Must Know High School Biology - Kellie Ploeger Cox 2019
Answer Key
1
Chemical Bonds and Reactions
1. This is an example of an intermolecular bond because it is occurring between two separate molecules (in this case, two water molecules) and is due to the slightly positive side of one molecule attracting the slightly negative end of a different molecule.
2. Both laws of thermodynamics apply to this example. Breaking up glucose and converting it into another form of energy describes the first law (energy cannot be created, but it can be converted from one form to another). The second law (every energy transformation includes some energy lost as heat) is why our metabolism provides body heat.
3. A reaction that breaks bonds is called a catabolic reaction because energy is released.
4. a. This is a noncompetitive inhibitor, meaning it binds at a site other than the active site. b. The effects of the inhibitor cannot be overcome by adding more substrate because when the non-competitive inhibitor binds to the enzyme, it changes the shape of the active site.
5. A covalent bond involves sharing two electrons between two atoms.
6. Reaction A requires an input of energy; it creates products that contain more energy than reactants. Reaction B has a negative delta G.
7. Enzymes are catalysts because they speed up reactions. They’re biological because all enzymes are proteins that are made by cells.
8. A catabolic reaction is one where bonds are broken and energy is released. An enzyme helps this to occur because when it grabs onto the substrate, it squeezes the molecule and puts stress on the covalent bonds, making them easier to break.
9. Both allosteric enzymes and non-competitive inhibitors rely on a substance binding to the enzyme at a site other than the active site. The difference, however, is when something binds to that other site it can either decrease or increase the activity of an allosteric enzyme. A non-competitive inhibitor can only decrease the activity.
2
Properties of Water
1. Two water molecules involved in a hydrogen bond:
Remember to draw your hydrogen bond as a dashed line and NOT a solid line (solid lines are reserved for covalent bonds).
2. When a water molecule forms a hydrogen bond with another water molecule, this is called cohesion; when a water molecule hydrogen bonds to another substance, it is called adhesion.
3. Water becomes less dense when it freezes (which is why your ice floats on the surface of your drink).
4. Water resists quick changes in temperature and doesn’t easily evaporate (or freeze); this is due to water’s high heat capacity.
5. It is only a partial charge (either negative or positive) because electrons are not evenly distributed throughout the molecule. If you said a full positive or negative charge, you would mean a complete loss (or gain) of an electron, which would make a water molecule an ion (which it’s not!).
6. The polarity of a water molecule (as described in the previous question) allows the partially positive side of one molecule to attract the partially negative side of another molecule. It’s like two magnets sticking to one another.
7. Cohesion creates surface tension because the water molecules are holding on to one another.
8. The ice floating on the surface acts as insulation to the water below, keeping it warmer. Furthermore, if ice sank, entire bodies of water would freeze solid in the winter, killing all life within it.
9. When water melts or boils, the hydrogen bonding between individual water molecules are being broken apart. The covalent bonds within individual water molecules do NOT break; otherwise, when you boil water, you would be producing oxygen and hydrogen gases (you’re not!).
3
Macromolecules
1. A polymer is a large macromolecule (term for large molecule) that is composed of repeating subunits called monomers.
2. The creation of a peptide bond between two amino acids (via a dehydration reaction):
3. a — 3; b — 5; c — 1; d — 6; e — 4; f — 7; g — 2
4. DNA is double stranded; RNA is single stranded.
DNA contains the sugar deoxyribose; RNA has ribose.
DNA has the base thymine; RNA has uracil.
DNA remains in the nucleus; RNA travels out to the cytoplasm.
5. First of all, to ensure the width of DNA remains constant, a pyrimidine (single-ringed base) must always pair with a purine (double-ringed base). Furthermore, both adenine and thymine like to form two hydrogen bonds, whereas guanine and cytosine like to form three hydrogen bonds.
6. a — 2; b — 5; c — 6; d — 3; e — 1; f — 4; g — 7
7. If a protein had a stretch of hydrophilic amino acids, you would expect them to face toward water.
8. When you digest starch, hydrolysis enzymes work at breaking up the starch polymer.
9. Primary structure is the sequence of individual amino acids in a protein. The statement describing that one subunit is composed of 141 amino acids and the other subunit is composed of 146 amino acids is describing the primary structure.
Secondary structure is a regular folding pattern (alpha helix or a beta-pleated sheet) created by hydrogen bonding between the amino acids. It was not mentioned in this description.
Tertiary structure is the 3D shape a protein makes when it starts to fold up into a particular pattern, based on the R groups of the individual amino acids. There is no direct mention of what each individual protein looks like when it’s folded up.
Quaternary structure is achieved when there are two or more separate proteins involved in the final overall structure. Hemoglobin has four total protein subunits; this is referring to its quaternary structure.
4
Overview of the Cell
1. The main structural component of the cell membrane is the phospholipid.
2. Archaea prokaryotes tend to live in environments of extreme pH, salinity, and heat. This is a remnant of their ancient ancestry because the first cells to evolve on Earth had to survive in the extreme conditions present on the planet 3.5 billion years ago.
3. A. phospholipid, B. protein, C. hydrophilic phosphate group, D. hydrophobic tails, E. phospholipid bilayer
4. Transmembrane proteins span the entire width of the membrane (from the outside of the cell all the way to the inside of the cell), thus providing a tunnel through which things can move.
