjunction during

A. meiosis

B. mitosis

C. cytokinesis


DIAGRAM 14.3


D. plasmolysis

10. Heredity diseases are disease that can be transmitted from parent to their offspring.

Which of the following is not the heredity disease?

A. Haemophillia

B. Colour blindness

C. Muscular dystrophy

D. Malaria

14.3 Understanding Genes and Chromosomes

11. Diagram 14.4 shows a part of the DNA structure.

Figure 1

Which of the following represents J,K,L and M ?

J K L M

A Deoxyribose

sugar

Nitrogenous

base

Phosphate

group

Phosphate

group

B Phosphate

group

Deoxyribose

sugar

Nitrogenous

base

Nitrogenous

base

C Phosphate

group

Nitrogenous

base

Deoxyribose Phosphate

group

D Nitrogenous

base

Deoxyribose

sugar

Nitrogenous

base

Phosphate

group


DIAGRAM 14.4


12. The uses of DNA fingerprint are

I. to identify the parent of someone

II. to solve criminal cases

III. to identify diseases inherited

IV. to identify a baby claimed by different mothers

A. I and IV only

B. II and III only

C. I, II and III only

D. I, II, III and IV

13. Which of the following is not true about DNA fingerprinting?

A. It is a print pattern from the right thumb

B. It can be made from a blood sample, hair or semen sample

C. It is used in forensic work

D. It can be used to confirm animal pedigrees

14. What are the advantages of introducing genetic engineering in agriculture?

I. Increase the food yield

II. Improve the food quality

III. Increase in resistance to pests and diseases

A. I only

B I and II only

C. II and III only

D. I, II, III

15. Which of the following is not the objective of the Human Genome Project?

A. Determine the sequence of all the base pairs found in the DNA of the human

genome

B Make map showing the exact locations of genes for major sections of human

chromosomes

C. Produce linkage maps where inherited traits can be tracked over generations

D. To determine the sex of foetus



Structured Items


1. Diagram 14.5 shows a cross between two varieties of pea plants in which tall pea plant

with red flowers is crossed with short pea plant with white flowers. Both plants are pure

bred.

(a) (i) Complete the alleles in Figure 1. [2 marks]

(ii) State the type of cross shown in Figure 1. [1 mark]

Dihybrid cross

(b) (i) State the genotype of plants in F1 generation. [1 mark ]

TtRr

(ii) State the phenotype of plants in F1 generation [1 mark]

Tall pea plant with red flowers


Parent

DIAGRAM 14.5

Key :

T – tall

t – short

R– red flowers

r – white flowers

F1 generation

Gametes

T

R

T

R

t

r

t

r

T

R

T

R

T

R

T

R

t

r

t

r

t

r

t

r

t

r

T

R


(c) If crossing over between genes takes place in the individuals of F1 generation

during meiosis, draw the possible gametes produced from the F1 generation.

[2 marks]

(d) Table 14.1 shows an incomplete Punnett’s of crossing made between a

heterozygous plant for tallness and the red flowers and a homozygous

recessive plant for the same traits.

Gametes TR Tr tR tr

tr TtRr Ttrr ttRr ttrr

(i) Complete Table 1 to show the various offspring produced. [2

marks]

(ii) Explain the results. [3 marks]

- A plant which is heterozygous for tallness and the red

flowers has the genotype TtRr. The plant produces four

types of gametes, TR, Tr,tR and tr.

[1]

- A plant which is homogygous recessive has the genotype

ttrr. The plant produces a type of gametes, tr.

[1]

- When the plant is crossed with another plant at random to

produce plants with TtRr, Ttrr, ttRr and ttrr as their

genotypes.

[1]

(iii) What is the percentage of the offspring produced for short pea plant with

white flowers? [1 mark]

25%


T

r

T

R

t

r

t

R

TABLE 14.1


14.2 Inheritance

2. Diagram 14.6 shows the pedigree of a family for the haemophillia disease.

Haemophollia is a hereditary sex-linked disease cause by a recessive allele found on

the X chromosome.

Hassan Anis

Akmal Adam Esah

Key

Male normal XH XH, XHY

Male haemophillia Xh Xh, Xh Y

Female normal XH Xh

Female haemophillia

(a) (i) What is Anis genotype? [1 mark]

Xh Xh

(ii) Explain how her genotype is determined. [2

marks]

The allele for haemophillia is recessive. Both alleles must be

recessive to produce the phenotype of haemophillia.

(b) (i) What is the genotype of Hassan? [1 mark]

XHY


DIAGRAM 14.6


(ii) Explain how his genotype is determined. [2

marks]

Male have only one X chromosome. To show the normal phenotype,

Hassan must have the dominant allele in the X chromosome.

(c) Esah has a heterozygote genotype, while Adam is a haemophillia patient.

