1. Note: The discussion here is a very general overview of topics
Genetics
that MIGHT come out in the UPCAT. Your discussion of these
topics in Bio3 will definitely be more formal and more Molecular Dogma
detailed. Also, the more complicated problems here will most Chemical Biology
probably NOT come out in the UPCAT, but the basic topics
from which they stem might. Cell Division

3. G1
What determines the sex of an unborn child?
The chromosome content of the sperm
The chromosome content of the ovum
Diet of the father before sexual intercourse
The number of days between ovulation and fertilization

4. • 2 varieties of sex chromosomes in humans: X & Y
– XX = female, XY = male
• The two sex chromosomes segregate during meiosis, and each
gamete receives one
– In the ovaries: In the testes:
X X
X X
X Y
X Y
• At the moment of conception:
X X
X X X Y
X Y
Female Male

5. G2
In pea plants, spherical seeds (S) are dominant to
dented seeds (s). In a genetic cross of two plants that
are heterozygous for the seed shape, what fraction of
the offspring would be spherical seeds?
1/4
1/2
3/4
1

6. In pea plants, spherical seeds (S) are dominant to
dented seeds (s). In a genetic cross of two plants that
are heterozygous for the seed shape, what fraction of
the offspring would be spherical seeds?
Alleles (alternative form) of the gene • Heterozygous: has one of
(segment of DNA) for seed shape each kind of allele (Ss)
• Dominant: trait is expressed all the • Homozygous: has two of the
time same kind of allele
o SS or Ss  spherical o Homozygous dominant: SS
• Recessive: trait is expressed only if the o Homozygous recessive: ss
organism has 2 copies of the allele
o ss  dented

7. In pea plants, spherical seeds (S) are dominant to
dented seeds (s). In a genetic cross of two plants that
are heterozygous for the seed shape, what fraction of
the offspring would be spherical seeds?
the parents both have the genotype (genetic makeup) Ss and
are both spherical (the phenotype, or physical and observable
expression of the genotype)
during meiosis, the alleles for seed shape are segregated as
follows, for both male and female parents:
S
S
s
s

8. In pea plants, spherical seeds (S) are dominant to
dented seeds (s). In a genetic cross of two plants that
are heterozygous for the seed shape, what fraction of
the offspring would be spherical seeds?
• During fertilization, the possible combinations of gametes
displayed in the table below, called a Punnett square:
S s Gametes from parent 1
S SS Ss
Possible genotypes of offspring
s Ss ss
Gametes from parent 2

9. In pea plants, spherical seeds (S) are dominant to
dented seeds (s). In a genetic cross of two plants that
are heterozygous for the seed shape, what fraction of
the offspring would be spherical seeds?
• The resulting ratios of genotypes and phenotypes are:
Genotype Ratio Phenotype Ratio
SS 1/4
Spherical seed (SS or Ss) 3/4
Ss 2/4
ss 1/4 Dented seed 1/4
4/4 4/4
• Take note that these are always the resulting ratios of a
monohybrid cross
mono: dealing w/ a single trait
hybrid: parents are heterozygous

10. G3
A man with type A blood marries a woman with type
B blood. Their child has type O blood. The genotypes
of the father and the mother, respectively, are
AB; BB
AA; BB
AO; BO
AO; BB

11. • The ABO blood type is controlled by a single gene with three
alleles: IA, IB, and i*. IA and IB are codominant; IA and IB are
both dominant over i.
• The following are the genotypes of the different ABO blood
group phenotypes:
Phenotype Genotype
Type A IAIA or IAi
Type B IBIB or IBi
Type AB IAIB
Type O ii
* For simplicity, IA is sometimes represented as A, IB as B, and i
as O

12. A man with type A blood marries a woman with type
B blood. Their child has type O blood. The genotypes
of the father and the mother, respectively, are…
Individual Phenotype Genotype
Father Type A IAIA or IAi
Mother Type B IBIB or IBi
Child Type O ii
You can construct a Punnett square using the
different combinations of parental genotypes to see
which combination will yield a type O child.