5. If a cell suddenly was exposed to much colder temperatures, the cell membrane would become too stiff. In order to counteract the stiffness, it would increase the percentage of unsaturated phospholipids. Unsaturated phospholipids have double bonds in the tails of the fatty acid chains, which make them unable to pack tightly together. Because they can’t pack tightly, this increases the fluidity (and helps to counteract the stiffness caused by cold temperatures).
6. Prokaryotic cells (bacteria) are the smaller and simpler cell types that lack organelles.
7. Characteristics of life: a. made of cells, b. reproduces and creates progeny, c. has a heritable genetic code (DNA), d. uses materials and undergoes metabolic processes, e. requires energy to survive, f. strives to maintain constant internal conditions (homeostasis), g. responds to its environment, h. evolves
8. A single phospholipid is amphipathic, meaning the phosphate head group portion of the molecule is polar and loves water, whereas the two fatty acid chains are nonpolar and hate water.
9. If a cell lived in very hot conditions (such as a volcanic deep-sea vent), its phospholipids would be composed of more saturated fatty acids in order to prevent the cell membrane from becoming too fluid. This is because saturated fatty acids form straight chains (without kinks and bends) and can pack tightly together. If the phospholipids of the phospholipid bilayer can pack tightly together, that makes the cell membrane more solid (which is good in really hot conditions).
10. List three possible roles for cell membrane proteins, and provide a brief description of each:
1. Transmembrane proteins can be used for transport, providing a pathway from the outside of the cell into the interior of the cell. Aquaporin is a specific example, and it provides a pathway for water molecules, increasing their movement into cells.
2. Proteins embedded in or on the cell membrane can be enzymes and play important roles in cell metabolism.
3. Some proteins are receptors and can latch onto a chemical signal on the outside of the cell. Once bound, it “transmits” the signal into the cell (this is called the process of signal transduction).
11.
a. Cell membrane Y
b. Nucleus N
c. Flagella Y
d. DNA Y
e. Cell wall Y
f. Ribosomes Y
g. Mitochondrion (an organelle) N
h. Nucleoid Y
12. An advantage to prokaryotic cells’ simplicity is it enables them to reproduce very quickly. Furthermore, their quick reproduction helps them to evolve quicker!
13.
a. An atom
b. A molecule
c. A virus ← Nope, not alive (because a virus is not made of a cell!)
d. An amoeba ← A single-celled amoeba is definitely alive!
e. A tree
14. The cell membrane is how a cell can create internal conditions different from its surroundings. The semipermeability of the membrane helps to keep certain things in the cell (and keep other things out of the cell).
5
Eukaryotic Cells and Their Organelles
1. Organelles create compartments within the cell, which can be used for microenvironments (creating conditions different from other areas of the cell).
2. A cell with a high amount of rough endoplasmic reticulum would be specialized in exporting large amounts of proteins (the rER makes proteins destined to be exported from the cell).
3.
4. Both mitochondria and chloroplasts have their own DNA, their own ribosomes, and they can reproduce on their own (independently of the cell).
5. A = outer membrane; B = inner membrane space; C = inner membrane; D = matrix
6. The glucose needed for cellular respiration in an animal cell comes from food/eating, whereas the glucose needed for cellular respiration in a plant cell comes from photosynthesis/the sun.
7. Plant cells have cells walls, central vacuoles, and chloroplasts (animal cells do not have these things).
8. The lysosome forms a space in which the pH is much lower (more acidic) than other locations within the cell. It also contains a bunch of digestive enzymes that prefer to function in these acidic conditions.
9. Fill in the blanks with the correct organelles: When a protein is made for export, it is created by ribosomes stuck onto the surface of the rough endoplasmic reticulum.
10. False! Only plant cells have chloroplasts, but both plant and animal cells have mitochondria.
11. A = outer membrane; B = thylakoid; C = granum; D = stroma
12. Turgor pressure occurs when the central vacuole of a plant cell fills with water. As it swells, it presses against the surrounding cell wall. This provides structure and rigidity to the plant cell (turgor pressure!).
6
Cells and Energy Transformation
1. The energy in ATP is released by cleaving the bond between: c. The third phosphate group and the second phosphate group
2. Cellular respiration is the process where cells convert the stored energy of glucose into the useable form, ATP.
3. Cellular respiration is a catabolic reaction because energy released as a molecule of glucose is broken down into smaller molecules of carbon dioxide.
4. The overall purpose of fermentation is to re-create NAD+ from NADH in order to keep glycolysis running in the absence of oxygen. The purpose of fermentation is NOT to create lactic acid or ethanol.
5. In photosynthesis, the light reactions create ATP (a source of energy) and NADPH (a source of hydrogens).
6. The high concentration of hydrogen ions are stored in the inner membrane space of the mitochondrion, or the thylakoid space of the chloroplast.
7. If adenosine triphosphate is analogous to a battery, the fully charged form is ATP and the dead battery form is ADP.
8. In glycolysis, two ADP are converted into two ATP, two NAD+ are reduced to form two NADH, and glucose is oxidized into two molecules of pyruvate.
9. The role of oxygen is to grab the low-energy electron at the end of the electron transport chain. Once the oxygen also combines with a couple hydrogen ions, it turns into water.
