Genetics
Molecular Dogma
Chemical Biology
Cell Division
Note: The discussion here is a very general overview of topics
that MIGHT come out in the UPCAT. Your discussion of these
topics in Bio3 will definitely be more formal and more
detailed. Also, the more complicated problems here will most
probably NOT come out in the UPCAT, but the basic topics
from which they stem might.
What determines the sex of an unborn child?
The number of days between ovulation and fertilization
The chromosome content of the sperm
The chromosome content of the ovum
Diet of the father before sexual intercourse
G1
• 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:
• At the moment of conception:
X
X
X
X
X
Y
X
Y
X XX
XX Y
X
Y
Female Male
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
1/2
1/4
3/4
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?
Alleles (alternative form) of the gene
(segment of DNA) for seed shape
• Dominant: trait is expressed all the
time
o SS or Ss spherical
• Recessive: trait is expressed only if the
organism has 2 copies of the allele
o ss dented
• Heterozygous: has one of
each kind of allele (Ss)
• Homozygous: has two of the
same kind of allele
o Homozygous dominant: SS
o Homozygous recessive: ss
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
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
S SS Ss
s Ss ss
Gametes from parent 1
Gametes from parent 2
Possible genotypes of offspring
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:
• Take note that these are always the resulting ratios of a
monohybrid cross
Genotype Ratio Phenotype Ratio
SS 1/4Spherical seed (SS or Ss) 3/4
Ss 2/4
ss 1/4 Dented seed 1/4
4/4 4/4
mono: dealing w/ a single trait
hybrid: parents are heterozygous
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
AO; BB
AO; BO
AA; BB
AB; BB
G3
• 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:
* For simplicity, IA is sometimes represented as A, IB as B, and i
as O
Phenotype Genotype
Type A IAIA or IAi
Type B IBIB or IBi
Type AB IAIB
Type O ii
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…
You can construct a Punnett square using the
different combinations of parental genotypes to see
which combination will yield a type O child.
Individual Phenotype Genotype
Father Type A IAIA or IAi
Mother Type B IBIB or IBi
Child Type O ii
• (Additional info) Immunological aspects of the ABO blood
group:
Recipient phenotype Donor phenotype(s)
Type A
• has A antigens
• produces antibodies vs. B antigens
A or O
Type B
• has B antigens
• produces antibodies vs. A antigens
B or O
Type AB (universal recipient)
• has both A and B antigens
• does not produce antibodies to either A or B
antigens (because they have both)
A, B, AB, or O
Type O (universal donor)
• does not have either A or B antigens;
• produces antibodies vs. both A and B antigens
O
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
G4
• 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
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.
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:
pp
pp pp
pp
Parents Offspring
PP x Pp 2 PP: 2 Pp
Pp x Pp 1 PP: 2 Pp: 1 ppThese results indicate that R and S
are both heterozygous.
pp
pp pp
pp
Parents Offspring
PP x pp All Pp
Pp x pp 2 Pp: 2 pp
At this point, you can already
answer the question, though I
encourage you to try to figure out
the genotype of individual Q.
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:
Pp Pp
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?
5'-TACACCTTA-3'
5'-UACACCUUA-3'
3'-TACACCTTA-5'
3'-UACACCUUA-5'
DNA X Protein
MD1
• 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.
• DNA (a polymer) carries genetic information via the sequence of its
nucleotides (its monomers).
• Nucleotides are made of
building blocks themselves:
a nitrogenous base,
a pentose (5-carbon) sugar,
and a phosphate group.
• Nitrogenous bases:
– Adenine (A)
– Thymine (T)
(or uracil (U), in RNA)
– Cytosine (C)
– Guanine (G)
Nitrogenous
base
Nucleoside
O
O
O
O P CH2
5’C
3’CPhosphate
group Pentose
sugar
O
• 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
• 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.
• 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.
• 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
Which component is NOT DIRECTLY involved in the
process known as translation?
ribosomes
RNA
DNA
mRNA
MD2
• Translation is the RNA-directed synthesis of a polypeptide.
• 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.
This means that RNA (in particular,
messenger RNA (mRNA), which was
produced by transcription) is the
template for making protein.
Proteins consist of one or more
polypeptides, which are chains of
amino acids.
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-Arg-Thr-Ser
Tyr-Arg-Thr-Asp
Tyr-Arg-Thr-Ser
Tyr-Gly-Ser-Leu
* 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.
MD3
• The genetic code is made of triplets of nucleotides and the amino acids for which they code.
• During translation, our cells (ribosomes, in particular) “read” the mRNA produced earlier by translation in sets of three nucleotides (codons). As the ribosome encounters a certain codon, the corresponding amino acid is added to the growing polypeptide.
