Biology 30
Genetics Exam Review
Omit Part A: Cell Division
1. Define the following terms:
a. Chromatin – loosely packed DNA during interphase
b. Chromosome – DNA that is coiled and packed into strands during replication
c. Gene – a section of DNA that codes for a specific protein
d. Dyad – two strands of chromosomes joined by a centromere
e. Tetrad – four strands of DNA (2 dyads) found in meiosis
f. Centromere – structure that holds sister chromatids together
g. Monosomy – condition where an individual only inherits one chromosome from a
pair
h. Trisomy – condition where an individual inherits three chromosomes instead of one
pair
i. Homologous pair – two chromosomes that consist of the same kinds of information
j. Sister Chromatids – two identical copies of a specific chromosome
k. Synapsis – during prophase, the tetrads are intertwined
l. Haploid – cell (gamete) with one set of chromosomes
m. Diploid – cell with two complete sets of chromosomes (zygote)
n. Reduction division – division where the number of chromosomes is halved
o. Cytokinesis – splitting of the cytoplasm during cell division
p. Crossing over – two homologous chromosomes exchange DNA
2. Describe briefly, what is happening at each of the following stages of cell
division.
Interphase – Cell processes occur, DNA replication, protein synthesis
Prophase – Nuclear membrane disappears, nucleolus disappears, chromosomes
are formed, cell is preparing to divide
Metaphase – Chromosomes line up along the equator, preparing to split evenly to
opposite poles
Anaphase – Chromosomes pull back to the opposite poles of the cell
Telophase – Nucleolus reappears, nuclear membrane reforms, chromosomes
unravel, cell cytoplasm splits into two.
3. Compare and contrast mitosis and meiosis using the following table:
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Initial number of chromosomes
Number of chromosomal pairs
Number chromosomes after first division
Number of dyads after first division
Tetrads present?
Crossing over?
Number of divisions
Number of chromosomes after second division
Stage when nucleus / nucleolus reforms?
Daughter cells haploid or diploid?
Mitosis
4
2
4
0
no
no
1
n/a
telophase
diploid
Meiosis
4
2
2
2
yes
yes
2
2
Telophase II
haploid
4. Describe what a karyotype chart is and what it shows.
A karyotype is a chart that organizes the chromosomes into pairs from largest to
smallest, with the sex chromosomes shown last. It is used to identify genetic
anomalies such as non-dysjunctions, chromosomal insertions and deletions.
5. Describe what non-dysjunction is. Give 3 examples of disorders resulting from
non-dysjunctions.
Non-dysjunction is the unequal separation of chromosomes during anaphase I or
anaphase II in meiosis. It could also occur in mitosis. The resulting daughter cells
have either one extra or one missing chromosome. This results in some characteristic
genetic disorders.
Down’s syndrome – trisomy of chromosome 21
Turners syndrome – monosomy of the X chromosome
Kleinfeldter’s syndrome – trisomy with XXY
Part B: DNA and Protein Synthesis
1. Describe the function /structure of the following:
a. Nitrogen bases – single (T,C) or double ring bases (A,G) that carry the genetic code
b. Backbone – consists of deoxyribose sugars connected to the nitrogen bases and
phosphates
c. Hydrogen bonding – holds the DNA bases together with double or triple bonds, also
responsible for the spiral
d. Codons – 3 bases of mRNA that code for one amino acid
e. Anti-codons – 3 bases of tRNA that are attached to one amino acid
f. Ribosomes – where proteins synthesis occurs, the amino acids are strung together
here
g. Transcription – a complementary mRNA strand is formed form the DNA and sent
to the ribosomes in the cytoplasm
h. Translation – tRNA reads the mRNA and drops off the amino acids in the correct
order to create a protein.
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2. Compare and contrast DNA and RNA in the table below:
DNA
RNA
deoxyribose
Ribose
Type of sugar
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1
Number of Strands
nucleus
cytoplasm
Where found
T,A,C,G
U,A,C,G
Nitrogen Bases
3. For the following strand of DNA, answer the following questions:
TACAGGCTACATTAACCCGAGTATTTA
a. What is the complementary strand?
ATGTCCGATGTAATTGGGCTCATAAAT
b. Which enzymes are required for replication?
Helicase,DNA polymerase, proofreading enzymes,ligase
c. Give the mRNA sequence to the original strand
AUGUCCGAUGUAAUUGGGCUCAUAAAU
d. How many amino acids will be in this protein? What is the exact sequence?
9 meth – ser – asp – val – iso – gly – leu – iso - asp
e. What will the tRNA units look like?