10. Photosynthesis is an anabolic reaction because energy is required in order to build a molecule of glucose from smaller molecules of carbon dioxide.
11. The Calvin cycle creates glucose from carbon dioxide. In order to do so, it needs a source of hydrogens (provided by NADPH) and energy in order to create new covalent bonds (provided by ATP). Both NADPH and ATP are provided by the light reactions occurring on the thylakoid membranes.
7
Cell Transport
1. In order for a cell to actively transport a substance across its cell membrane, there needs to be a protein pump embedded in the cell membrane, and the energy (ATP) needed to make it run.
2. Diffusion simply means the movement of a substance from an area of high concentration to low concentration. Osmosis is the diffusion of water, specifically. If in order for a molecule to diffuse across a cell membrane it needs a protein to act as a tunnel, it is referred to as facilitated diffusion.
3. The process of endocytosis is active transport because energy is required to change the shape of the cell in order to engulf a food particle.
4. A cell living in hypotonic conditions must deal with osmosis occurring into the cell.
5. The paramecium would suddenly be in hypertonic conditions, which means water would diffuse out of the cell (and the cell would most likely die).
6. Ψs = —iCRT
7. Facilitated diffusion would help the passive movement of glucose molecules into the cell. Facilitated diffusion provides a tunnel through which the glucose molecules could diffuse.
8. It is bad to disrupt a cell’s chemical gradients because the cell would no longer be able to do work. For example, if there were no longer hydrogen ion gradients in a mitochondrion, the cell would no longer be able to generate ATP.
9.
a. A plant cell floating in pure water: hypotonic.
b. A protist cell normally living in a freshwater lake transferred to a marine environment: hypertonic.
c. A bacterium in a solution that has the same water potential as inside the bacterial cell: isotonic.
10. Some cells that live in hypotonic conditions have evolved to rely on an organelle called a contractile vacuole to pump out the extra water that diffuses inward.
11. The overall water potential is the sum of the solute potential plus the pressure potential: Ψ = Ψs + Ψp
Ψp = 0 (because it is in an open container)
Ψs = —2.0 MPa
Therefore, overall Ψ = —2.0 MPa
8
Signal Transduction: Cell Communication
1. Protein signals rely on signal transduction because the proteins themselves cannot pass through the cell membrane.
2. In order for the receptor to do its job, it needs to relay the signal into the interior of the cell. The protein does so by changing the shape of the inside portion of the receptor protein. If the receptor did not span the membrane, it could not do this.
3. c, a, d, b
4. The transduction step amplifies the message as soon as it enters the cell. This ensures that even a small signal (only a few signal molecules) is enough to elicit a huge response.
5. Reception, transduction, response.
6. When the receptor binds to the signal, the inside region of the receptor protein changes shape. This moves the message (but not the actual chemical signal) into the cell.
7. c, d, b, a
8. Neurons (the cells of the nervous system) use ligand-gated ion channels in order to transmit nerve impulses from one cell to the next.
9. Protein kinase enzymes play an important role in signal transduction because they amplify/increase the message by creating a phosphorylation cascade.
9
DNA and RNA: Structure of Nucleic Acids
1. From smallest to largest: nucleotide, gene, chromosome, genome
2. The upright portion of the DNA backbone is composed of alternating sugar (deoxyribose) and phosphate portions of the nucleotides. This backbone of the DNA is very strong because it is linked by covalent bonds. The two strands of the double helix hold onto one another because of hydrogen bonding between the nitrogenous bases of complementary nucleotides.
3. The width of DNA would be inconsistent because both cytosine and thymine are the smaller pyrimidine bases, whereas guanine and adenine are larger purine bases.
4. RNA nucleotides contain the sugar ribose (DNA contains deoxyribose).
RNA is single stranded instead of double stranded, like DNA.
RNA has the nitrogenous base uracil instead of DNA’s thymine.
5. DNA sequence: ACTGACA
Complementary RNA sequence: UGACUGU
6. First, a single-ringed pyrimidine (T and C) must pair with a double-ringed purine (A and G). Secondly, adenine and thymine both create two hydrogen bonds, whereas guanine and cytosine both create three hydrogen bonds.
7. What is the name of the monomer of DNA? nucleotide. Which part of the monomer is the basis for the genetic language? the base (nitrogenous base).
8. DNA sequence: GGACACTT
Complementary DNA sequence: CCTGTGAA
9. A strand of DNA is wound around special proteins called histones. This mixture of DNA and protein is called chromatin.
10
Replication of DNA
1. Before a cell divides, it must replicate its DNA. This ensures each resulting daughter cell has a full copy of chromosomes.
2. The double helical structure of DNA is antiparallel. This means that the 3′ end of one strand (which has a hydroxyl/—OH group hanging off of it) must line up with the 5′ end of the complementary strand (which has a phosphate group hanging off of it).
3. The lagging strand needs multiple RNA primers.
4. Free nucleotides are added to a growing strand of DNA through a dehydration reaction, which creates a new phosphodiester bond.
5. In order for DNA polymerase to attach a nucleotide through a dehydration reaction, there needs to be a free —OH (hydroxyl) group available; the 3′ end of a DNA molecule has a free hydroxyl group.
