Biology practice questions & answers

Question 1

Mendelian genetic

Answer 1

_ If the phenotype skips 1 generation, then it is recessive

_ Mother to son transmission is the hallmark of X-linked disorder

Question 2

Autosomal inheritance

Answer 2

In autosomal inheritance, each offspring receives one allele from the father and one from the mother.

Question 3

Autosomal recessive inheritance

Answer 3

In an autosomal recessive disorder, the offspring has to receive two disease alleles to exhibit the disease phenotype.
A carrier of an autosomal recessive disorder possesses one diseased allele and one normal allele.

Question 4

What charge does amino acids with acidic side have at physiologic pH ~7.4

Answer 4

Amino acids with acidic side groups carry a net negative charge at physiologic pH.

Question 5

A polypeptide with a net positive charge at physiologic pH (~7.4) most likely contains amino acids with R groups of what type?

Answer 5

Basic R group

Question 6

The amino acids in hemoglobin (or any protein) uniformly have which of the following configurations?

Answer 6

All biologically produced amino acids are in the L configuration.

Question 7

Which of the following properties of a protein is least likely to be affected by changes in pH?

Answer 7

Primary structure is the amino acid sequence of a protein plus the peptide bonds joining them together.
Change in pH are unlikely to alter amino acid sequence or break peptide bonds.
The primary structure of a protein is least likely to be affected by changes in pH.

Question 8

Hydrogen bonding between separate subunits of DNA polymerase is an example of which of the following?

Answer 8

1 degree structure through 3 degree structure have to do with individual polypeptides.
Hint #2
4 degree structure refers to the global three dimensional arrangements found in multi-subunit proteins.
Hint #3
Hydrogen bonding between separate subunits of DNA polymerase is an example of 4 degree structure.

Question 9

The unique cyclic structure of which of the following amino acids plays a central role in the formation of alpha helices and beta sheets?
Please choose from one of the following options.
Lysine
Arginine
Valine
Proline

Answer 9

Alpha helices and beta sheets are two three dimensional motifs that regularly appear in local segments of amino acids.
Hint #2
Proline has a unique cyclic structure which differentiates it from the other common amino acids.
Hint #3
Proline plays a central role in the FORMATION of alpha helices and beta sheets. While proline’s unique structure may also disrupt both alpha helixes and beta sheets, it’s ability to make sharp turns facilitates the FORMATION of both structures, with proline commonly being found at the beginning of alpha helices or at the turns in beta sheets.

Question 10

Electrophoretic separation of leucine from a protein sample would be least effective at which of the following pH values?
Please choose from one of the following options.
2.4
1.4
0.4
7.4

Answer 10

Leucine has an aliphatic side chain.
Hint #2
At physiological pH, leucine exists as a zwitterion.
Hint #3
Electrophoretic separation of leucine from a protein sample would be least effective at pH 7.4.

Question 11

Electrophoretic separation at pH 6 of a sample of polypeptide 1 (mw 100) polypeptide 2 (mw 200) and polypeptide 3 (mw 400) would result in which of the following?
(Note: the isoelectric point of each polypeptide occurs at pH 6)
Please choose from one of the following options.
Polypeptide 3 would move the farthest
Polypeptide 1 would move the farthest
Polypeptide 2 would move the farthest
None of the polypeptides would move

Answer 11

The isoelectric point for a polypeptide is the pH at which the molecule does not have a net charge.
Hint #2
Electrophoretic separation depends on the existence of a negative net charge.
Hint #3
None of the polypeptides would move at pH 6.

Question 12

Phosphoglucomutase, which catalyses the formation of glucose-6-phosphate from glucose-1-phosphate, is best classified as which of the following enzyme type?
Please choose from one of the following options.
Kinase
Oxidoreductase
Isomerase
Lyase

Answer 12

Enzymes are classified according to the sort of reaction they catalyze.
Hint #2
glucose-6-phosphate is an isomer of glucose-1-phosphate.
Hint #3
Phosphoglucomutase is an isomerase.

Question 13

Which of the following molecules cannot be classified as an enzymatic cofactor?
Please choose from one of the following options.
Heme
Valine
Mg2+​​
Flavin adenine dinucleotide (FAD)

Answer 13

A cofactor is a nonorganic molecule whose presence is necessary for the proper function of an enzyme.
Hint #2
Valine is an organic molecule, namely an amino acid.
Hint #3
Valine cannot be classified as an enzymatic cofactor.