UAC AGG CUA CAU UAA CCC GAG UAU UUA
meth ser
asp
val
iso gly leu iso
asp
4. Draw a flowchart showing the process in protein synthesis.
DNA-- transcription - mRNA - Ribosomes- translation - tRNA - amino
acids - protein
5. What is meant by “one gene one protein”?
Each gene codes for a different protein. A number of proteins together often produce a
particular trait
6. Draw a flowchart illustrating the process of gene cloning.
Donor gene - cut with restriction enzymes and remove - Host DNA  cut with
restriction enzyme to open - introduce donor gene into host- supply DNA ligase -
gene is inserted, foreign protein can be made
7. Explain what restriction enzymes are and why they are important.
Restriction enzymes cut at palindromes and leave sticky ends behind. They are important
because they recognize only specific sites and allow us to remove specific genes. The sticky
ends are important because we can splice in other pieces of DNA that was cut by the same
restriction enzyme. The sticky ends match.
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8. What is genetic fingerprinting?
Genetic fingerprinting involves cutting the DNA into a series of fragments and identifying
those fragments on a gel electrophoresis. Each individual has a specific DNA barcode or
signature. This can be used to prove identity of individuals and to prove paternity.
9. What is DNA amplification and why is it used?
DNA amplification is the process where DNA is replicated for therapeutic, cloning or
investigative purposes. Enzymes and heat are added in alteration to produce a polymerase
chain reaction which can make many copies of DNA in a short period of time.
10. Describe 4 types of genetic mutations:
1. Translocation – a gene is removed from one part of the chromosome and re-located at
another position
2. Deletion – a single or multiple bases are removed, may cause a frame shift in the way the
DNA is read
3. Addition – extra bases may be inserted in the DNA strand, may cause a frame shift
4. Inversion – a gene is removed and re-inserted backwards, this does not change the
number of bases, but changes the order
Part C: Mendelian Genetics
1. Define / describe each of the following terms:
a. Gene – functional unit of DNA that codes for a protein that produces a trait
b. Allele – a specific form of one gene
c. Dominant – an allele that is expressed over other alleles present
d. Recessive – an allele that is only expressed in the homozygous state
e. Co-dominant – two alleles that are expressed together
f. Homozygous – two alleles are identical
g. Heterozygous – two alleles are different
h. Genotype – the combination of alleles / genes an individual posesses
i. Phenotype – what an individual looks like
j. Di-hybrid – an individual who is heterozygous for two alleles
k. Sex-linked – a gene found on the X or Y chromosomes
l. Autosomal - a gene found on a chromosome other than X or Y
2. For Mexican hairless dogs, the hairless condition is dominant to hairy. A litter of
8 pups is found to consist of 6, which are hairless, and two, which are hairy.
What are the genotypes of the parents?
Aa x Aa
3. Palomino horses are known to be caused by the interaction of two different genes.
The Cr in the homozygous condition produces a chestnut (red) horse while the Cm
in the homozygous condition produces a cream color called cremello.
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Heterozygous horses are palomino (reddish bodies with cream manes and tales).
What would the expected ratios in the F1 generation be if a cremello horse was
mated with a palomino? What if a palomino was mated with a palomino?
a) ½ palomino, ½ creamello b) 1 red: 2 palimino: 1 creamello
4. In blood they alleles for types A and B are co-dominant, but both are dominant
over the type O allele. The Rh factor is separate from the ABO blood group and
RH+ is dominant to Rh-. What are the possible phenotypes of the offspring in the
mating between a woman who is blood type AB and heterozygous Rh+ with a
man who is heterozygous for type A blood and is also heterozygous for the Rh
factor?
A+, AB+, AB-, A-, AB-
5. Colorblindness is a sex-linked recessive trait. What would the phenotypes and
genotypes of the offspring be if a colorblind man mates with a heterozygous
normal sighted woman?
Phenotypes: 1 normal female: 1 colorblind female: 1 normal male: 1 colorblind male
Genotypes: 1 XAXa: 1XAy: 1XaXa: 1XaY (A= normal, a =colorblind)
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6. Identify the type of inheritance in the pedigree chart below. Identify the
genotypes for each individual listed.
Autosomal Recessive
7. Use the following chart of crossover frequencies to plot a gene map.
Gene combination
Recombination Frequencies
VF/B VF/C B/C B/S VF/S C/S
2.5% 3.0% 5.5% 5.5% 8.0% 11.0%
C-----3----VF---2.5---B----------5.5-----------------S
8. Identify the following processes that involved biotechnology and genetics. What
is each used for? How is each done?
Recombinant DNA – DNA that comes form two different individuals. This allows
one individual to make new proteins it couldn’t make before
Cloning – is the process of replicating specific genes or replicating specific
individuals
Gene Therapy – is the process where an individuals DNA is altered for therapeutic
purposes. A defective gene is replaced with a functional gene. The gene is often
introduced by viral or bacterial vectors
Ames Test – is a test used to determine whether a substance is likely to be
carcinogenic (cancer causing). Bacteria that require histidine as a nutrient are
exposed to a potentially carcinogenic substance, then grown in the absence of
histidine. If they survive, a mutation has occurred allowing them to produce histidine
and the substance is likely mutagenic and possible carcinogenic.
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