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Principle S Of Genetics Question Paper
Principle S Of Genetics
Course:Bachelor Of Environmental Science
Institution: Pwani University College question papers
Exam Year:2012
PWANI UNIVERSITY COLLEGE
A CONSTITUENT COLLEGE OF KENYATTA UNIVERSITY
UNIVERSITY EXAMINATIONS 2008/2009 ACADEMIC YEAR
2ND
YEAR
1ST SEMESTER EXAMINATION FOR THE DEGREE
OF STREAM:BACHELOR OF ENVIRONMENTAL SCIENCE & BACHELOR
OF AGRICULTURE & ENTERPRISE DEVELOPMENT
AST 201:
PRINCIPLES
OF GENETICS
ENS 234:
ENVIRONMENTAL GENETICS
END SEMESTER
:
I
TIME: 3
HOURS
INSTRUCTIONS
1.Answer ALL
questions in Section
A
and ANY
TWO
from section B
2.Marks for each question are indicated in brackets
3.Electronic calculators can be used and NOT mobile phones.
SECTION A
-
(40 MARKS)
Q.1)
a)What is the significance of mitosis?
(3 mks)
b)Give the differences in Mechanism and in the results between mitosis and
meiosis.
(5 mks)
c)
Explain the chromosomal behaviour during the following stages of cell
division;
i)
Mitotic anaphase
(2 mks)
ii) Zygotene
(2 mks)
iii)Telophase I(2 mks)
d)Name two ways in which meiosis increases genetic variability (2 mks)
Q.2)
a)State the law of independent assortment
(1 mk)
b)
Explain the following terms used in genetics;
i)Linkage group
(1 mk)
ii)Multiple allelism
(1 mk)
iii)
Epistasis
(1 mk)
iv)Non-disjunction
(1 mk)
2
v)
Test cross
(1 mk)
b)
Flower colour in sweet pea plants is determined by two allelomorphic pairs
of genes (R,r and S, s). If at least one dominan
t gene from each
allelomorp
hic pair is present, the
flowers are purple. All other
genotypes are white. If two purple flowered plants, each having the
genotype Rr Ss, are crossed, what will be the phenotypic ratio of the
offsprings?
(4 mks)
Q.3)
A homozygous purple flowered short
-
stemm
ed plant was crossed with a
homozygous red
-
flowered log stemmed plant and the F
1
phenotypes had purple
flowers and short stems. When the F
1
generation was test crossed with a
double homozygous recessive plant the following progeny were produced; 52
purple flower, short stem; 47 purple flower long stem; 49 red flowers, short stem
and 45 red flower, long stem. Explain these results fully and show your working
.
(6 mks)
Q.4)
In maize the genes for coloured seed and full seed are dominant to
the genes for
colourless seed and shrunken seed. Pure breeding strains of the double
dominant variety were crossed with the double recessive variety and a test cross
of the F1 generation produced the following results.
Coloured, full seed
380
Colourless, Shrunken seed
396
Coloured, shrunken seed
14
Colourless, full seed
10
Calculate the distance in map units between the genes for coloured seed an
seed shape on the chromosomes.
(4 mks)
Q.5)
a)Differentiate reverse mutation from suppressor multation
(2 mks)
b)Write out the base sequence of the
m
RNA
formed from a DNA strand
with the following
sequence.
A T G T T C G A G T A C C A T G T A A C G
(2 mks)
SECTION B
-
(30 MARKS)
-
ANSWER ANY TWO QUESTIONS
Q.6)
a)
In Cats, the genes controlling the coat colour are carried on the X
–
chromosome and are codominant. A black
–
coat female mated with a
ginger
–
coat male produced a litter consisting of black male and tortoise
shell female kittens. What is the expected F
2
Phenotypic ratio? Explain
your results.
(7 mks)
b)
i)
Explain using appropriate genetic symbols, the possible blood
groups of children whose parents are both heterozygous, the father
being blood group A and the mother blood group B
(6 mks)
ii)If these parents have non
-
identical twins, what is the probability
that both twins will have blood group A?
(2 mks)
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