13. • (Additional info) Immunological aspects of the ABO blood
group:
Recipient phenotype Donor phenotype(s)
Type A A or O
• has A antigens
• produces antibodies vs. B antigens
Type B B or O
• has B antigens
• produces antibodies vs. A antigens
Type AB (universal recipient) A, B, AB, or O
• has both A and B antigens
• does not produce antibodies to either A or B
antigens (because they have both)
Type O (universal donor) O
• does not have either A or B antigens;
• produces antibodies vs. both A and B antigens

14. G4
The diagram below shows the family tree of a family with
phenylketonuria (PKU). PKU is a disease that is expressed in
homozygous recessive individuals.
Which of the following correctly describes the genotype of
individuals in the family tree?
A. P and Q are heterozygous
B. P and Q are homozygous dominant
C. R and S are homozygous dominant
D. R is homozygous dominant, S is heterozygous dominant

15. • A pedigree is a family tree describing the interrelationships
of parents and children across the generations
• Legend:
–  is a ;  is a
–   indicates a mating, with offspring listed below in
their order of birth
–  shaded symbols stand for individuals with the trait
being traced

16. pp
pp pp
pp
If we let P = dominant allele of the gene associated with the occurrence of PKU, and
p = recessive allele, this means that all shaded symbols have a genotype of pp.
In determining the unknown genotypes in a pedigree, it is usually helpful to focus
first on matings that produce homozygous recessive offspring, since this can narrow
down your choices for the parents’ genotypes.

17. pp
pp pp
pp
Let’s focus first on the mating of R and S. Since they are both unaffected, and they
have an affected offspring (which means one or both of them must carry but not
express the recessive allele) their possible genotypes are P_ x P_. By constructing
Punnett squares of the possible genotypes of the parents, we can get these results:
Parents Offspring
PP x Pp 2 PP: 2 Pp
These results indicate that R and S
Pp x Pp 1 PP: 2 Pp: 1 pp
are both heterozygous.

18. pp
pp pp Pp Pp
pp
Next, let’s focus on individual P (unknown genotype) and its mate (genotype pp).
Individual P’s possible genotypes are PP or Pp (can you figure out why?). Again,
constructing Punnett squares for the possible mating combinations would yield the
following results:
Parents Offspring At this point, you can already
PP x pp All Pp answer the question, though I
encourage you to try to figure out
Pp x pp 2 Pp: 2 pp the genotype of individual Q.

20. MD1
The diagram below shows the steps in protein synthesis.
If the DNA sequence used to produce X is
5’-ATGTGGAAT-3’, what will be the sequence of X?
3'-UACACCUUA-5'
3'-TACACCTTA-5'
5'-UACACCUUA-3'
5'-TACACCTTA-3'
DNA X Protein

21. • The flow of genetic information, summarized in the diagram
below, is usually referred to as the central dogma of
molecular biology.
Notice that this diagram is
very much similar to the
diagram in the question.

22. • DNA (a polymer) carries genetic information via the sequence of its
nucleotides (its monomers).
• Nucleotides are made of
building blocks themselves: Nucleoside
a nitrogenous base,
Nitrogenous
a pentose (5-carbon) sugar, base
and a phosphate group.
• Nitrogenous bases: O 5’C
– Adenine (A) O P O CH2
O
– Thymine (T)
O
(or uracil (U), in RNA)
Phosphate
3’C
– Cytosine (C) group Pentose
sugar
– Guanine (G)

23. • The DNA molecule consists of two nucleotide chains that spiral
around an imaginary axis, forming a double helix.
• The nitrogenous bases of one
chain are linked to those of the
other chain through hydrogen
bonds. The two chains are said
to be complementary. The
base-pairing rules are
as follows:
– adenine with thymine
(or uracil, in RNA)
– cytosine with guanine

24. • The central dogma illustrates
how information in DNA is
used to make proteins.
• The nucleotide sequence in
DNA (or simply, the DNA
sequence) is used as a basis to
make an intermediate
molecule, RNA. This is the
process of transcription.
• The RNA sequence is then
used to come up with a
sequence of amino acids,
which eventually make up a
protein. This is the process of
translation.

25. • Going back to the question: If the DNA sequence used to produce X
is 5’-ATGTGGAAT-3’, what will be the sequence of X?
• We are thus being asked to make an RNA sequence given this DNA
sequence.
• Before we start, though, we must know that when an RNA
molecule is made, it elongates from the 5’ end to the 3’ end
(5’3’). Thus its DNA template must be the strand that goes from
3’5’, which means that we need to know the complementary
strand of the DNA sequence given above.
Given strand: 5’-ATGTGGAAT-3’
Complementary strand: 3’-TACACCTTA-5’
This will be the template for the RNA
sequence we are being asked of.

