GENETICS

PAST KCSE QUESTIONS ON TOPICS

1. The figure below is a structural diagram of a portion from a nucleic acid strand.

            — S — P — S — P — S — P — S 

                 C     G             U              C

(a)       Giving a reason, name the nucleic acid to which the portion belongs.                                                                                                                                 ( 2 marks)

Name  _________________________

Reason _________________________

(b)       Write down the sequence of bases of a complimentary strand to that shown above                                                                          ( 1 mark)

2. State two structural differences between ribonucleic acid ( RNA) and

deoxyribonucleic acid ( DNA)                                               ( 2 marks)

3. Name a disorder of human blood that is caused by mutation ( 1 mark)
4. State the function of deoxyribonucleic acid (DNA) molecule ( 1 mark)
5. Give a reason why it is only mutation in genes of gametes that influence evolution ( 2 marks)
6. In an experiment, red flower were crossed with plants with white flower. All the plants in the F1 generation had pink flowers.

(a)       Give a reason for the appearance of pink flower in the F1

generation                                                                   ( 1 mark)

(b)       If the plants from F1 generation were selfed, state the phenotype ratio of the F2 generation                                       ( 2 marks)

7. State two characteristics that researchers select in breading programmes.

( 2 marks)

8. Give an example of sex- linked trait in humans on; ( 2 marks)

Y chromosome __________________

X chromosome __________________

9. In an experiment, a variety of garden peas having a smooth seed oat was crossed with a variety with a wrinkled seed coat. All the seeds obtained in the F1 had a smooth seed coat. The F1 generation was selfed. The total number of F2 generation was 7324.

(a)       Using appropriate letter symbols, work out the genotype of the F1 generation.                                                                  ( 4 marks)

(b)       From the information above, work out the following for the F2 generation

(i) Genotype ratio                                                       ( 2 marks)

(ii) Phenotype ratio                                                    ( 1 mark)

(iii) Wrinkled number                                                            ( 1 mark)

10. In a certain plant species, some individual plant may have white, red or pink flower. In an experiment a plant with white parent plant were pure lines. All the plants from F1 generation were pink. Using letter R to represent the gene for red colour and letter W for white colour;

(a)       Work out the genotype of F1 generation                   ( 3 marks)

(b)       If the plants from F1 generation were selfed, what would be the phenotypic ratio of the F2 generation?                      ( 3 marks)

(c)       What is the genetic explanation for the absence of plants with red and white in the flower F1 generation?                ( 2 marks)

11. In a breeding experiment, plants with red flower were crossed. They produced 123 plants with red flowers and 41 with white flowers.

• Identify the recessive character. Give a reason
• What were the genotypes of the parent plants that give rise to the plants with red and white flowers?
• If the white flowers were selfed, what would be the genotypes of their offspring?

12. (a) Name two disorders in humans caused by gene mutation

( 2 marks)

(b)       Describe the following chromosomal mutations

(i) Inversion                                                                ( 2 marks)

(ii) Translocation

(c)       In mice the allele for black fur is dominant to the allele for brown fur. What     percentage offspring would have brown fur from a cross between heterozygous black mice and brown mice? Show your working. Use letter B to represent the allele for black colour.                                                                                             ( 4 marks)

13. (a)       What is meant by the term allele?                             ( 1 mark)

(b)       Explain how the following occur during gene mutation

(i)        Deletion                                                          ( 1 mark)

(ii)       Inversion                                                         ( 1 mark)

(c)       What is a test- cross?                                                 ( 1 mark)

14. In maize the gene for purple colour is dominant to the gene for white colour. A pure breeding maize plant with purple grains was crossed with a heterozygous plant.

• (i) Using letter G to represent the gene for purple colour, work out

the genotypic ratio of the offspring                           ( 5 marks)

(ii) State the phenotype of the offspring                          ( 1 mark)

(b) What is genetic engineering?                                           ( 1 mark)

15. Define the following terms as used in genetics.

(i)        Alleles

(ii)       Genotype

(iii)      Phenotype

16. A farmer mated his dark red cow with a white bull. The cow gave birth to a light red calf

• State why the calf is light red and not dark red or white
• If a light red bull is mated with a dark red cow, work out using appropriate letter symbols the probability of getting  a light  offspring

17. (a) What is meant by linked genes?

(b) (i) In fruit flies (Drosophila) the gene for red eyes ® is dominant  over

the one for white – eye (r). If a true breeding white – eyed male, all the offspring will be red eyed. However, if a true – breeding white- eyed female is mated with a true- breeding red- eyed male, all the female offspring will be red – eyed. Explain this apparent contradiction.

(ii)             Work out the ratio of the expected phenotypes if a red- eyed female offspring from the cross- described in (i) above  is mated with red- eyed males.

