Share this:

 12. Genetics Questions

1.  A woman with blood group A gave birth to twins both having blood group AB.

 Determine the genotype of:

 a) Father

 b) Mother

2.  50 black mice and 50 white mice were released into an area inhabited by a pair of owls. After four

 months, the mice in the area were recaptured and only 38 of the black mice and 9 of the white

 mice were remaining.

 a) How would this observation be explained ?

 b) Name the theory of evolution that supports the results in (a) above.

3.  State three mechanisms that prevent self pollination in a flower that has both male and female


4.  (a) Distinguish between complete and incomplete dominance  

 (b) State two sources of variation

5.  Part of one strand of a DNA molecule was found to have the following base sequence.

  G – T – C – A – G – T

 (a) What is the sequence on m-RNA strand copied from this DNA portion?

(b) State two roles of DNA molecule.

6.  State three ways by which plants compensate for lack of ability to move from one place to


7.  A student mixed a sample of urine from a person with Benedict’s solution and heated, the colour

changed to orange.

(a) What was present in the urine sample?  

(b) What did the student conclude on the health status of the person?

(c) Which organ in the person may not be functioning properly?  

8.  Differentiate between continuous and discontinuous variations  

9.  Members of the same species of organism tend to differ due to variation. State three causes of

  variation in organisms

10.  Identify the type of gene mutations represented by the following pairs of words:-

 (i) Shirt instead of skirt

 (ii) Hopping instead of shopping  

 (iii) Eat instead of tea  

11.  A DNA stand has the following base sequence:  GCCTAGATCAC
 What is the sequence of the :  (i) Complementary DNA strand?  

(ii) M-RNA strand coped form this DNA strand

12.  The figure below represents the distribution of height of pupils in a school







(a) Name the type of variation represented by the curve

(b) Outline two possible causes of variation in height of individuals in man

13.  a) Wekesa and Wanjiku who are siblings are both normal as their parents but have a hemophilic

brother. Give the Genotype of their parents.  

 b) i)What are linked genes?

  ii) What do you understand by the phase a test cross?

14.  There are at least 205 known sex – linked recessive disorder  

 a) Name any two of them.

 b) State a reason why sex – linked recessive why traits tend to effect the male child.

 c) State why if a mother has the trait all her sons will have it

15.  The table below is a representation of a chromatide with genes along its length. It undergoes

mutation to appear as shown below:

Before mutation







After mutation







a) Name the type of chromosomal mutation represented  

b) Name one mutagenic agent

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

a) Giving a reason, name the nucleic acid to which the portion belongs  

 b) Write down the sequence of bases of a complementary DNA strand  

17.  In an experiment, plants with red flowers was crossed with plants with white flowers.  

 All the plants in the F1 generation had pink flowers.

 a) Give a reason for the appearance of pink flowers in the F1 generation

 b) If plants in F1
were selfed, state the phenotypic ratio of the F2 generation

 c) Explain;  i) Why women should drink extra milk during pregnancy  

ii) A pregnant women might want to urinate more often in late pregnancy    

18.  State the meaning of the following terms giving an example in each case:

 (a) Sex-linked genes

 (b) Multiple alleles

19.  In a certain breeding experiment, a plant species with red flowers was selfed. It produced 119

 red flowered and 41 white flowered offsprings.  

 (a) Using letter R to represent allele for the red flowers, state the genotype of the red flowered

 parent plant

 (b) Determine the phenotypic ratio of red and white flowered plants. Show your working  

20.  Give an example of a sex-linked trait in human on:  (i) Y – Chromosome  

 (ii) X – Chromosome

21.  Explain why growth of long hair on the pinnae of the ears in human occurs in males only  

22.  Explain why prophase 1 of meiosis contributes towards genetic variation in living  


23.  A pure Red flowered plant was crossed with a pure white flowered plant. All the F1 generation

 plants had pink flowers.

 (a) Give an explanation for the absence of Red and white flowered plants in the F1 generation.  (b) If the F1 generation pea plants were selfed, state the phenotypic ratio of the F2 generation


24.  (a) Name a genetic disorder due to gene mutation that affects the malpighian layer of the

skin in man.  

 (b) Give two functions of the fluid produced by sebaceous glands.  

25.  (a) Define the term “Gene mutation.”

 (b) Name the genetic disorders that result from gene mutation in human beings.