Illustrate by using a schematic diagram to show their children genotype and

phenotype probability. [4

marks]

Adam Esah

(d) What is the probability of their son to get haemophilia? [1mark]

25%

(e) What is the meaning of hereditary sex-linked? [2marks]

The genes on the sex chromosomes which are not involve in sex

determination


XhY XH Xh

Xh Y Xh XH

Y XhY XH

Xh

Xh XH

Xh Xh

Meiosis



3. Diagram 14.7 shows the molecular arrangement of part of a DNA molecule.

(a) (i) Name the basic unit that made up the DNA molecule. [1mark]

Nucleotide.

(ii) Circle on the diagram above to show a basic unit of DNA molecule.

[1mark]

(iii) Name the component that made up the basic unit named in (a)(i).

[1mark]

A deoxyribose sugar, a nitrogenous base and a phosphate group.

(b) Describe the structure of the DNA molecule. [1mark]

A DNA molecule consist of two polynucleotide strand coiled together to

form double helix. The two strands are anti parallel.


DIAGRAM 14.7


(c) The DNA carries the genetic codes that determine the characteristic of

organism. Which part of the molecules that form the genetic codes?

[1mark]

The sequence of nitrogenous bases.

(d) Complete Diagram 14.8 below to show the organization of a chromosome .

[5marks]


Chromosome

Nucleotide

Bes nitrogenous

Phosphate group

Deoxyribose sugar

Polynucleotide

DNA

DIAGRAM 14.8


Essay Questions


1. (a)

Explain the above statement by using the monohybrid cross between a homozygous

dominant long-winged Drosophilia is mated with a homozygous recessive vestigal-

winged Drosophila.

[10 marks]

- if a homozygous long-winged Drosophilia is mated with a vestigial-winged

Drosophilia, all the F1 offspring are found to be long-winged

[1]

- if two of the F1 flies are allowed to mated with each another,

[1]

- a mixture of long-winged flies and vestigial-winged flies are produced

[1]

- in the ratio of 3 : 1

[1]

Parents : Long-winged X

Vestigial-

winged

Genotypes : LL X ll

Meiosis process

Gametes : All L All l

Fertilisation :

F1 Genotypes : L l

F1 phenotypes : All long-winged


Mendel’s First law :

The characteristic of an organism is determined by a pair of genes located

at the same locus of a pair of homologous chromosomes. Only one of the

pair of genes can be present in a gamete.


When F1 flies are allowed to mated each another

F1 Genotypes : L l X L l

Gametes : L l L l

F2 genotypes : LL L l L l ll

F2 phenotypesLong-

winged

Long-

winged

Long-

wingedVestigial-winged

Ratio 3 1

Diagram = [3]

- in this case, the gene for long-winged, L, is dominant [1]

in the production of gametes, genes from homologous chromosomes

come together [1]

and then segregate into separate gamete [1]

therefore, one gamete receive one of a pair of alleles [1]

in the F1 generation, the gametes produced receive either L or l

[1]

[max

10]

(b) A dihybrid cross between two varieties of oil palm, Dura and Pasifera, produces

a new variety, Tenera. Tenera has better quality fruits compared to Dura and

Pisifera as shown in Table 14.2.

Oil palm varietyCharacteristics of the trait

Genotype Phenotype

Dura hhFFThin husk

Thick flesh



Pisifera HHffThick husk

Thin flesh

TeneraThick husk

Thick flesh

Table 14.2

Key :

H – represents dominant allele for thick husk

F – represents dominant allele for thick flesh

Using a schematic diagram, explain why self-crossing between the Tenera variety

does not produce offspring of the same quality as their parents.

[10 marks]

Parents : Dura X Pisifera

Genotypes : hhFF X HHff

Meiosis process

Gametes : All hF All Hf

Fertilisation :

Genotypes F1 :

F1 phenotypes :

Genotypes F1 : HhFf

Meiosis

process:

Gametes :

F2 generation

Gametes HF Hf hF hf

HF HHFF HHFf HhFF HhFf

Hf HHFf HHff HhFf Hhff

hF HhFF HhFf hhFF hhFf


[5]

[3]

HhFf

Allow F1 plants to self-pollinate

HhFf

All thick husk, thick flesh

HF Hf hF hfHF Hf hF hf

1

1

1

1

1

1


hf HhFf Hhff hhFf hhff

Just 9/16 x 100% = 900/16 % = 56¼ % @ = 56.25% of F2 generation that produced

have a phenotype thick husk and thick flesh the same quality as their parents. [1]

14.2 Inheritance

2. (a) Mala and Siva is a married couple with four girls. Siva always blamed his wife

because of not give birth a baby boy. As a consultant at LPPKN, please consult

these couple to understand the fact related to the sex determination of babies.

Draw a schematic diagram to explain your answer.

[10marks]

- The human females have two identical sex chromosomes called X

chromosomes. [1]

- The human males have one X chromosomes and one smaller Y

chromosomes. [1]

- The chromosomes X and Y determine the sex of individual, whether

boy or girl. [1]

- During the formation of gametes in female, the sex chromosomes

separate, all the female eggs produced contain one X chromosome. [1]

- The human male produces two types of sperm: sperm with one X

chromosome and sperm with one Y chromosome. [1]

- If the sperm with X chromosome fertilizes the ovum, the child is a

female. [1]

- If the sperm with Y chromosome fertilizes the ovum, the child is a male.