6.
a. The enzyme helicase unwinds and unzips DNA in order for replication to occur.
b. The enzyme primase must first synthesize a small piece of RNA in order for another enzyme to begin adding DNA nucleotides to the growing strand.
c. Once DNA is unzipped, single-stranded binding proteins stick to the unpaired bases in order to keep the two strands apart.
d. The enzyme ligase glues together fragments of newly synthesized DNA by creating new phosphodiester bonds.
e. The bulk of the work of creating a new strand of DNA is done by DNA polymerase, the enzyme responsible for pairing up complementary nucleotides.
7. First, in order for DNA polymerase to be able to create two identical DNA molecules, the original DNA must be double stranded. Otherwise, if the original DNA was a single strand, the newly synthesized strand would be a complementary (not identical) copy. Secondly, double-stranded DNA is much more stable than a single-stranded piece.
8. b. One strand from the original DNA molecule and one created from new nucleotides
9. False. Immediately after the DNA replication (and before the cell splits into two daughter cells), another DNA polymerase will replace the RNA primers with DNA.
11
Gene Expression and Differentiation
1. The process of differentiation is when an undifferentiated stem cell begins to selectively express certain genes and turns into a specific tissue type.
2. The promoter is a sequence in the DNA right before a gene. The enzyme RNA polymerase needs to land right before a gene to be transcribed, and the landing strip is the promoter.
3. General transcription factors are small proteins that bind to the promoter and help RNA polymerase to land properly. If the cell needs to create a lot of mRNA, however, it needs the help of specific transcription factors called activators. Unlike the general transcription factors, these specific transcription factors bind upstream of the promoter at sequences in the DNA called enhancer regions.
4. By adding methyl groups to [DNA/histones], gene transcription will be [increased/decreased]. If, instead, acetyl groups are added to [DNA/histones], gene transcription will be [increased/decreased].
5. General transcription factors bind at the TATA box within the promoter, whereas specific transcription factors (activators) bind at a specific combination of distal control elements in the enhancer region.
6. The trp operon is a repressible operon, meaning it is normally transcribing the genes for the synthesis of tryptophan. This suggests that the bacterium normally does NOT have tryptophan available in its environment, because by default, it is creating the enzymes for the synthesis of the amino acid.
7. For each of the following, indicate whether the statement is true or false:
a. During transcription, the entire chromosome is unwound and unzipped: False. Only the portion of the chromosome that contains the gene being expressed is unzipped.
b. The process of transcription involves the enzyme RNA polymerase moving down the entire chromosome and creating a piece of mRNA: False. Yes, the enzyme is RNA polymerase, but no, it is not transcribing the entire chromosome (refer back to part “a”).
c. For a given gene, only one side of the DNA double helix contains the correct code for protein synthesis: True. This is called the template strand.
8. Transcription factors are small proteins that first land on a gene’s promoter, and then help RNA polymerase to align properly on the promoter.
9. A. enhancer region; B. activator proteins; C. RNA polymerase; D. promoter; E. gene; F. DNA bending protein
10. mRNA: CG AUG ACU AGC UGG GGG UAU UAC UUU UAG
Protein: MET THR SER TRY GLY TYR TYR PHE [stop codon]
11. The lac operon is a(n) [inducible/repressible] operon because it is normally not transcribing the genes for lactose digestion.
12.
12
The Cell Cycle and Mitosis
1. Mitosis is specifically focused on division of the nucleus, or more specifically, the proper division of the replicated chromosomes. Cytokinesis occurs after mitosis, and is when the entire cytoplasmic contents of the cell is split in half.
2. Fill in the blanks: After the S phase, each chromosome is now composed of two genetically identical sister chromatids and they are attached at the centromere.
3. There are three checkpoints that a dividing cell must pass through: the G1 checkpoint, the G2 checkpoint, and the M checkpoint (during mitosis).
4. The cell is currently in the G2 phase of the cell cycle. Each of the chromosomes have already been replicated in the S phase (as indicated by the presence of two sister chromatids attached to one another), and there are two centrosomes (organelles are copied during G2).
5. Absolutely. There are many instances when a cell should NOT divide and instead will exit the cell cycle during the G1 checkpoint. A cell will then be in the G0 phase, and instead of actively dividing, will instead just do its cellular thing (whatever that may be).
6. During G1, the cell grows in size and performs normal metabolic functions for the given cell type. During S, DNA will be replicated (synthesized) in preparation for cell division. In G2, the cell will copy its organelles to ensure the two daughter cells have the correct assortment. Finally, the cell divides into two, first by mitosis (division of the chromosomes), and then cytokinesis (division of all the cytoplasm).
7. Any genetic differences between daughter cells after mitosis are considered to be mutations. False! The goal of mitosis is to create two genetically identical daughter cells.
8. If a cell fails a checkpoint but has already copied its DNA, it may undergo apoptosis in order to prevent a defective cell from dividing (and possibly becoming cancerous).
9. The cell is currently in metaphase. The middle dividing line (the metaphase plate) has one sister chromatid on either side. This ensures when the cell splits in two, each resulting daughter cell will receive one copy of each chromosome.
10. Growth factor
13
Meiosis
1. Meiosis creates cells with half the DNA of the parent cell. Specifically, each gamete contains only one set of chromosomes.
2. False. The only cells in your body that can undergo meiosis are the special cells in the ovaries and testes that produce the eggs and sperm.