26.   (i) What are mutations

(ii) Name two mutagens

27.  A section of a DNA strand contains the following sequence CGGATAC

 (a) Write the;  (i) Complementary DNA strand  

   (ii) MRNA strand

(b) Name the site for protein synthesis in a cell  

28.  In a certain bird species, red flight feathers is controlled by gene R while white flight feathers is

 controlled by gene r. The heterozygous condition Rr results into pink flight feathers. The two

genes are also sex linked and transmitted on x-chromosome.

 a) By use of fusion lines, find the genotypes of across between a male with pink flight

  feathers and a female with white flight feathers

 b) Which type of dominance is illustrated here?  

 c) i)Identify the nucleic acid whose base sequence is shown below:


  ii) Give a reason for your answer in c (i) above

iii) If the nucleic acid was involved in protein synthesis, how many amino acids would be

present in the protein synthesized? Give a reason

29.  Study the genetic chart below showing the inheritance of the gene responsible for haemophilia

Image From EcoleBooks.comin a family.









Image From




a) Write the genotype of individuals A, B, F

 b) A member of this family labelled F marries a haemophiliac male. What will be the phenotypic

ratio of the offspring? Show your workings

c) Other than the condition stated above, state any other two common genetic disorders that

result from gene mutation.

30.  Haemophilia is due to a recessive gene. The gene is sex-linked and located on X chromosome.

 The chart below represents the offspring of parents who are phenotypically normal for









 (i) What are the parental genotypes?

Explain your answer in (i) above

 (ii) Work out the genotypes of the offspring  

31.  A cross between a red-flowered and a white flowered plant produced only pink –flowered

F1 plants

 (a) There was neither a red nor white –flowered F1 plants. Explain  

 (b) The F1 offspring were selfed to get F2 generation. Using appropriate letter symbols, work

  out the genotypes of F2 generation

(c) Give the genotypic and phenotypic ratios of F2 generation  

(d) Distinguish between dominant and recessive genes

32.  A true-breeding purple maize variety was cross-pollinated with a true-breeding yellow

maize variety.

 The offspring produced all purple fruits.  

 The plants grown from these F1 grains were interbred among each other.

 A typical cob of F2 generation is shown below:

Image From  The yellow fruits are shaded while the purple ones are un-shaded.





(a) (i) In terms of flowers only, state why it is easier to work out genetic crossings using maize

  (ii) Count separately the yellow and purple grains and therefore find the rations of purple

grains to yellow grains

 (b) Using appropriate symbol, work out a genetic cross for F2 generation

 (c) From the above information, give the dominant gene

 (d) State two practical applications of genetics in identity determination  

33.  The figure below is a pedigree chart showing incidence of albinism which is transmitted through a

 recessive gene-a. Study and answer questions that follow;













 (a) Write down the genotype of persons 1 and 2. Give a reason for your answer

 (b) Giving your reason state the most likely genotype of person 3  

 (c) The cross between person 15 and 16 represents mating between first cousins. Comment

  why it is not advisable for close relatives to marry

 (d) Apart from albinism name two other effects of gene mutation  

34.  The table below shows results of test to determine blood groups of persons Y and Z.A tick ()

 Represents, agglutination while a cross (x) represents no agglutination;


Test with

antibody (a)

Test with

antibody (b)

Test with

Rhesus antibody

Blood group

Y- (male)






X- (female)






 (a) Fill the blank space in table to show the blood group of the persons Y and Z

 (b) In order to investigate the inheritance of Rhesus factor, work out a cross between a male

with Rh+
and female with Rh .Let D represent the presence of Rhesus factor and d to

represent the absence of the Rhesus factor

 (c) Determine the genotype of the cross in (b) above.  

 (d)Which of the children can donate blood to their mother?

35.  Describe the behavioural adaptations of animals to temperature  

36.  In man blood group inheritance is controlled by multiple alleles in which allele A is co dominant

 to allele B. a woman laterozygous for blood group A married a man heterozygous for blood

 group B

 a) State the genotype of both parents

 b) Using a pun net square, show the genotypes of F1 generation  

 c) State one application of knowledge of blood group inheritance in man

 d) The nitrogenous bases in nucleic acids are Adenine (A), cytosine(C), Guanine (G),

Thiamine (T) and uracil (U). Input of a molecule of DNA the sequence of bases is CTT.