Question 14

In order to analyze the catalytic effect of two different enzymes on the same chemical reaction, it is best to compare which of the following quantities?
Please choose from one of the following options.
The differences in enthalpy between the reactants and products
The differences in entropy between the reactants and products
The differences in free energy between the reactants and products
The difference between transition state energies

Answer 14

Analyzing the catalytic effect of two different enzymes on a chemical reaction requires comparison between the reaction kinetics.
Hint #2
Entropy, enthalpy, and free energy are all thermodynamic quantities.
Hint #3
In order to analyze the catalytic effect of two different enzymes on the same chemical reaction, it is best to compare differences in transition state energies.

Question 15

The active site of an enzyme E, which catalyzes a reaction X, is partially denatured: which of the following quantities associated with X is most likely to be affected by the partial denaturation of E compared to the native form of E?
Please choose from one of the following options.
the equilibrium constant K​eq
​​ 
the boltzmann constant k​B
​​ 
the heat of reaction ΔH
the rate-constant k

Answer 15

Denaturation of an enzyme will alter the kinetics of a reaction that it catalyzes.
Hint #2
The kinetics of a reaction can be characterized by a rate-constant.
Hint #3
Denaturation of E is most likely to affect the rate-constant associated with X.

Question 16

The induced fit model of enzyme binding states that which of the following molecules alters the enzyme active site to more closely match the shape of the substrate?
Please choose from one of the following options.
A cofactor
A coenzyme
An allosteric effector
The substrate

Answer 16

A substrate for an enzyme is the reactant of the reaction that the enzyme catalyzes.
Hint #2
An enzyme and a substrate must come into close physical proximity for binding to occur, and such proximity can introduce physical forces that alter the shape of the enzyme.
Hint #3
The induced fit model of enzyme binding states that the substrate itself alters the enzyme active site.

Question 17

Which of the following enzyme types catalyzes the formation of a single bond between two substrates through the elimination of H2O
Please choose from one of the following options.
Ligase
Oxidoreductase
Isomerase
Hydrolase

Answer 17

Classes of enzymes are named after the sort of reaction that they catalyze.
Hint #2
A ligase catalyzes the “ligation” of two molecules via the formation of a covalent bond.
Hint #3
A hydrolase catalyzes a reaction opposite to the sort catalyzed by a ligase: the hydrolysis of a covalent bond.

Question 18

Local conditions can affect the specificity of an enzyme for its substrate, and thus the enzymes catalytic ability: which of the following alterations would most likely not affect an enzyme in this manner?
Please choose from one of the following options.
Increased temperature
Increased substrate concentration
Increased concentration of H+
Increased concentration of OH-
​​

Answer 18

The specificity of an enzyme for its substrate is partially governed by the shape of the enzyme active site.
Hint #2
An enzyme active site is stabilized by non-covalent intramolecular interactions.
Hint #3
Non-covalent intramolecular interactions are affected by heat and pH, amongst other things.
Hint #4
Alterations in substrate concentration would not affect the specificity of an enzyme for its substrate.

Question 19

Given that ∆G’° for the reaction S⇋P is negative in the direction of S→P, reaction equilibrium favors the formation of which substance?
Please choose from one of the following options.
S
The reaction does not proceed in either direction
The reaction proceeds equally in both directions
P

Answer 19

A negative standard free energy change (∆G’°) indicates the free energy of the ground state product formed from the reaction (P) is lower than that of S.
Hint #2
The negative standard free energy change of this reaction suggests that energy is released as S is converted to P.
Hint #3
The reverse reaction, therefore, would require energy to form S from P.
Hint #4
Reactions proceed spontaneously in the direction of the formation of low energy products from high energy reactants.
Hint #5
Given that ∆G’° for the reaction S⇋P is negative in the direction of S→P, reaction equilibrium favors the formation of P.

Question 20

How would the addition of a catalyst to the reaction S⇋P affect the difference between the free energies of S and P in their ground states (∆G’°)?
Please choose from one of the following options.
The change in ∆G’° cannot be predicted without more information
∆G’° would become less negative
∆G’° would not change
∆G’° would become more negative

Answer 20

A catalyst functions by lowering the activation energy required for a reaction to proceed.
Hint #2
The overall energy change from the beginning to the end of a reaction is not affected by the addition of a catalyst.
Hint #3
A catalyst affects reaction rate, but does not alter the position or direction of equilibrium.
Hint #4
The addition of a catalyst to the reaction S⇋P would not affect the difference between the free energies of S and P in their ground states, therefore ∆G’° would not change.