3. The specialized cells in the ovaries and testes that produce the gametes start off as diploid cells. Once they divide by meiosis, they produce the haploid gametes (egg or sperm). When an egg and sperm fuse in sexual reproduction, it creates a diploid zygote cell.
4. Nothing. The sister chromatids are genetically identical, so if segments switched between the two, there would be no difference.
5. The purpose of sexual reproduction is to create new genetic combinations in the offspring. When each individual creates different variations of egg or sperm, this ensures their offspring will have their own unique genetic makeup.
6. The cell that divides by meiosis to create the gametes is diploid because it contains two sets of chromosomes (referred to as a diploid cell). The resulting gametes (either egg or sperm) contain one set of chromosomes and are called haploid cells.
7. Synapsis occurs when the homologous chromosomes clump together. Also, the pairs of homologous chromosomes (called tetrads) line up in the middle during metaphase I and are split up into separate cells during anaphase I.
8. When the two pairs of homologous chromosomes segregate into separate cells during anaphase I, the resulting daughter cells are haploid (because they contain only one of each homologous chromosome pair).
9. Meiosis creates gametes with different genetic combinations through the process of crossing over (when segments of two homologous chromosomes swap with one another) and independent assortment (the homologous chromosomes line up in a different pattern during metaphase I).
14
Mendelian Genetics
1.
1. e
2. b
3. f
4. a
5. g
6. c
7. d
2. In this example, we would use R to indicate the dominant red allele and r for the recessive ebony allele. The heterozygous red-eyed fly has a genotype of Rr, and the ebony-eyed fly must be rr.
The percentage of offspring with the recessive ebony eyes (rr) would be ½, or 50%.
3. If the male cat is heterozygous for both traits, his genotype must be BbSs. The female shows both recessive phenotypes, so her genotype must be bbss. When determining the odds that the kitten will have white, long fur (bbss), create two separate Punnett squares to determine the odds of the kitten having either white fur or long fur.
In regards to fur color, the odds of producing a white kitten are 50%, or ½. Now set up a second Punnett square looking only at the fur length:
In regards to fur length, 25% (or ¼) of the kittens would have long fur. The question is asking you about the odds of producing a white kitten with long fur, so using the rule of multiplication, multiply the odds of producing each of these traits independently:
½ × ¼ = ⅛ … there’s a 1 in 8 chance of these two cats producing a white, long-haired kitten!
4. The man is XCY (he has the gene for normal color vision), and you know that the woman must be XCXc because her father is color-blind (and the father provides the Xc chromosome to his daughter).
Of their four offspring, they could produce a colorblind boy (¼ chance).
5. The purple allele is dominant because it masks the white allele in the second generation (they are all purple).
6. False: It will always result in all the offspring having the dominant phenotype (because all the offspring are heterozygous and the dominant allele will be the “winner”).
7. The white rooster (recessive phenotype) is heterozygous for a large comb, so his genotype is bbLl. The heterozygous hen is BbLl. In regards to feather color:
The odds of creating a chicken with white feathers is ½.
In regards to comb size:
There is a ¾ chance that the offspring will have a large comb. Therefore, the odds of creating a chick with both white feathers and a large comb is ½ × ¾ = ⅜.
8. The color-blind man has the genotype XcY. The woman must be XX. Therefore:
In order for a daughter to be a carrier, they must have one allele for color blindness (XcX). There is a 100% chance that their daughter will be a carrier for color blindness.
15
The Theory of Natural Selection
1. Evolution is change over time. Natural selection is the mechanism that explains how it happens.
2. The current belief at the time was Earth was only a thousand years old, and thus not enough time existed to allow evolution to occur. The geologists, however, provided the insight that Earth was actually billions of years old, providing plenty of time for the slow process of natural selection. Furthermore, Earth itself could change during that huge amount time.
3. The principle of inheritance of acquired characteristics believes that traits acquired during one’s lifetime could then be passed on to the offspring. This isn’t true, because a trait must be genetic for it to make it to the next generation.
4. An adaptation is a helpful trait found in a population that is the product of natural selection.
5. A species of woodpecker that evolved to possess brightly colored head feathers is an example of microevolution. The ancestral woodpecker that gave rise to many different species better adapted for their particular environment is an example of macroevolution.
6. a—4; b—3; c—2; d—1
7. In natural selection, the environment selects the most-fit phenotype. In artificial selection, humans select the most-fit phenotype.
8.
a. Any beneficial genetic mutation will be passed on to the next generations. False. Only mutations in the gametes (eggs and sperm) are able to be passed on to the next generation.
b. Populations, but not individuals, are able to evolve. True.
c. Once a most-fit phenotype is selected through evolution, the population’s adaptations are “fixed” and will no longer change. False. An adaptation that was considered beneficial at one point in a population’s lifetime can change. If the environment changes, so do the adaptations.
9. First, there needs to be variation in a population (individuals within a population are born with genetic differences). There are too many offspring for the environment to support. Due to the limited resources, there is a lot of competition amongst members of a population, and any individual that has a trait that makes it better able to outcompete others will end up surviving. These most-fit individuals are the ones left to reproduce and pass on their specific genes to the next generation. This results in a larger percentage of the next generation having these “most-fit” alleles.