Using the letters A, C, G, T, U where appropriate, write down the base sequence in;

i) Corresponding part of the complementary strand of DNA molecules

ii) Corresponding part in mRNA

iii) A change in the DNA molecules caused the base sequence in the triplets to change from

CTT to CAT. State one factor which could have caused the change

37.  In an investigation plants with red flowers were crossed with plants with white flowers. All the

plants in the F1 generation had pink flowers when the F1 plants were crossed, he counted 480

 plants in F2 generation

 (a) Using appropriate letter symbols, work out the cross between the F1 plants to get the F2


 (b) Give the phenotypic and genotypic ratios for the F2 generation  

  Phenotypic ratio

Genotypic ratio

 (c) How many plants in the F2 generation had pink flowers? (show your work)

38.  In an experiment, a black mouse was mated with a brown mouse. All the off springs in F1

generation were black. The off springs grew and were allowed to mate with one another. The total

number of F2 generation offspring were 96.

 (a) Using letter B to denote the gene for black colour. Work out the genotype of the F1 generation.

(Use a punnet square)  

 (b) State the following for the F2 generation

  (i) Genotypic ratio  

(ii) Phenotypic ratio  

  (iii) The total number of brown mice


39.  (a) Distinguish between Homologous structures and analogous structures. Give an example

in each case.  

Homologous structures


Analogous structures


 (b) Explain why parasites develop resistance to certain drugs after a long time of exposure.

(c) (i) What is non— disjunction?

(ii) Give one example of a genetic disorder associated with non-disjunction  .  



 12. Genetics Answers

1.  a) BB;

 b) AA;


2.  a) Black mice are better adapted camouflage with the environment hence less are eaten by the

owls compared to the white mice which are easily seen;

 b) Theory of natural selection;


3.  – Heterostly – stigma located above anthers;

 – Self sterility or incompatibility – pollen grain from the same plant do not germinate

 – Protandry – Male parts mature before female parts;

 – Protogyny – Female parts mature before male parts;


4.  (a) Complete dominance is when an alliele completely surprises another intermediate fruits;

Incomplete dominance is when heterozygous organisms show an intermediate trait;  

(b) Genetic recombination’s of alleles reading to variations; Independent assortment of


Random fusion of gametes; mutations;

Environment (may either enhance or suppress expression of a gene);


5.  (a) C-A – G – U – C _ A

 (b) – Stones genetic information (in a coded form);

– enables transfer of genetic information unchanged to daughter cells through replication);

– Translates genetic information into characteristic of an organism 9thorugh protein synthesis);


6.  Ability to pollinate; response to stimuli (tactic) nastic or tropics); Ability to exploit localized nutrients an ability to photosynthesize; Ability to disperse seeds/fruits, propagation;


7.  (a) Glucose;

 (b) The person was a sufferer of diabetes mellitus;

 (c) Pancrease;


8.  Continuous variation shows gradation in characteristic with intermediate; discontinuous shows distinct characteristics between organisms with no intermediate groupings;


9.  -mutation;

  -intermixing of genes already in the population through sexual reproduction recombination;

-crossing over during prophase of meosis I

-interdependent assortment of chromosomes, during metaphase of meosis I


10.  i) Substitution;  

 ii) Deletion;

  iii) Inversion;


11.  i) C G G A T C T A G T G;

 ii) C G G A U C U A G U G;


12.  a) Continuous  

 b) Nutrition/ environment; genes;


13.  a) Father XHY  ;  

  Mother XHXh  

 b)  i) Genes found in the same chromosome and usually transmitted together;

ii) Across to determine an unknown genotype involving use of a recessive parent;


14.  a) Colour blindness; haemophilia;

  Sickle cell anaemia;

 b) Part of X chromosome has homologous portion on the Y chromosome therefore if the X has

the recessive trait, it will show on the male phenotype

 c) The son inherits the X chromosome from the mother while the daughter inherits the

X chromosome from the father;


15.  (a) Inversion  ;

  • ustard gas;
  • ionizing radiation;
  • gamma rays;
  • X- rays ;


16.  (a) Ribonuclei acid /RNA

  – Because it has uracil / presence of uracil;



 / G C A G



17.  (a) Due to co-dominance /partial dominance/incomplete dominance/(Acc. equal dominance)

 (b) Red: 2Pink : white – 1: 2:1 (Acc. 1RR: 2RW: 1WW) mark as a whole;

 (c) Why women should drink extra milk;

(i) Bore formation for infants  ;

(ii) pressure on bladder by the enlarging uterus;


18.  a) Genes which are located on the sex- chromosomes and therefore are transmitted along with


Example Haemophilia; colour blindness;

 b) Where more than two genes control a particular characteristic/ trait;

Example ABO blood group system;


19.  a) Parental Genotype Rr, Rr  ;

 b) Red: white;

119/41; 41/41;

2.90: 1

3: 1;