Question 21

For the enzyme-catalyzed reaction E + S ⇋ ES ⇋ E + P, what equation defines the rate at which ES is formed? ([Et​ ]= total enzyme concentration, [ES]= enzyme-substrate complex concentration, [S]=substrate concentration, [P]= product concentration, k1
​​ =rate constant for ES formation from E and S, k-1 = reverse reaction rate constant)
Please choose from one of the following options.
k1[S]
​k-1S
k1([Et]-[ES])[S]
k-1[S]

Answer 21

The rate of a reaction is dependant upon the concentration of reactants.
Hint #2
The rate of the reaction is equal to the rate constant times the concentration of reactants.
Hint #3
As ES is formed, the amount of unbound E and S decreases.
Hint #4
[Et] - [ES] is equal to the concentration of unbound enzyme, [E].
Hint #5
The rate at which ES is formed is k1([Et]-[ES])[S]

Question 22

Problem
If the enzyme-catalyzed reaction E + S ⇋ ES ⇋ E + P is proceeding at or near the V{max}​​ of E, what can be deduced about the relative concentrations of S and ES?
Please choose from one of the following options.
[S] is vanishingly low, [ES] is vanishingly low
S is abundant, [ES] is at its highest point
S is abundant, [ES] is vanishingly low
[S] is vanishingly low, [ES] is at its highest point

Answer 22

At V_{max}
​max, the reaction is proceeding at a rate which is equal to the physiologic rate at which the enzyme can function.
Hint #2
At V_{max}, the enzyme is fully saturated, and no more free enzyme is available to bind the substrate and catalyze the reaction.
Hint #3
If all of the enzyme, E, is bound and operating, it must all exist in the ES state.
Hint #4
In order to maintain V_{max}
​max, there must be sufficient substrate for the enzyme to bind, so [S] must be high enough to sustain this saturated state.

Question 23

What relative values of Km and kcat would describe an enzyme with a high catalytic efficiency?
Please choose from one of the following options.
Low Km and high kcat
​​High K_m and high k_cat
​​High K_m and low k_{cat}
Low K_m and low k_{cat}

Answer 23

An enzyme with a high catalytic efficiency could be described by having a Low K_m and high k_{cat}

Question 24

The lagging strand of a DNA molecule undergoing replication reads 3’-CGCATGTAGCGA-5’. What is the code of the DNA that is the template for the complementary leading strand of this segment?
Please choose from one of the following options.
5’-GCGTACATCGCT-3’
5’-CGCATGTAGCGA-3’
3’-CGCATGTAGCGA-5’
3’-GCGTACATCGCT-5’

Answer 24

The leading strand is complementary to the lagging strand, i.e. 5’-GCGTACATCGCT-3’.
Hint #2
The newly synthesized strands of DNA (the leading and lagging strands) are complementary to their template strands.
Hint #3
The template strand for the leading strand is complementary to the leading strand. The newly synthesized leading strand is also complementary to the lagging strand.
Hint #4
As the template leading strand and the newly synthesized lagging strand are both complementary to the newly synthesized leading strand, the template leading strand and the newly synthesized lagging strand will have the same sequence. As the newly synthesized lagging strand’s sequence was given as 3’-CGCATGTAGCGA-5’, this is the sequence of the template leading strand, and the correct answer.

Question 25

In what direction is DNA synthesized when catalyzed by either DNA polymerase or reverse transcriptase?
Please choose from one of the following options.
DNA polymerase catalyzes the synthesis of DNA in the 3’→5’ direction, while reverse transcriptase catalyzes the synthesis of DNA in the 5’→3’ direction
Both DNA polymerase and reverse transcriptase catalyze the synthesis of DNA in the 5’ to 3’ direction
Both DNA polymerase and reverse transcriptase catalyze the synthesis of DNA in the 3’ to 5’ direction
DNA polymerase catalyzes the synthesis of DNA in the 5’→3’ direction, while reverse transcriptase catalyzes the synthesis of DNA in the 3’→5’ direction

Answer 25

During both transcription and DNA replication, DNA templates are read from 3’→5’.
Hint #2
Reverse transcriptase is ‘reverse’, in that it catalyzes the synthesis of DNA from RNA.
Hint #3
Reverse transcriptase still reads its template from 3’→5’.
Hint #4
Both DNA polymerase and reverse transcriptase catalyze the synthesis of DNA in the 5’ to 3’ direction.

Question 26

In a bacteria possessing the lac operon, which of the following occurs when glucose is low and lactose is abundant?
Please choose from one of the following options.
Beta-galactosidase acetylases the operon
Lactose permease cleaves lactose to glucose and galactose
Lactose metabolism is increased by lactose binding to the operon
Transport of lactose into the cell is enhanced

Answer 26

The lac operon contains genes necessary to enhance lactose metabolism.
Hint #2
Cleavage of lactose is carried out by the beta-galactosidase encoded for by lacZ, not by lactose permease.
Hint #3
When glucose is low but lactose is readily available, the lac operon is activated, enhancing transport of lactose into the cell as well as lactose utilization.