16
The Evidence for Evolution
1. The more similar the cytochrome C sequence, the closer the relation. In order from most closely related to least related: mouse, donkey, carp, corn, euglena.
2. Tail, limb buds, and pharyngeal arches
3. A butterfly wing and a bird wing are examples of analogous structures because they do not arise from a common ancestor.
4. The leaves of a cactus have been modified into spines, and the Venus flytrap’s “mouth” is also a modified leaf. The spines and the “mouth” are examples of homologous structures in plants.
5. The lower the sediment layer, the older the fossilized species.
6. A previous question listed the percent similarities between different organisms and a human:
We are mammals, so it makes sense that the two mammals on the list—a mouse and a donkey—would be our closest relatives. A carp (fish) isn’t a mammal, but it is a vertebrate, so definitely related to us. A plant (a multicellular eukaryote like us) is next, leaving the single-celled protist (Euglena) as our most distant relative.
7. What is the relationship between homologous and vestigial structures? A vestigial structure is a homologous structure in a species that is no longer functional.
8. What is the driving force behind the development of analogous body structures? The two unrelated organisms have similar features because they live in a common environment. Their analogous body part was an adaptation for this common environment.
9. Hutton’s theory of gradualism stated that geological structures were formed through slow changes over long periods of time.
17
Microevolution: Evolution of Populations
1. (1) No mutations allowed, (2) only random mating, (3) the population needs to be large, (4) no gene flow is allowed, and (5) no natural selection.
2. Recall that you should always start by finding the frequency of the recessive (q) allele. Tan is the recessive phenotype (bb) and 40% of the snails are tan, then bb = q2 = 0.4. Take the square root of 0.4 to get q = 0.63. By applying the Hardy-Weinberg equation that summarizes the two alleles (p + q = 1), you can determine that p equals 1 — 0.63 = 0.37. The first question asks what percentage of snails are heterozygous. You know from the second equation that 2pq equals the percentage of heterozygous critters, so by plugging in your p and q values … 2(0.63)(0.37) = 0.47. The number of heterozygous snails would be 118 (rounded) snails. The number of homozygous dominant individuals would be 251 × (p2), or 251 (0.37)2 = 35 (rounded) homozygous dominant snails.
3. If a population suddenly suffers a drastic reduction in numbers due to a catastrophic lava flow, the remaining population will eventually repopulate. The reestablished population has very little genetic diversity due to the bottleneck effect.
4. This is an example of disruptive selection. The most-numerous medium-smelly bugs are removed, and the two extremes (slightly smelly and extremely smelly) are now the most-fit survivors and will become the most-represented phenotypes.
5. Write the Hardy-Weinberg equation that tallies the different alleles for a given gene: p + q = 1. Next, write the equation that summarizes all the given genotypes possible: p2 + 2pq + q2 = 1. The homozygous dominant genotype is represented as p2, the homozygous recessive genotype is q2, and the heterozygous genotype (both pq and qp) is 2pq.
6. 
7. Evolution in a population that occurs because of a chance change in allele frequency is referred to as genetic drift. For example, a small number of individuals may leave their original population and start their own. If these colonizing individuals happen to carry an assortment of alleles that are not similar to their original population, their new population will have a different allele frequency. This is an example of founder effect.
8. Stabilizing selection reinforces the current most-fit phenotype by removing organisms with the extreme phenotypes.
18
Macroevolution: Evolution of Species
1. Since the beetles’ environments are different colors, there will be different phenotypes that are considered “most fit.” For example, on the side with dark soil and not much grass, those beetles that happen to have a darker exoskeleton would be more camouflaged and have a better chance at surviving (not being eaten). This would lead to a change in the gene pool to have a higher percentage of dark-colored beetles. The opposite would occur on the side with more grass coverage; the beetle population would evolve to have a higher percentage of green-colored exoskeletons. If these two populations of beetles are no longer able to interbreed because brown beetles only choose to mate with other brown beetles (and green beetles mate only with green beetles), speciation has occurred!
2. One species of flower can be fertilized by another species of flower, but once the egg and sperm fuse, their differing chromosome numbers prevents the formation of a viable zygote. This is an example of offspring viability, a type of post-zygotic barrier that prevents the production of offspring between two different species.
3. About 200 years ago, an ancestral population of flies laid their eggs on tree fruits called hawthorns. Once domestic apple trees were introduced, some flies in the population chose to lay their eggs on apples, instead. The maggots that developed in the hawthorn fruit grew into adult flies who preferred to mate with other hawthorn-reared flies; maggots from apples developed into flies who preferred to mate with other apple-grown flies. This is an example of behavioral isolation, a type of pre-zygotic barrier that prevents the production of offspring between two different species.
4. Both terms refer to a genetic change in a species over time. Microevolution focuses on a change occurring in the gene pool of a specific species, whereas macroevolution looks at the bigger picture of how different species arise from an ancestral species (due to changes in their gene pools).
5. In order for speciation to occur, there needs to be some sort of barrier that divides the current population’s gene pool, thus interrupting the gene flow between the two resulting groups. Next, the isolated gene pools need to change/evolve in a way that renders them unable to interbreed if brought back together.
6. One species of field cricket mates in the spring, whereas a second species of field cricket mates in the fall. This is an example of temporal isolation, a type of pre-zygotic barrier that prevents the production of offspring between two different species.