26. • Going back to the question: If the DNA sequence used to produce X
is 5’-ATGTGGAAT-3’, what will be the sequence of X?
• Now that we know what the sequence of the template DNA is, we
can put together the RNA that will result during transcription:
Template strand: 3’-TACACCTTA-5’
RNA after transcription: 5’-AUGUGGAAU-3’
• Note that the resulting RNA sequence is the same as the DNA
sequence initially given to us, except that the T’s were replaced
by U’s.
• Note, also, that the correct answer is not among the choices given in the problem.
Sorry about that :P

27. MD2
Which component is NOT DIRECTLY involved in the
process known as translation?
mRNA
DNA
RNA
ribosomes

28. • Translation is the RNA-directed synthesis of a polypeptide.
This means that RNA (in particular, Proteins consist of one or more
messenger RNA (mRNA), which was polypeptides, which are chains of
produced by transcription) is the amino acids.
template for making protein.
• Recall from your lessons on cell structure and function the
functions of ribosomes.
• Given these information, you will then be able to answer
the previous question.

29. MD3
Based on the genetic code below, which amino acid sequence is
specified by the mRNA sequence UAUCGCACCUCAUAG when the
first triplet (UAU) is used as the start of the reading frame?*
Tyr-Gly-Ser-Leu
Tyr-Arg-Thr-Ser
Tyr-Arg-Thr-Asp
Tyr-Arg-Thr-Ser
* The PDF version of this reviewer
only has tripeptides as choices for A
and B. Please use the choices as
listed here. Sorry about that.

30. • The genetic code is made of
triplets of nucleotides and the DNA Gene 2
molecule
amino acids for which they Gene 1
code. Gene 3
• During translation, our cells DNA strand
3 5
(ribosomes, in particular) (template) A C C A A A C C G A G T
“read” the mRNA produced
TRANSCRIPTION
earlier by translation in sets of
three nucleotides (codons). As mRNA 5
U G G U U U G G C U C A
3
the ribosome encounters a Codon
certain codon, the TRANSLATION
corresponding amino acid is Gly
Protein Trp Phe Ser
added to the growing Amino acid
polypeptide.

31. • Given the mRNA sequence,
separated into codons:
UAU CGC ACC UCA UAG
• The first codon: UAU
– Locate the 1st nucleotide, U,
on the leftmost column of the
genetic code to the right
– Then locate the 2nd nucleotide,
A, on the top row
– Then locate the 3rd nucleotide,
U, on the rightmost column
– The amino acid found at the
intersection of those three
nucleotides is the amino acid
corresponding to that codon.

32. • Given the mRNA sequence,
separated into codons:
UAU CGC ACC UCA UAG
• If you work out the rest of the
sequence, you should come up
with: Tyr-Arg-Thr-Ser
• The stop codons – UAA, UGA, and
UAG – are chemical signals to the
cell that indicate the end of
translation

33. MD4
Based on the genetic code below, identify a possible sequence of
nucleotides in the DNA template for an mRNA coding for the
polypeptide sequence Phe-Pro-Lys.
AAA-GGG-UUU
TTC-CCC-AAG
TTT-CCA-AAA
UUU-CCC-AAA

34. • Following the previous discussion, we now work our way
back with this problem.
• First, let us list the codons that
code for the given amino acids:
Amino acid Codons
Phe UUU, UUC
Pro CCU, CCC, CCA, CCG
Lys AAA, AAG

35. Amino acid Codons
• Since we have many choices
for the middle amino acid (aa), Phe UUU, UUC
let us focus first on the 1st and Pro CCU, CCC, CCA, CCG
3rd aa’s.
Lys AAA, AAG
• A Phe-Pro-Lys tripeptide will result from an RNA sequence that
– starts with either 5’-UUU or 5’-UUC
– ends with either AAA-3’ or AAG-3’
• This means that its DNA template would have a sequence such as
- 3’-AAA or 3’-AAG
- TTT-5’ or TTC-5’

36. • With these options for the end sequences:
- 3’-AAA or 3’-AAG
- TTT-5’ or TTC-5’
We can eliminate the obviously wrong answers from the options
given to us. Notice, too, that the eliminated choices below could
have also been discarded at the start because of the uracils in
their sequence (DNA does not have uracil).
A. AAA-GGG-UUU C. TTT-CCA-AAA
B. TTC-CCC-AAG D. UUU-CCC-AAA
• Now that you’re left with just options B and C, you can now look at
the possible sequences for the 2nd aa in order to choose your final
answer. Can you figure out why the BEST answer is C?