Question 27

The effect of allolactose is:
Please choose from one of the following options.
Dissociation of RNA polymerase from DNA, terminating transcription of genes regulated by the lac operon
Enhancement of binding of RNA polymerase to DNA, increasing gene transcription
Conformational change in the repressor protein, inducing the lac operon
Repression of the lac operon when glucose is present

Answer 27

Allolactose is not involved in the relationship of RNA polymerase and DNA.
Hint #2
Nor does allolactose deactivate the lac operon when glucose is present.
Hint #3
Allolactose engenders a conformational change in the repressor protein, which causes this protein to disassociate from the lac operon, allowing for transcription.

Question 28

Deacetylation of histones has which of the following effects?
Please choose from one of the following options.
Uncoiling of histone structure, allowing it to be accessed by transcriptional machinery
Uncoiling of histone structure, preventing it from being accessed by transcriptional machinery
Coiling of the histone structure, preventing it from being accessed by transcriptional machinery
Coiling of the histone structure, allowing it to be accessed by transcriptional machinery

Answer 28

Acetylation causes uncoiling of the histone structure; deacetylation causes re-coiling.
Hint #2
When the histone structure is coiled, it cannot be accessed.
Hint #3
Therefore deacetylation causes coiling and prevents transcriptional machinery from accessing the genetic material.

Question 29

Methylation of CpG islands is essential to which process(es)?
Please choose from one of the following options.
Reversible inactivation of genes as cells terminally differentiate
Enhancing cell differentiation into different organ systems with distinct functions
Enhancing oncogenic potential of rapidly-differentiating cells
Stable silencing of DNA as cells terminally differentiate

Answer 29

CpG methylation diminishes the oncogenic potential of cells.
Hint #2
CpG island methylation does not have a role in the differentiation of cells.
Hint #3
However, once cells are done differentiating, their DNA needs to be turned off, or silenced. CpG methylation permanently, or stably, silences the DNA of cells that have completed the process of differentiation.

Question 30

What is the role of an activator?
Please choose from one of the following options.
Displaces a repressor protein from an operon, allowing RNA polymerase to proceed in transcription
Causes conformational changes in the DNA, bringing promoter regions into proximity with enhancer regions
Binds to an operon and transcribes RNA
Enhances interaction between RNA polymerase and the promoter

Answer 30

An activator does not interact with repressor proteins.
Hint #2
Nor does the activator perform the actual act of transcription.
Hint #3
The activator’s role is to enhance the interaction between the promoter region and RNA polymerase, usually by binding a promoter-proximal element.

Question 31

In the ABO blood group system in humans, if a person of type-B blood has children with a person of type-AB blood, what blood types could their children have?
Please choose from one of the following options.
Type-AB, type-A, and type-B
Type-A and type-B
Type-B and type-AB
Type-AB, type-A, type-B, and type-O

Answer 31

A person of type-B blood could be homozygous (BB) or heterozygous (BO). The question intentionally leaves this ambiguous.
Hint #2
One way to answer this question is to draw Punnett squares. To account for both possibilities of the genotype of the type-B blood person, you will have to draw two Punnett squares- one for an ABxBB cross and one for an ABxBO cross.
Hint #3
In an ABxBB cross, the possible genotypes are AB and BB. For an ABxBO cross, the possible genotypes are AB, AO, and BO.
Hint #4
Since the A and B alleles are codominant, the possible blood types of the children are type-AB, type-A, and type-B.

Question 32

A gene for corn has two alleles, one for yellow kernels and one for white kernels. Cross pollination of yellow corn and white corn results in ears of corn that have an approximately even mix of yellow and white kernels. Which term best describes the relationship between the two alleles?
Please choose from one of the following options.
Incomplete dominance
Genetic recombination
Chimerism
Codominance

Answer 32

Genetic recombination refers to the exchange of genetic information between two DNA molecules.
Hint #2
Chimerism is the presence of multiple distinct genomes within one organism, caused by the merging of two fertilized eggs. Although this is a possible explanation for corn with multiple colors, it is unlikely because chimerism is very rare in nature.
Hint #3
Incomplete dominance is the blending of alleles, resulting in a phenotype that is in between the two extremes. (This one is close, but there’s a term that better describes the relationship of the alleles).
Hint #4
Codominance refers to traits that are both expressed at the same time in heterozygotes (individuals with one copy of each allele). In this case, both the yellow kernels and the white kernels are expressed in the phenotype.

Question 33

Suppose that in barley plants, the allele for tall stalks is dominant over short stalks and the allele for wide leaves is dominant over thin leaves. What would be the best way to determine the genotype of a barley plant with a tall stalk and wide leaves?
Please choose from one of the following options.
Perform a testcross with a barley plant that has a short stalk and thin leaves
Perform a testcross with a barley plant that has a tall stalk and wide leaves
Perform a testcross with a known heterozygous barley plant
Perform a testcross with a barley plant that has a tall stalk and thin leaves

Answer 33

By breeding the original barley plant with a plant that has both recessive alleles, we can determine whether or not recessive alleles are present in the original barley plant based on whether or not offspring with the recessive trait are produced. Thus, we would want to breed the original barley plant with a barley plant that had a short stalk and thin leaves.