[1]

Phenotype Thick husk Thick husk Thin husk Thin husk

Thick flesh Thin flesh Thick flesh Thin flesh

Ratio 9 : 3 : 3 : 1

1


- There is an equal chance of mother to give birth a baby boy or a baby

girl. [1]

Diagram = [3]

(b) Dina has a problem to get a second child with the history of three times baby

stillborn. After consulted by a specialist, she found that she is facing with the

problem of Rh factor. As a specialist explain to Dina to understand her problem.

[10marks]

- The Rh factor is referring to the antigen found on the surface of some

red blood cells. [1]

- The people who have the Rh factor are said to be Rh positive and

those who not are called Rh negative. [1]

- The allele Rh+ is dominant to the allele Rh-. [1]

- In the case of a Rh- mother who become pregnant by a Rh+

heterozygous father has 0.5 probability of having a child who is Rh +.

[1]

- During her late pregnancy or during birth, some of the fetal blood cells

enter the mother’s blood. [1]

- The mother’s blood will produce anti-Rh antibodies. [1]

- The harmful effect does not show during the first pregnancy, her first

child is safe. [1]

- In the next pregnancy, the antibody level built up in the maternal blood

stream. [1]



- Her antibodies may cross the placenta and cause agglutination and

haemolysed of the fetal red blood cells. [1]

- The baby may be stillborn or may die within a few days after birth. [1]


1. Diagram 14.9 show three different molecules which are found in the DNA.

(a) Describe the structure of the DNA. [4 marks]

- A nucleotide is formed when the three molecules, phosphate, sugar

and base are linked together. [1]

- One nucleotide joins another nucleotide through condensation

forming a dinucleotide [1]

- Many nucleotides joint together forming a long double strand

polynucleotide chain. [1]

- Sugar and phosphate molecules alternate to form the backbone of

the nucleotide chain [1]

- Base molecules project out sideways from the sugar molecules [1]


Phosphate Base Sugar

Diagram = [3]

[Max10]

DIAGRAM 14.9


- Two polynucleotide chains twist to form a double helix structure of

DNA [1]

Max = [4]

(b)

The transfer of genes or DNA plays an important role in genetic

engineering. For instance, gene from humans can be transferred to

bacterial cells. Once inside the bacteria, these human gene control

the bacterial cells. This causes the bacteria to synthesise something

they would not normally make in nature. For example, bacteria can

produce human insulin.

Discuss the advantages and disadvantages of genetic engineering.

[6 marks]

Advantages of genetic engineering:

- Enables the mass production of various types of products

through the development of new strains of crops and

livestock. [1]

- ensures that food sources are readily available at cheaper

cost [1]

- enables the mass production of medical and pharmaceutical

products [1]

- can help solve environmental problems such as oil spills by

using genetically engineered bacteria [1]



Any 3 = [3]

- Disadvantages of genetic engineering:

- the introduction of foreign genes in microorganisms leads to

the creation of harmful pathogens. [1]

- New species of crops that are produced can cause the

original species to become extinct. [1]

- Genetically modified food may contain an animal gene which

may be objectionable to vegetarian. [1]

- transgenic crops can colonise and displace the natural plant

population [1]

- the side effects of eating food from genetically modified

organism are still uncertain. [1]

any 3 = [3]

(c) Describe briefly on the important of DNA fingerprinting and human genome

project to mankind. [10 marks]

- DNA fingerprinting can be used for identification purposes in solving

criminal cases [1]

- for example, DNA samples from blood , skin, hair or semen left by a

criminal at the scene of crime can be analysed. [1]

- to identify the parent of someone [1]

- to test potential organ donors for compatibility with a particular patient

[1]

- to examine the relationship among human populations [1]

- to detect human genetic diseases and cancer [1]

- to confirm the genotypes of animals and plants in agriculture [1]

max = [5]

The importance of human genome project:

- A genome is the total genetic content of any cell in an organism. It

consists of all the genes on all the chromosomes. [1]

- human genome project aims to map the position of genes on the

chromosome and determine the sequence of bases in the DNA. [1]

- Identification of defective genes and hence the opportunity to offer

early treatment. [1]



- Identification of genes which confer a susceptibility to certain diseases

and so enable individuals to take preventive measures. [1]

- Prediction of proteins that the genes produce, giving an opportunity to

design appropriate drugs to enhance or inhibit the activities of these

protein. [1]

- Discovering the function of all the genes in the human genome will

produce exciting new information. This should help us understand

more about how body works, and how to prevent and care diseases. [1]

max = [5]


ch 14 inheritance teacher

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blood group ab b

human blood group

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blood group o

possible blood group

meiosis b

haemophillia b

brown hair