Zool 221: Animal Genetics And Evolution Question Paper

Zool 221: Animal Genetics And Evolution 

Course:Bachelor Of Education Science

Institution: Kabarak University question papers

Exam Year:2010



KABARAK UNIVERSITY
UNIVERSITY EXAMINATIONS
2009/2010 ACADEMIC YEAR
FOR THE DEGREE OF BACHELOR OF EDUCATION SCIENCE
COURSE CODE: ZOOL 221
COURSE TITLE: ANIMAL GENETICS AND EVOLUTION
STREAM: SESSION IV
DAY: THURSDAY
TIME: 2.00 – 4.00 P.M.
DATE: 12/08/2010


INSTRUCTIONS:
1. Answer ALL Questions in Section A.
2. Answer TWO Questions only in Section B
Section A: Answer ALL Questions (40 Marks)
1. a) Explain meaning of the following terms: (6 mks)
i) Biological evolution
ii) Barrh bodies
iii) Epistasis
b) State the following: (4 mks)
i) Mendel’s Law of Dominance
ii) Mendel’s Law of Segregation

2. Use Figure 1 to answer the following question;
a) Explain how you would distinguish identical and fraternal twins in a family tree.
(4 mks)
b) In cats, yellow and black coat colours are determined by b and B alleles, respectively.
Heterozygous condition gives brown coat, and genes are sex-linked. State the
chances of getting a brown-coat male from a cross of black male and brown female?
(6 mks)
3. a) Explain natural selection as driver of evolution. (6 mks)
b) Describe the process of speciation. (6 mks)

4. Write explanatory notes on:
a) ZW sex determination system. (4 mks)
b) Cytoplasmic inheritance (4 mks)

Section B: Answer TWO Questions Only (30 Marks)
5. Discuss the evidence of organic evolution. (15 mks)
6. Discuss the evolutionary lineage of human. (15 mks)
7. a) Albinism in humans is inherited as a simple recessive trait. For the following
families, determine the genotypes of the parents and offspring.
i) A normal male and albino female have six children, all normal.

ii) A normal male and albino female have six children, three normal and three
albinos. (8 mks)
b) The table below shows results of Mendel’s monohybrid cross of two sheep. Calculate
X
2
value, hence test the hypothesis that the observed variations is largely due to chance.
Let P=0.05 Observed Expected
With wool 882 885
Without wool 299 295 (7 mks)


Figure 1: Human pedigree of night






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