Question 34

A cross between two pea plants with genotypes PpLl and PpLl results in an F1 generation that is 25% PPLL, 50% PpLl, and 25% ppll. Which reason most likely explains why other possible genotypes are not present?
Please choose from one of the following options.
The loci of the genes are on different chromosomes.
The genes underwent independent assortment
The loci of the genes are close together.
Crossing over occurred between chromosomes.

Answer 34

Notice that a big P is always paired with a big L and a small p is always paired with a small l.
Hint #2
This atypical distribution of genotypes breaks Mendel’s second law of independent assortment among different genes.
Hint #3
The P gene and the L gene are inherited as a pair because crossing over is unlikely to happen between the P and L genes–why might this be? It may be helpful to draw the locations of the genes on a chromosome.
Hint #4
If the genes are inherited as a pair, we know that the loci (location on a chromosome) of the genes are close together.

Question 35

The X-linked recessive trait of color-blindness is present in 5% of males. If a mother who is a carrier and father who is unaffected plan to have 2 children, what is the probability the children will both be male and color-blind?
Please choose from one of the following options.

Answer 35

Don’t get distracted by the 5% number in the question stem. You actually don’t need this information to answer the question.
Hint #2
There is a 50/50 chance each child will be male. The probability of having 2 males is .5 * .5 = .25 = 25%.
Hint #3
Each male receives only 1 X-chromosome and it must come from the mother since the Y chromosome can only come from the father. Since the mother is a carrier, there is a 50/50 chance she will pass on the gene for color-blindness to each male child and a 25% chance she will pass it on to both.
Hint #4
Combining the last two hints, we know there is a 25% chance of having two sons and a 25% chance of two sons being colorblind. To find the probability of both these events happening we multiply them together: 25% * 25% = ¼ * ¼ = 1/16 = 6.25%.

Question 36

What of the following provides the best evidence that DNA is the genetic material?
Please choose from one of the following options.
Biomolecular composition of chromosomes
Mechanism of semi-conservative DNA replication
Transformation using heat-inactivated bacteria
Presence of DNA in all cells

Answer 36

Chromosomes contain both DNA and proteins.
Hint #2
While DNA is found in all cells, the same could be said of proteins and carbohydrates.
Hint #3
Heating of bacteria denatures proteins.
Hint #4
If heat-inactivated bacteria can transform (transfer genetic information to) other bacteria, protein cannot be the genetic material. Thus, DNA must be the genetic material.

Question 37

Which of the following statements about synaptonemal complexes is not true?
Please choose from one of the following options.
They form between sister chromatids
They act as a support structure
They are made up of functional RNA and protein
They form during synapsis

Answer 37

Synaptonemal complexes are often described as the scaffold upon which recombination takes place.
Hint #2
Synaptonemal complexes form between the two structures that will undergo genetic recombination – homologous chromosomes, or sister chromatids
Hint #3
Like most functional components of the cell, synaptonemal complexes are made up highly specialized proteins, but no RNA is involved.

Question 38

A color-blind man marries a woman with no family history of color-blindness. What is the likelihood that they have a color-blind daughter?
Please choose from one of the following options.
0%
25%
50%
100%

Answer 38

Hint #1
Color-blindness is a recessive, sex-linked trait.
Hint #2
Sex-linked traits are most often found on the X chromosome.
Hint #3
If the mother has no family history of color-blindness, it is likely that she is not a carrier for the allele.
Hint #4
If the mother can only pass on color-vision alleles, there is 0% chance of a daughter (2 X chromosomes) to be color-blind.

Question 39

Which of the following population’s coat/fur/feather color gene is most likely to be in Hardy-Weinberg equilibrium?
Please choose from one of the following options.
A flock of seagulls that feeds on one type of fish
Lab rats being bred by scientists
A pack of wolves in a zoo
A flock of moths that must blend in with their environment to survive

Answer 39

For a population to be in Hardy-Weinberg equilibrium, mating must be random.
Hint #2
Hardy-Weinberg equilibrium is only achieved if natural selection is not occurring.
Hint #3
A flock of seagulls that eats only type of fish may be under selection for traits related to catching that type of fish, but that would likely not have an effect on the seagulls’ coat color gene, which thus should be in Hardy Weinberg equilibrium.