19
Phylogeny and Vertebrate Evolution
1. Humans and chimpanzees evolved from a common ancestor.
2.
a. Node 3
b. Species A
3.
4. Species E is most closely related to species D.
5. Lungs and four limbs
6.
a. Salmon
b. Node X
20
Plant Structure and Transport
1. Dermal tissue is the skin of the plant. It can protect the plant against water loss by secreting a waxy coating. Ground tissue composes most of the “filling” of the plant and is specialized for photosynthesis and storage. The vascular tissue is the circulatory system of the plant, responsible for moving glucose and water around the plant’s body.
2. Parenchyma cells are the “typical” plant cell, whose main roles are photosynthesis and storage. Collenchyma cells provide structure without restraining growth. Sclerenchyma cells are specialized for structure and are no longer able to elongate.
3. The upper, sunny side of the leaf is composed of a palisade layer that is tightly packed with cells filled with chloroplasts. A leaf has openings in the bottom to allow carbon dioxide gas to enter (required for photosynthesis). The gas is able to move through the leaf tissue into the palisade layer because it can freely travel through the airspaces in the spongy layer. The vascular tissue (both xylem and phloem) forms the veins of the leaf, providing easy access to water and a place to deposit the glucose.
4. The root cells actively move ions from the soil into the root’s xylem. This lowers the water potential in the root xylem (compared to the water potential in the surrounding soil). Water will move from [higher water potential] → [lower water potential], so water will passively flow from the soil into the root. This provides a bit of “push” up the body of the plant.
5. Both phloem and xylem are components of the plant’s vascular tissue. Phloem moves downward, transporting photosynthetically derived glucose from the leaves down toward the roots. Xylem, on the other hand, moves upward, transporting water taken in to the roots up toward the rest of the plant.
6.
a. Has two cell walls (S)
b. Found in strings of celery because they provide structure without restraining growth (C)
c. Has lignin in its cell walls (S)
d. Primary function is photosynthesis (P)
e. The reason a pear has a gritty feel (S)
7. The process of transpiration is the major force that pulls water from the roots up a plant’s body. As water evaporates from the top of the plant, it pulls a chain of water molecules up the xylem. Each water molecule creates a chain through the process of cohesion, and the entire chain sticks to the sides of the hollow tracheids and vessel elements through the process of adhesion.
21
Plant Reproduction
1. It has both sperm-producing anthers and egg-producing ovaries.
2. In a flower, the anther produces the pollen grains, each of which carries sperm.
3.
4. Once fertilization occurs, the ovary develops into fruit, and the ovule develops into the seed coat (protecting the embryo). Since the embryo is stuck inside a protective coating, it needs a packed lunch to keep it alive until the embryo grows into a photosynthesizing plant; this food source is the endosperm.
5. Imperfect flowers cannot self-fertilize because an imperfect flower has ONLY eggs or sperm.
6. Pollination is when a pollen grain lands on the stigma. Fertilization is when the egg in the ovary is fertilized by one of the two sperm that travels down the style through the tube created by the third cell in the pollen grain, called a tube nucleus. The second sperm will fertilize the polar nuclei, turning into food for the embryo, called endosperm.
7. [outermost] petals → ovary → ovule → egg [innermost]
8. The fruit itself does not help protect the embryo (that is the seed coat’s job). The fruit, however, may be eaten by an animal that passes the indigestible seeds through its poop, thus helping to spread the seeds (and embryos) throughout the ecosystem.
22
Introduction to Animals
1. Animals are defined as heterotrophic organisms, meaning that they can only obtain their nutrients by eating.
2. Nematodes and arthropods
3. Flatworms’ flatness creates a high surface-area-to-volume ratio. This allows oxygen to easily diffuse from the environment and permeate their tissues. Therefore, they do not need a circulatory system to transport oxygen around their bodies.
4. Based on the phylogenetic tree, the phylum that includes echinoderms/sea stars is the closest relative of our phylum, the chordates.
5. True.
6. Nervous tissue and muscle tissue.
7. They have radial symmetry and, instead of a brain, they possess a nerve net.
8. Arthropods are the most numerous (it includes all insects!).
23
Animals Need Homeostasis
1. Maintaining a stable internal environment prevents dangerously wild swings in physiological conditions (such as temperatures, pH, salt levels, etc.).
2. A negative feedback loop always counteracts a stimulus to bring conditions back to a set point. Positive feedback, on the other hand, increases the stimulus in order to increase the response even further.
3.
4. The process of maintaining proper levels of water and solutes in the blood is referred to as osmoregulation. The millions of small nephrons in the kidneys are able to adjust the amount of water reabsorbed back into the bloodstream by changing the number of aquaporin proteins that line the collecting ducts. The chemical signal that leads to an increase in the numbers of these transport proteins is antidiuretic hormone (ADH).
5. When maintaining homeostatic levels of glucose in the bloodstream, the hormone insulin is responsible for lowering blood glucose levels. One way it removes free glucose from the bloodstream is by stimulating the liver to take in glucose and form the storage polymer, glycogen.
6. Uric acid is a solid, so when it is produced by the developing embryo that is stuck in the egg, it will form a solid precipitate (instead of toxifying the fluid environment).