37. MD5
Which of the following is NOT true of a codon?
It consists of three nucleotides.
It may code for the same amino acid as another codon does.
It never codes for more than one amino acid.
It is the basic unit of the genetic code.
You can easily answer this by studying the genetic code from the
previous problems.

38. MD6
Corn plants have been genetically modified to contain a gene
which can produce toxins to kill pests of the plant. Which of the
following arguments is NOT a valid reason for opposing the
widespread use of this genetically modified crop?
Beneficial insects may inadvertently be killed.
The inserted gene could cause mutation to occur on other parts of
the plant chromosome.
The inserted gene could become incorporated into the genome of
weed species growing nearby.
The modified chromosomes could cause the chromosomes in the
cell in the body of the consumer to mutate.
Statements A and C describe possible scenarios. Thus your choices are down
to B or D, both of which have little or no scientific evidence of happening (at
least not yet).

40. CB1
Some substances were isolated from a chimpanzee that directly
played roles in body structure, immunology, and oxygen
transport. What is the classification of these substances?
lipids
carbohydrates
proteins
nucleic acids
Carbohydrates and lipids are mainly sources of energy, while nucleic
acids store genetic information. Proteins are called the workhorses of
the body, since they perform various functions such as the ones
mentioned in this problem.

41. CB2
Given the table below, what are substances Y
and Z?
fats; amino acids
carbohydrates; amino acids
fats; fatty acids and glycerol
proteins; monosaccharides
Digestive enzyme Substance digested Resulting simple
molecules
Amylase Polysaccharides Disaccharides
(W) Proteins (X)
Lipase (Y) (Z)

42. • The name of the enzyme – lipase – indicates that the enzyme acts
on lipids. Thus the question is: what are the building blocks of
lipids?
• The following table lists the important molecules found in living
organisms, as well as their building blocks. Note that only
carbohydrates, proteins, and nucleic acids are considered polymers
(long molecules consisting of many similar building blocks called
monomers). Lipids consist of two types of smaller molecules.
Biomolecule Building blocks
Carbohydrates Monosaccharides
Proteins Amino acids
Nucleic acids Nucleotides
Lipids 1 glycerol + 3 fatty acids

43. CB3
A cell may use lipids to
speed up the chemical reaction.
form a barrier between the cell and its environment.
provide structural support.
hydrolyze food molecules.

44. • Let’s analyze which biomolecules are being referred to by
each option:
A. speed up the chemical reaction.
 these are enzymes, which are proteins
B. form a barrier between the cell and its environment.
 the barrier refers to the cell membrane, which
consists of a lipid bilayer
C. provide structural support.
 proteins such as collagen and elastin are known to
provide structural support in animals
D. hydrolyze food molecules.
 again, these refer to proteins

45. CB4
Which of the following best describes the difference
between the structure of carbohydrates and fats?
Carbohydrates Fats
A. Do not contain the element Contain the element nitrogen
nitrogen
B. Ratio of H atoms to O atoms is No fixed ratio of H and O
2:1 atoms
C. Made of glucose molecules Made of fatty acids and
joined together glycerol joined together
D. No fixed ration of H to O Ratio of H atoms to O atoms is
atoms 2:1

46. • All you need to answer the question is to recall the
structure of carbohydrates and fats:
– Carbohydrates: C(H2O)n
– Fats or lipids: 1 glycerol
+ 3 fatty acids

47. CB5
Which of the following is NOT a property of ALL
enzymes?
Enzymes catalyze catabolic reactions only.
Enzymes have active sites of specific shapes.
Temperature and pH affect the rate of enzyme reactions.
Enzymes are required in very small quantities.
You can answer this easily 

48. CB6
Scientists can create new genes and place them in
another organism’s cells. What is this technique?
DNA transfer
DNA mutation
cloning
recombinant DNA technology
You can answer this easily  The definitions of these terms are in
Campbell.

49. CB7
Which is NOT true about the structure of
eukaryotic DNA?
It has double helical structure.
It is made up of two complementary strands.
It contains nucleotides as its basic structural unit.
It is composed of phosphates, ribose, and nitrogenous bases.
You can answer this easily 

50. CB8
Which of the following statements about
nucleic acids is correct?
In a given sample of DNA the amount of thymine is equal to the
amount of cytosine.
Both DNA and RNA are assembled from nucleoside triphosphates.
The bases and sugar molecules found in DNA and RNA are the
same.
All of the above are correct.