Question 40

Sickle-cell anemia is an autosomal recessive genetic disorder whose carriers have a genetic advantage in surviving malaria. If 42% of the population is malaria resistant, but not anemic, what is frequency of the sickle-cell allele?
Please choose from one of the following options.
9%
49%
21%
30%

Answer 40

If the two allelic frequencies are p and q, then p + q = 1, and p^{2+}
​2+
​​ start superscript, 2, plus, end superscript 2pq + q^2
​2
​​ start superscript, 2, end superscript = 1
Hint #2
The population that is malaria resistant but not sick is represented by 2pq
Hint #3
If 2pq = .42, then pq = .21. Since p + q = 1, p and q must be .3 and .7. Only 30% is available as a choice, so that must be the allelic frequency of the sickle-cell allele.

Question 41

Which type of lethal, inheritable allele is most likely to persist in a population?
Please choose from one of the following options.
Recessive
Codominant
Incompletely Dominant
Dominant

Answer 41

A lethal allele can persist through carriers in the population.
Hint #2
Carriers are generally heterozygous (1 disease allele; 1 healthy allele).
Hint #3
A lethal, inheritable allele is most likely to persist in a population if it is recessive.

Question 42

A small population of flies that has an even distribution of black-bodied and tan-bodied flies suffers a drought in which almost the entire population dies. As the population is recovering, the next generation has twice as many black-bodied flies as tan-bodied flies. What most likely occurred within the population?
Please choose from one of the following options.
Inbreeding
Natural selection
Adaptation
The bottleneck effect

Answer 42

Inbreeding occurs when mates have a large proportion of shared ancestry. The small population of flies that was left after the drought did not necessarily share the same ancestry.
Hint #2
We do not know whether or not the drought selected for black flies and tan flies differently. Instead, the drought likely killed flies randomly.
Hint #3
Sometimes when the population becomes very small, one trait can randomly be present in larger proportions than another trait in the remaining individuals.
Hint #4
If we consider the possibility that the flies were killed randomly, two of the answers become less fitting for the situation.
Hint #5
If the flies were killed randomly, neither natural selection nor adaptation would be happening. Furthermore, natural selection and adaptation take many generations. On the other hand the bottleneck effect, which describes a situation where a population becomes very small due to an environmental stress and loses much of its genetic diversity, fits this situation whether or not the drought killed black and tan flies randomly.

Question 43

Consider a population of wildflowers that can be either purple or white. On average, 25% of flowers (regardless of color) survive until reproduction. Which reason could explain the changes in frequency of purple and white flowers shown in the table?
Generation 1 Generation 2 Generation 3
# of Purple Flowers 96 121 152
# of White Flowers 213 176 149
Please choose from one of the following options.
A drought has caused the flower population to fall.
Purple flowers have a lower level of fecundity than white flowers.
A pollinator is more likely to visit a purple flower than a white flower.
Stabilizing selection is causing the number of white and purple flowers to equalize.

Answer 43

The data in the table shows that over three generations the number of purple flowers is increasing, whereas the number of white flowers is decreasing.
Hint #2
The question stem tells us that the same percentage of white and purple flowers successfully reproduce.
Hint #3
Stabilizing selection describes a process in which extremes of a trait (being very tall or very short for example) are selected against and an intermediary trait (like being average height) is selected for. This does not appear to be happening in this scenario.
Hint #4
A high fecundity means an organism is able to produce more offspring. In this case, it is possible that purple flowers have a HIGHER fecundity, but probably not the other way around.
Hint #5
One way purple flowers might have higher fecundity would be if a pollinator is more likely to visit a purple flower than a white flower.

Question 44

Which of the following forms of reproductive isolation are pre-zygotic?
I. Hybrid Inviability
II. Gametic Isolation
III. Zygote Mortality
IV. Temporal Isolation
Please choose from one of the following options.
I, II, and IV
II and IV
I and III
IV only

Answer 44

Pre-zygotic reproductive isolation refers to barriers that prevent two organisms from mating and occur before a zygote is successfully formed by the combination of a sperm and an egg.
Hint #2
Hybrid inviability occurs when a hybrid is born, but cannot live to a reproductive age.
Hint #3
Gametic isolation refers to the inability of two gametes (i.e. a sperm and egg) to join.
Hint #4
Zygote mortality refers to the inability of a zygote to survive embryological development.
Hint #5
Temporal isolation is when organisms are separated by a factor of time, such as day and night or summer and winter, and therefore cannot reproduce.
Hint #6
So, which of these occur before formation of the zygote? The answer is gametic isolation and temporal isolation, II and IV.