18. (a) Explain the term  variation with reference to the study of genetics.

(b)       Using relevant examples distinguish between discontinuous variation and   continuous variation

(c)       What is the importance of genetic variation?

(d)       Describe one example where genetic variations has helped a species to survive

19. The diagram below shows the base sequence of part of a nucleic acid stand. Observe it and answer the questions that follow

G         T          T          A         G         C         T          G         A

(a)       What do the letters G, T , C  and A represent?

(b)       Giving your reasons state whether it is part of DNA or an RNA strand.

• Show the complementary DNA strand
• Show the complimentary RNA strand

20. In human couples the sex of a baby is determined by the man. Explain this statement.

CHAPTER NINE

PHOTOELECTRIC EFFECT

Photoelectric effect was discovered by Heinrich Hertz in 1887. Photoelectric effect is a phenomenon in which electrons are emitted from the surface of a substance when certain electromagnetic radiation falls on it.

Metal surfaces require ultra-violet radiation while caesium oxide needs a visible light i.e. optical spectrum (sunlight).

DEMONSTRATING PHOTOELECTRIC EFFECT

 Using Neutral Plates

Procedure

• Set up the apparatus as shown in figure 9
• Direct UV-radiation towards plate A. Observe what happens to the galvanometer.
• Place the glass barrier in between the source and plate A and again observe what happens.

Observation

When ultraviolet radiation is allowed to fall on metal plate A, the galvanometer shows deflection.

When the barrier is introduced so that the radiation is cut off, the galvanometer shows no deflection .

Explanation

When ultraviolet radiation energy falls on a metal surface, some electrons absorb this energy and are dislodged from the surface.

The deflection of the galvanometer indicates that electrons are emitted at plate A and attracted by the plate B, causing a current to flow.

The glass plate, however, cuts off the ultraviolet radiation.

(b) Using Charged Electroscope

Procedure

• Direct ultraviolet radiation from a mercury vapour lamp onto the zinc plate and observe the divergence of the leaf when the electroscope is positively charged, see figure

Observation

For the positively charged electroscope, the leaf divergence remains the same. However, for the negatively charged electroscope, the leaf divergence decreases.

Explanation

When the zinc plate is irradiated with ultraviolet radiation, electrons are emitted from its surface.
The photoelectrons emitted from the positively charged zinc plate do not escape due to attraction by the positive charge on the plate and the leaf divergence therefore remains the same.

The photoelectrons emitted from the negatively charged zinc plate are repelled and the electroscope becomes discharged as a result of which the leaf divergence decreases.

If a sheet of ordinary glass (which absorbs ultraviolet radiation) is introduced between the negatively charged zinc plate and the ultraviolet source, the leaf divergence remains same.

Light Energy and Quantum Theory

Max Planck in 1901 came up with the idea that light energy is propagated as small packets of energy.

Each packet is called a quantum of energy (plural: quanta).

 In light, these discrete amounts of energy are called photons.

According to Planck, the energy E possessed by one photon is given by; E = hf, where, h is Planck’s constant equal to 6.63 x 10^-34 J s and f the frequency of the radiation.

c=fλ

Hence

Thus  where c is the velocity of the radiation in vacuum and λ is the wavelength.

Since c and h are constant, a radiation of longer wavelength has lower energy

Terms used in photoelectric effect

Work function W₀

A minimum amount of work is needed to remove an electron from its energy level so as to overcome the forces binding it to the surface.

This work is known as the work function with units of electron volts (eV). One electron volt is the work done when one electron is transferred between points with a potential difference of one volt; that is,

1 eV = 1 electron × 1 volt

1 eV = 1.6 × 10^-19 × 1 volt

• eV = 1.6× 10^-19 Joules (J)

Threshold frequency f₀

This is the minimum frequency of the radiation that will cause a photoelectric effect on a certain surface. The higher the work function, the higher the threshold frequency.

Threshold Wavelength λ₀

This is the maximum the maximum wavelength beyond which no photoelectric emission will occur.

Work function W₀=hf₀

For any radiation of frequency f less than f₀, hf will be lower than W0 and emission will not occur.

When the frequency of the radiation is f_o‘ hf₀ = W₀ (the work function), and emission occurs.

When the frequency of the radiation f > f₀, hf > W₀ and the excess energy in this case appears as the kinetic energy of the emitted electron.

Factors affecting the photoelectric effect

1. Intensity of the incident radiation– the rate of emission of photoelectrons is directly proportional to the intensity of incident radiation.
2. Type of the metal/Work function of the surface– photoelectrons are emitted at different velocities with the maximum being processed by the ones at the surface.

• Energy of the incident radiation– the cut-off potential for each surface is directly proportional to the frequency of the incident radiation.