20.  (i) Y – Chromosome-hairy pinna, pre-mature boldness;  ; (any one)  

 (ii) X – Chromosome- haemophilia (bleeders disease); colour blindness; (any one)


21.  The Gene that determine the growth of long hair on pinna is sex linked and an Y-chromosomes; V hence can only be inherited by males as a single gene and it expresses itself out phenotypically


22. Due to crossing over: that results in exchange of genetic materials between homologous







23. (a) Co dominance/ incomplete dominance:

(b) 1 Red flowered; 2 pink flowered; I white flowered: for ratio for phenotype)


24.  (a) Albinism;  

 (b) Makes skin supple;

– Kills bacteria/ a mild antiseptic;


25.  – Change in base sequence of the DNA;   1.


26.  (i) Sudden and spontaneous change in structure of chromosome and DNA which is inherited

(ii) Chemical ionizing radiations, Uv light, extreme temperature or some virus  


27.  (a) GCCTATG – DNA  


(b) Ribosome;


28.  a) Parental phenotype Pink flight feathers  X  White flight feathers

  Parental genotype XRxr X   XrY


  Parental gametes





  F1 genotypes

 b) incomplete;

 c)  i) Ribonucleic acid;

ii) has uracil base;

ii) – 3;


29.  A – XhY;

 B – XHY;

 F – XHXh


XH Xh X XhY;

Image From








(c) Albinism; sickle cell anaemia; colour blindness; chondrodystrophic dwarfism;


30.  (i) Father  Mother


Since father cannot have the recessive gene ad fail to be affected. The mother must be a carrier

on her second X chromosome for a male son to be haemophiliac.

 (ii) Parental phenotypes mother carrier, father normal

 Parental genotypes XHXh  XHY










31.  a) the two genes that control flower colour ,that is the gene for red flowers and the

one for white are codominate;  b) F₁ phenotype pink flowers pink flowers

 F₁ genotype RW X RW

  Gameter R W R W


 F₂ genotypes RR RW RW WW

  F₂ phenotypes red

Flowered pink white flowered


c)genotypic ratio= 1RR:2RW:1WW/RR:RW:WW=1:2:1

 Phenotypic ratio=1 red flowered:2 pink flowered:1white flowered

 Notes: i) there must be cross on genotype

ii) gameter should be circled

d) recessive gene expressed it self only underlined homozygous condition while dominant gene expresses it self in both homozygous and heterozygous conditions;


32.  (a) (i) Male and female flowers are separate hence cross pollination is made possible.  

 (ii) 1 Yellow : 3 Purple

Rej.: 15 yellow : 45 Purple





Image From





Image From EcoleBooks.comImage From

(b) Let letter T represents purple maize grain

Image From   Let letter t represent yellow maize grain

Image From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From











(c) Gene for purple grain;

 (d) (i) Finger prints are used to identify criminals;

  (ii) Blood groups are used to settle parental disputes;


33.  a) Aa; Aa; because of one child – 4

 b) AA; Aa; because of cross between parent 1 and 2

 c) Lethal genes easily inherited;

 d) Sickle celled anaemia; colour blindness; haemophilia


34.  (a) Y – Blood group A+

  Z – Blood group B


Parental phenotypes

Rhesus (+ve) Rhesus (-ve)

Parental phenotypes

Rh Rh X Rhd Rhd;





Parental gametes




F1 offspring;

 c) All Rhesus positive/ all RhDRhd

 (d) None


35.  – By keeping their mouth open/panting; to lose heat over surface area of the tongue by evaporation;

-Basking; to gain heat by conduction;

– Shivering; to generate heat through increased metabolism;

– Physical activity (e.g. running); to generate heat through metabolism;

– Hibernation; to increase metabolism;

– Putting on warm clothes when it is cold; to retain the heat energy;

– Reduction of physical activity; to reduce the metabolic rate;

– Migratory behaviour to cooler environment; to reduce the body temperature;

– Moving into water when it is hot; to cool the body;

– Staying around fire place; to gain heat by convection;

– Taking hot drinks; to warm the body;

36.  a) Parental genotypes  

i) Woman/ OX –  AO

ii) Man/ O – BO











c) Cases of disputed paternity settlement

  – Determining compatible blood groups in blood transfusion

 d)  i) Corresponding complementary DNA strand  GAA;

ii) Corresponding RNA  CUU

 iii) Nitrates/ sulphites/ hydroquinone/ gamma/ beta/ alpha/ x-rays/ UV light/ hydrogen peroxide