Question 45

A certain disease in humans follows a simple inheritance pattern through a gene locus with only two alleles: D (the dominant non-disease allele) and d (the recessive disease allele). If the frequency of the recessive allele is 10% in a population of 100,000, how many people would you expect to be healthy carriers of the recessive allele? (Assume Hardy-Weinberg equilibrium.)
Please choose from one of the following options.
36,000
18,000
9,000
24,000
Hint #1

Answer 45

For this question we will need to use the Hardy-Weinberg equation: p^2
​2
​​ start superscript, 2, end superscript + 2pq + q^2
​2
​​ start superscript, 2, end superscript = 1 (you’ll want to have this memorized for test day). In this case, p will be the frequency of the dominant allele, D, and q will be the frequency of the recessive allele, d.
Hint #2
The first goal is to solve for 2pq since this represents the frequency of the heterozygous genotype and therefore the percentage of people that are carriers.
Hint #3
From the question stem, we know that 10% of people in the population have the disease allele, which tells us that q = 0.1.
Hint #4
Another important Hardy-Weinberg equation is p + q = 1
Hint #5
Using the above equation, we know that if q = 0.1 then p = 0.9. Therefore 2pq = 2(0.9)(0.1) = 0.18
Hint #6
To find the answer, we just need to multiply the frequency of the heterozygotic genotype (0.18) by the size of the population (100,000): 0.18*100,000 = 18,000

Question 46

Which term describes alterations in the composition of a gene pool due to migration of individuals between different populations?
Please choose from one of the following options.
Assortive mating
Directional selection
Gene flow
Genetic drift

Answer 46

Genetic drift refers to changes in the frequency of alleles in a gene pool, simply due to chance.
Hint #2
Assortive mating is when mates are not chosen randomly, but instead certain individuals are selected over others possibly because of phenotype or location.
Hint #3
Directional selection occurs when one extreme of a trait is favored over another and the population starts to express that extreme phenotype more often.
Hint #4
Gene flow refers to alterations in the composition of a gene pool due to migration of individuals between different populations.

Question 47

Which of the following scenarios represents disruptive selection within a population?
Please choose from one of the following options.
Oak trees with medium-sized leaves are evolutionarily favored, as large leaves suffer from dehydration and small leaves don’t receive sufficient sun.
As pollution creates turbidity in a lake, brighter fish have greater reproductive success, causing the population to become more brightly colored on average.
Heterozygous pink flowers are selected against while homozygous red and homozygous white flowers are selected for.
A population of snakes is split by the construction of a large road, causing speciation of the divided populations.
Hint #1

Answer 47

Disruptive selection occurs when two extreme phenotypes on a continuum are selected for.
Hint #2
Oak trees with medium-sized leaves being evolutionarily favored over those with large or small leaves would be an example of stabilizing selection (the opposite of disruptive selection).
Hint #3
If a population of snakes is split by the construction of a large road, it would likely cause speciation, but this is different than disruptive selection.
Hint #4
Pollution creating turbidity in a lake and causing brighter fish to have greater reproductive success would be an example of directional selection.
Hint #5
The best example of two extremes being favored is heterozygous pink flowers being selected against while homozygous red and homozygous white flowers are selected for.

Question 48

Researchers are seeking evidence to confirm a close evolutionary relationship between two extant species. Which of the following pieces of evidence would be LEAST likely to help the researchers demonstrate this relationship?
Please choose from one of the following options.
The identification of several analogous structures between the species
The sequence of a ribosome-coding gene from each species
A fossil that shares traits with both species
The discovery of similar stages of embryonic development

Answer 48

The question asks the test taker to identify which piece of evidence would NOT support the hypothesis that two extant (living) species have a close evolutionary relationship.
Hint #3
A fossil that shared traits with both species would support the idea that the two species have a recent common ancestor.
Hint #4
One tool evolutionary biologists use to determine relationships between species is their embryonic development. Similar embryonic stages would support the researchers’ hypothesis.
Hint #5
Another tool for evolutionary biologists is DNA. The most commonly sequenced genes for testing evolutionary relationships are ones that are essential to almost all organisms, such as the genes for ribosomes.
Hint #6
Analogous structures are structures that serve the same function, but come from a different evolutionary origin. Identifying these structures would be an argument AGAINST the researchers’ hypothesis.

Question 49

Vervet monkeys make noise to warn others about the approach of a predator. Even though the individual may stand out and be preyed upon while warning the others, this trait persists in the population because of which evolutionary concept?
Please choose from one of the following options.
Natural selection
Differential reproduction
Coevolution
Group selection
Hint #1

Answer 49

By making noise, the monkeys are sacrificing their individual fitness, but increasing the fitness of related population members.
Hint #2
Natural selection is the tendency for the most “fit” individuals to survive and reproduce, thus driving evolution. By sacrificing itself, the monkey does not increase its own fitness.
Hint #3
Coevolution occurs when two species evolve alongside each other and exert evolutionary pressures on one another. Coevolution doesn’t very well describe this self-sacrificing concept.
Hint #4
Differential reproduction is the idea that certain individuals are more likely to reproduce than others, and will therefore have their genetic traits expressed at a higher rate in future generations. Differential reproduction also doesn’t describe this self-sacrificing concept very well.
Hint #5
Group selection is the idea that traits that benefit the population as a whole will still be selected for even if they don’t directly help the individual expressing that trait to survive and reproduce. The trait of alarming other monkeys persists in Vervet monkeys because of this concept of group selection.