DNA
molecule
Gene 1
Gene 2
Gene 3
DNA strand
(template)
TRANSCRIPTION
mRNA
Protein
TRANSLATION
Amino acid
A C C A A A C C G A G T
U G G U U U G G C U C A
Trp Phe Gly Ser
Codon
3 5
35
• 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.
• 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
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.
UUU-CCC-AAA
TTT-CCA-AAA
TTC-CCC-AAG
AAA-GGG-UUU
MD4
• 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
• Since we have many choices
for the middle amino acid (aa),
let us focus first on the 1st and
3rd aa’s.
• 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’
Amino acid Codons
Phe UUU, UUC
Pro CCU, CCC, CCA, CCG
Lys AAA, AAG
• 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?
Which of the following is NOT true of a codon?
It is the basic unit of the genetic code.
It never codes for more than one amino acid.
It may code for the same amino acid as another codon does.
It consists of three nucleotides.
You can easily answer this by studying the genetic code from the
previous problems.
MD5
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?
The modified chromosomes could cause the chromosomes in the
cell in the body of the consumer to mutate.
The inserted gene could become incorporated into the genome of
weed species growing nearby.
The inserted gene could cause mutation to occur on other parts of
the plant chromosome.
Beneficial insects may inadvertently be killed.
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).
MD6
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?
nucleic acids
proteins
carbohydrates
lipids
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.
CB1
Given the table below, what are substances Y
and Z?
proteins; monosaccharides
fats; fatty acids and glycerol
carbohydrates; amino acids
fats; amino acids
Digestive enzyme Substance digested Resulting simple
molecules
Amylase Polysaccharides Disaccharides
(W) Proteins (X)
Lipase (Y) (Z)
CB2
• 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
A cell may use lipids to
hydrolyze food molecules.
provide structural support.
form a barrier between the cell and its environment.
speed up the chemical reaction.
CB3
• 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
Which of the following best describes the difference
between the structure of carbohydrates and fats?
Carbohydrates Fats
A. Do not contain the element
nitrogen
Contain the element nitrogen
B. Ratio of H atoms to O atoms is
2:1
No fixed ratio of H and O
atoms
C. Made of glucose molecules
joined together
Made of fatty acids and
glycerol joined together
D. No fixed ration of H to O
atoms
Ratio of H atoms to O atoms is
2:1
CB4
• 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
Which of the following is NOT a property of ALL
enzymes?
Enzymes are required in very small quantities.
Temperature and pH affect the rate of enzyme reactions.
Enzymes have active sites of specific shapes.
Enzymes catalyze catabolic reactions only.
You can answer this easily
CB5
Scientists can create new genes and place them in
another organism’s cells. What is this technique?
recombinant DNA technology
cloning
DNA mutation
DNA transfer
You can answer this easily The definitions of these terms are in
Campbell.
CB6
Which is NOT true about the structure of
eukaryotic DNA?
It is composed of phosphates, ribose, and nitrogenous bases.
It contains nucleotides as its basic structural unit.
It is made up of two complementary strands.
It has double helical structure.
You can answer this easily
CB7
Which of the following statements about
nucleic acids is correct?
All of the above are correct.
The bases and sugar molecules found in DNA and RNA are the
same.
Both DNA and RNA are assembled from nucleoside triphosphates.
In a given sample of DNA the amount of thymine is equal to the
amount of cytosine.
CB8
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
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
CB9
• 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
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.
CB
10
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?
All of the above statements are false.
Injuries to the spinal cord, heart, and smooth muscle cause serious
damage and take a very long time to heal.
Injuries to the spinal cord and heart cause serious and permanent
damage.
Damage caused by injuries to smooth muscle may be reversed
because of its capability to regenerate.
You can answer this easily
CD1
Which of the following stages of mitosis can be
viewed as the opposite of prophase if we consider
only the changes in the nucleus?
telophase
metaphase
interphase
anaphase
CD2
Recall the stages of mitosis:
The following statements regarding mitosis
and meiosis are true EXCEPT
Mitosis results in two genetically identical diploid daughter cells,
while meiosis results in four genetically nonidentical haploid
daughter cells.
The second meiotic division is similar to mitosis in that the
chromosome number is reduced.
Mitosis involves only one round of cell division, while meiosis
involves two rounds.
Replication of DNA occurs before nuclear division begins.
CD3
Recall the stages of meiosis (those of mitosis are on
a previous slide already).
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:
Cell Amount of DNA Type of Cell
I 2X
II 1.6X
III 0.5X
V X
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.
CD4
Here are some clues:
Cell Amount of DNA Type of Cell
I 2X
II 1.6X
III 0.5X
V X Try to figure
this out first.
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?
“Chance favors the prepared mind.”
- Louis Pasteur