37.  let R rep. gene for Red flowers

  W.rep gene for white flowers


a) parental phenotype Pink flowered Pink flowerd








red pink white

b) phenotypic ratio 1Red:2Pink:1White;

  genotypic ratio: 1RR:2RW:1WW;

c) 2 x 480;=240















Fl generation Award  for punnet Square and genotypes

(b) (i) IBB : 2Bb: lbb  

(1 mark for ratio, 1 mark Par genotype)


(ii) 3 B lack: I brown  

(iii) 24;

39.  (a) Homologous structures:

Structures of common embryonic origin modified to perform different functions;  

Example: Eye structure in man and octopus/ wings in birds and insects (I mark)

Analogous structures Example

(b) They undergo mutations: resulting in new forms that rcsis selection resistant to drugs;

(c ) (i) Failure of chromosomes to separate during anaphase I resulting in gametes

with an extra chromosome and others with less chromosomes: (I mark)

(ii) Downs syndrome / Klinefelters syndrome/ Turners syndrome: any I ( 1mk)

40.  a) Homologous structures have a common embryonic origin but are modified to  

Perform different functions; while analogous structures have different embryonic origin but are modified to perform similar functions;

 b) Nictitating membrane; post anal tail; body hair;

41.  a) Pentadactyl limb structure of mammals; beaks of birds; feet of birds;  

b) – Missing links between fossils because some parts or whole organisms were not fossilized

  – Some parts were distorted during fossilization hence may give wrong impression

of structures;

– Some structures have been destructed by geological activities;

42.  Camouflage is the conceal/ element of identity of an organism by resembling the color  

of the environment while mimicry is the imitation of non- living organisms to conceal identity

43.  Light energy splits water molecules; into hydrogen ions and oxygen atoms;

44.  (a) Caecum/ Rumen/ pauch;

 (b) Closes to prevent food from moving up the oesophagus;

45.  (a) – the soft bodied organisms fail to fossilize;

– Human activities interfere with fossilization;

Earth movements e.g. volcanic eruptions interfere with fossilization; (mark any first2 pts  

(b) – They resembled from neck downwards;

-They walked upright;

– The shape of the skull suggested they were able to speak;

46.  a i)vestigial structures are those structures that have ceased to be functional over along

period of time and hence reduced in sizes



  -coccyx or tail/tail bone;

  – Nictitating membrane/semi – lunar fold at the corner of the eye;

  -ear muscles

  – Body hair;

 b) Disease causing organism mutates; and became resistant;

47.  Struggle for existence –environmental pressure on the population in order to survive;  

Survival for the fittest-advantageous variations an individual possesses to make it survive;

48.   Secretion of antidiuretic hormone; rearbsorption of salts at the loop of Henle;

49.  -Divergent evolution refers to a situation where by organisms that are believed to have

had a common ancestral origin have homologous structures which have been modified to suit different environments;

50.  a) Allows survival of organisms with better qualities / traits / characteristics; eliminates

organisms with unfavorable characteristics/ traits;

 b) Divergent;

51.  Evidence does not support Larmarks theory

Acquired characteristics are not inherited/;

Inherited characteristics are found in reproductive cells

52.  (a) Vestigial structures

  (i) Are those structures that have ceased to be functional over a long period of time hence

reduced in size;

  (ii) Appendix/coccyx/tail/ nictitating membrane semilun fold at the corner of the

eye/caecum/ear muscles, body hairs;

 (b) Disease causing micro-organisms mutate and become resistant;

53.  a) The gradual emergence of complex life forms from pre-existing simple forms over along

period of time  ;  

 b) Nature selects those organisms with structures that are well adapted to survival in

the environment. These structures are passed to their offspring; organisms with structures

that are poorly adapted perish

54.  The insecticide kills most of the insects when introduced; those that survive; give rise to a new

generation of flies that are resistant to insecticide.

55.  – Most organisms especially soft-bodied ones do not form fossils;  

  – Most fossils have not yet been discovered;

  – Exposed fossils are usually destroyed by physical and chemical weathering;

– Earth movements e.g. volcanicity, earthquakes, tsunami do destroy fossils;

– Most animals are prayed upon;

56.  -Fossil records/paleontology  ;

-Comparative anatomy/taxonomy;

-Comparative embryology;

-Geographical distribution;

-Cell biology;

-Comparative cellulogy/immunology; (award 1st three 3mks)


Share this:

EcoleBooks | 12. Genetics Questions And Answers


Leave a Reply

Your email address will not be published. Required fields are marked *

Accept Our Privacy Terms.*