Question 50

As prehistoric humans moved to different regions of the globe, different populations began expressing different concentrations of melanin in their skin. This is an example of which evolutionary concept?
Please choose from one of the following options.
Genetic drift
Adaptive radiation
Convergent evolution
Parallel evolution

Answer 50

The question asks us to assign a term to the event of humans moving to different regions of the globe and developing different skin colors to better survive and reproduce in each new environment.
Hint #2
Convergent evolution occurs when two populations that don’t share a recent common ancestor start to resemble each other as they adapt to the same or similar environments (e.g. a dolphin and a fish). This is the opposite of what we’re looking for.
Hint #3
Parallel evolution occurs when two related populations evolve similarly for a long period of time as they experience similar environmental conditions. This is also not what we’re looking for.
Hint #4
Genetic drift refers to changes in the frequency of alleles in a gene pool, simply due to chance. This is also incorrect, as the changes were not by chance, but were directed by environmental pressures.
Hint #5
By the process of elimination, we’ve arrived at adaptive radiation. Adaptive radiation is when one lineage diverges into many others as isolated populations occupy different niches. This fits with the example of humans and skin color.

Question 51

In nature, restriction enzymes are meant to protect
Please choose from one of the following options.
Viruses from bacteria
Bacteria from other bacteria
Humans from bacteria
Bacteria from viruses

Answer 51

Restriction enzymes destroy unprotected DNA, which is usually foreign. Hosts can protect their DNA by methylation.
Hint #2
Viruses do not have restriction enzymes.
Hint #3
Restriction enzymes are found in bacteria as a defence against invading viruses

Question 52

A student was attempting to make cDNA that would code for protein Y. They began with an isolated and amplified sample of the target post-transcriptional mRNA, and all the tools required to complete the task. However, at the end of the experiment the student was left with a molecule containing ribonucleotides. What most likely went wrong?
Please choose from one of the following options.
The student forgot to add DNA polymerase
The student added RNA polymerase instead of DNA polymerase
The cloning vector already transcribed the DNA
The student forgot to add reverse transcriptase

Answer 52

DNA polymerase builds DNA using a DNA template.
Hint #2
RNA polymerase builds RNA using a DNA template.
Hint #3
The student most likely forgot to add reverse transcriptase, leaving the experiment stranded as mRNA.

Question 53

Problem
PCR (polymerase chain reaction) is an excellent method of generating copies of target DNA. If a single piece of double stranded DNA (dsDNA) is put into a PCR machine, how many dsDNA segments will there be after 3 rounds?
Please choose from one of the following options.
8 segments, with the 2 original strands paired
8 segments, with the 2 original strands on different segments
16 segments, with the 2 original strands paired
16 segments, with 2 original strands on different segments

Answer 53

Hint #1
During PCR, DNA segments are separated such that the polymerase can build complementary strands.
Hint #2
In an ideal experiment, each DNA segment is separated and replicated during each round.
Hint #3
After 3 rounds of PCR there should be 8 DNA segments, with the original segment split.

Question 54

Which procedure is least effective for detecting a specific DNA sequence in a sample?
Please choose from one of the following options.
Southern Blot
Polymerase Chain Reaction
Gel Electrophoresis
Microarray

Answer 54

Polymerase Chain Reaction uses specially constructed primers to amplify a specific DNA sequence. Thus, you can tell if the target sequence was in the sample based on if the reaction produces more nucleic acid.
Hint #2
A Southern blot is one of the most efficient method for detecting the presence of a specific DNA sequence in a sample. It works via hybridization of the target sequence to a probe. If the probe latches on, the target is there.
Hint #3
Gel Electrophoresis separates fragments based on size, and so is not nearly as specific as the other techniques mentioned above.

Question 55

The reaction A + B ⇆ C is an exothermic reaction; if an equilibrium mixture of A, B, and C is heated slowly, what would happen to the concentrations of each species?
Please choose from one of the following options.
All three concentrations increase
A and B decrease, C increases
All three concentrations decrease
A and B increase, C decreases

Answer 55

An exothermic reaction produces heat as a product.
Hint #2
If a product of a reaction is added to an equilibrium mixture of elements of the reaction, the equilibrium will shift towards reactants.
Hint #3
The concentration of reactants will increase if an equilibrium mixture of A, B, and C is heated slowly.