51. To evaluate the correctness of each statement, you need to recall a
few facts about nucleic acids.
A. In a given sample of DNA the amount of thymine is equal to the
amount of cytosine.
 base-pairing rules: A-T and C-G
B. Both DNA and RNA are assembled from nucleoside triphosphates.
 Nucleoside triphosphates are the ingredients of polynucleotide
synthesis
C. The bases and sugar molecules found in DNA and RNA are the
same.
 Clue: DNA, RNA

52. CB9
A scientist analyzed several DNA samples to determine the
relative proportions of purine (adenine and guanine) and
pyrimidine (cytosine and thymine) bases. Her data are
summarized in the table below:
Percentages of Nitrogenous Bases in Three Samples
Sample G C A T
I 35 35 15 15
II 40 10 40 20
III 25 25 25 25
Which sample(s) supports the base pairing rules?
A. Sample I only C. Sample I and II
B. Sample II only D. Samples I, II, and III

53. • If a sample supports the rules on base pairing (already
mentioned earlier), that would mean that the amounts of G
and C will be equal, and that the amounts of A and T will be
equal.
Note that none of the choices is valid. Again, sorry for that :P

54. CB
10 A scientist analyzed several DNA samples to determine the
relative proportions of purine (adenine and guanine) and
pyrimidine (cytosine and thymine) bases. Her data are
summarized in the table below:
Percentages of Nitrogenous Bases in Three Samples
Sample G C A T
I 35 35 15 15
II 40 10 40 20
III 25 25 25 25
If the scientist had analyzed mRNA rather than DNA, what percentage
of uracil would you expect to find in Sample II?
A. 10 C. 35
B. 25 D. 40
For this question, you just need to remember that in RNA uracil is present
rather than thymine.

56. CD1
Nerve cells and cardiac (heart) cells have long life spans and do not
undergo mitosis once they are formed. Smooth muscles cells also have
long life spans but are capable of undergoing mitosis. Which of the
following statements is FALSE regarding consequences of injuries to
the spinal cord, heart, or smooth muscle?
Damage caused by injuries to smooth muscle may be reversed
because of its capability to regenerate.
Injuries to the spinal cord and heart cause serious and permanent
damage.
Injuries to the spinal cord, heart, and smooth muscle cause serious
damage and take a very long time to heal.
All of the above statements are false.
You can answer this easily 

57. CD2
Which of the following stages of mitosis can be
viewed as the opposite of prophase if we consider
only the changes in the nucleus?
anaphase
interphase
metaphase
telophase

58. Recall the stages of mitosis:

59. CD3
The following statements regarding mitosis
and meiosis are true EXCEPT
Replication of DNA occurs before nuclear division begins.
Mitosis involves only one round of cell division, while meiosis
involves two rounds.
The second meiotic division is similar to mitosis in that the
chromosome number is reduced.
Mitosis results in two genetically identical diploid daughter cells,
while meiosis results in four genetically nonidentical haploid
daughter cells.

60. Recall the stages of meiosis (those of mitosis are on
a previous slide already).

61. CD4
Mature cells no longer divide, so they do not replicate their DNA. A
cell biologist found that there was X amount of DNA in a human nerve
cell. The biologist then measured the amount of DNA in four other
types of human cells; the results are recorded in the chart below.
Complete the chart by filling in the type of cell from these choices:
A. sperm cell
B. bone marrow cell just beginning interphase of the cell cycle
C. skin cell in the S phase of the cell cycle
D. intestinal cell beginning mitosis.
Cell Amount of DNA Type of Cell
I 2X
II 1.6X
III 0.5X
V X

62. Here are some clues:
A. sperm cell
 haploid
B. bone marrow cell just beginning interphase of the cell cycle
C. skin cell in the S phase of the cell cycle
D. intestinal cell beginning mitosis.
 Choices B, C, and D are all diploid cells. At each of the
stages of the cell cycle mentioned, is there any DNA
replication going on?
Cell Amount of DNA Type of Cell
I 2X
II 1.6X
III 0.5X
V X Try to figure
this out first.

63. “Chance favors the prepared mind.”
- Louis Pasteur