The graph of the frequency of the stopping potential against frequency of the radiation is a straight line as shown below

The graph is a straight line. From Einstein’s photoelectric equation;
Work done by stopping potential is given by ev.

By work energy theorem,
Substituting in Einstein’s photoelectric equation;
hf= hfo + eV_s

Therefore, e V_s = hf – hf₀

But hf₀ is equal to work function W₀.  Hence, the graph of Vs against f is straight line cutting the f-axis at f_₀.

The slope of the graph is    and the V_s. intercept is

Both Planck’s constant h and the work function W₀ can therefore be calculated from the graph.

Einstein’s photoelectric equation

When a photon strikes an electron, all its energy is absorbed by the electron.

Some of the absorbed energy is used to dislodge the electron from the metal surface while the rest appears as the kinetic energy of the emitted electron.

The energy transformation during photoelectric emission is thus summed as follows;

Examples

• The cut-off wavelength for a certain material is 3.310 × 10^-7 What is the cut-off frequency for the material?

Solution

Speed of light ‘c’ = 3.0 × 10⁸ m/s. Since f = c / λ, then  f = 3.0 × 10⁸ / 3.310 × 10^-7 = 9.06 × 10¹⁴ Hz.

• The work function of tungsten is 4.52 e V. Find the cut-off potential for photoelectrons when a tungsten surface is illuminated with radiation of wavelength 50 × 10^-7 m. (Planck’s constant, h = 6.62 × 10^-34 Js).

Solution

Frequency ‘f’ = c / λ = 3.0 × 10⁸ / 2.50 × 10^-7.

Energy of photon = h f = 6.62 × 10^-34 × (3.0 × 10⁸ / 2.50 × 10^-7) × (1 / 1.6 × 10^-19)  = 4.97 eV.

Hence h f_co = 4.52 e V. e V _co = 4.97 e V – 4.52 e V = 0.45 e V = 7.2 × 10^-20 J V _co = 7.2 × 10^-20 / 1.6 × 10^-19 = 0.45 e V.

• The threshold frequency for lithium is 5.5 × 10¹⁴ Calculate the work function for lithium. (Take ‘h’ = 6.626 × 10^-34 Js)

Solution

Threshold frequency, f _o = 5.5 × 10¹⁴ Hz, ‘h’ = 6.626 × 10^-34 Js

Φ = h f = 5.5 × 10¹⁴× 6.626 × 10^-34 = 36.4 × 10^-20 4. Sodium has a work function of 2.0 e V. Calculate

1. The maximum energy and velocity of the emitted electrons when sodium is Illuminated by a radiation of wavelength 150 nm.

• Determine the least frequency of radiation by which electrons are emitted. (Take ‘h’ = 6.626 × 10^-34 Js, e = 1.6 × 10^-19, c = 3.0 × 10⁸ m/s and mass of electron      = 9.1 × 10^-31 kg).

Solution

1. a) The energy of incident photon is given by h f = c / λ

= (6.626 × 10^-34 × 3.0 × 10⁸) / 1.50 × 10^-9 = 1.325 × 10^-18 J

K.E _max = h f – Φ = (1.325 × 10^-18) – (2 × 1.6 × 10^-19) = 1.0 × 10 ^-18 J (max. K.E of the  emitted electrons)

But K.E _max = ½ m v²_max. Therefore;

1.0 × 10 ^-18 = ½ × 9.1 × 10^-31 × V²_max

    V²_max = (1.0 × 10 ^-18 / 9.1 × 10^-31)^1/2 = 1.5 × 10⁶ m/s (max. velocity of emitted  electrons).

1. b) Φ = h f _co and f _o = Φ / λ, Φ = 2 × 1.6 × 10^-19 f_o = (2 × 1.6 × 10^-19) / (6.626 × 10^-34) = 4.8 × 10¹⁴ Hz ( threshold frequency of

the emitted electrons)

Applications of photoelectric effect

Photo-emissive cells– they are made up of two electrodes enclosed in a glass bulb (evacuated or containing inert gas at low temperature). The cathode is a curved metal plate while the anode is normally a single metal rod)

They are used mostly in controlling lifts (doors) and reproducing the sound track in a film. Photoconductive cells – some semi-conductors such as cadmium sulphide (cds) reduces their resistance when light is shone at them (photo resistors). Other devices such as photo-diodes and photo-transistors block current when the intensity of light increases.

Photo-conductive cells

Photo-conductive cells are also known as light dependent resistors (LDR) and are used in alarm circuits i.e. fire alarms, and also in cameras as exposure meters.

Photo-voltaic cell– this cell generates an e.m.f using light and consists of a copper disc oxidized on one surface and a very thin film of gold is deposited over the exposed surfaces (this thin film allows light). The current increases with light intensity.