7. The hypothalamus would increase ADH secretion, leading to an increased number of aquaporins embedded in the kidneys’ nephrons (this leads to more water being reabsorbed back into the bloodstream). The brain would also send signals of “thirsty!,” increasing water intake.
8. Land animals have kidneys that work more like marine fish because land animals are also battling water loss. The kidneys save water by reabsorbing it back into the bloodstream instead of allowing too much to be excreted in the urine.
24
Animal Digestion
1. The first stage of digestion is ingestion, when you take a bite of food. That food is broken up during the process of digestion, when large molecules are broken up into smaller components. The small components are more easily taken up into the bloodstream through the process of absorption. Finally, any indigestible material is passed out of the body through elimination.
2. Hydrolysis reactions break up macromolecules (protein, fats, nucleic acids, and large polysaccharides) into monomers.
3. The lining of the small intestine is highly folded, increasing its surface area. The higher the surface area, the faster the rate of nutrient absorption through the intestinal wall into the capillaries. The folds are called villi, each covered with cells with hair-like projections called microvilli.
4.
5. Mechanical digestion (such as chewing up food and bile’s emulsification of fat) is simply breaking up big pieces into little pieces; no covalent bonds are broken and no enzymes are used. Chemical digestion, on the other hand, is driven by enzymatic hydrolysis and covalent bonds are broken.
6.
1. k
2. j
3. e
4. f
5. i
6. g
7. h
8. a
9. c
10. d
11. b
7. The tiger is a carnivore, whose diet consists of mostly meat (and very little cellulose). Since the cecum plays a role in cellulose digestion, the tiger would have a small cecum. The rabbit, on the other hand, is an herbivore with an entirely plant-based diet. In order to help digestion, the rabbit would evolve to have a very large cecum.
8. Interesting poop trivia! About a third of your poop is composed of bacteria, and poop is brown because of pigments produced from the breakdown of red blood cells/hemoglobin/bile/bilirubin (any of these answers work!).
25
Animal Circulation and Respiration
1. The organism’s body must have a high surface-area-to-volume ratio.
2. A four-chambered heart has two separate ventricles separated by a septum. This prevents oxygen-poor blood in the right ventricle from mixing with oxygen-rich blood in the left ventricle.
3. The place of gas exchange between the respiratory system and the circulatory system occurs in the alveoli of the respiratory system and the surrounding capillaries of the circulatory system.
4. The partial pressure of oxygen would be higher than that of carbon dioxide in the bloodstream after it leaves the lungs (having dropped off CO2 and picked up O2), before it enters the oxygen-starved tissues.
5. When prokaryotic cells became too large, their SA:VOL ratio decreased. In order to increase their membrane surface area to compensate for their larger volumes, eukaryotic cells evolved to have compartmentalized interiors (organelles).
6. The mammalian left ventricle has a thicker wall than the right ventricle. The left ventricle needs to be stronger because it powers the blood through the entire body (the systemic circuit). The right ventricle only needs to power the blood out to the lungs (the pulmonary circuit).
7. The respiratory system brings in the oxygen that is picked up by the circulatory system. After the circulatory system moves the oxygen to the tissues, it picks up the waste carbon dioxide the tissues generated through cellular respiration. The circulatory system moves the carbon dioxide back to the lungs, where the respiratory system expels (exhales) it from the body.
8. Blood coming from the lungs and entering the tissues has a high oxygen partial pressure. The tissues have a high carbon dioxide partial pressure because the cells have been producing a lot of carbon dioxide through the process of cellular respiration. These partial pressures facilitate the movement of oxygen from the blood into the tissues.
26
Animal Neurons and Signal Transduction
1. A resting neuron has a higher concentration of K + on the inside of the cell and a higher concentration of Na + on the outside of the cell. The net charge on the inside of the cell is negative.
2. A neuron transmits a signal by generating an action potential, which is a wave of positive charge generated when sodium ions rush into the cell.
3. e-b-d-c-a-f
4. A neuron actively transports 3 Na+ ions out for every 2 K+ ions brought in. This creates a net movement of positive charges out of the cell. The inside of the neuron also has large proteins that carry an overall negative charge, further contributing to the negative interior.
5. g-f-b-d-a-e-c
6. When an action potential reaches the end of the presynaptic neuron, it must rely on the chemicals called neurotransmitters to cross the space between adjacent neurons (the synapse). Once the chemicals bind to the postsynaptic neuron, sodium channels will open and the action potential will continue.
27
The Animal Immune System
1. The first and second lines of defense are nonspecific, meaning the response occurs regardless of the antigen. Conversely, the third line of defense is specific for the particular invading antigen.
2. 1-E, 2-C, 3-H, 4-G, 5-D, 6-A, 7-B, 8-F
3. The cell-mediated response targets intracellular pathogens such as viruses and relies on T cells.
4. Once the particular lymphocyte (either B or T cell) is selected for by a particular antigen, it will produce two population of cloned cells: the active cells that take care of the immediate infection, and a second population of memory cells that last for years (even decades) awaiting a second invasion. If the same antigen presents itself again, the memory cells respond immediately to remove the invader before it has a chance to cause problems. Vaccines work in the same way, except the initial “infection” is only a part of the virus/bacterium, just enough to stimulate a lymphocyte response (but not itself able to cause the disease).
5. The other barriers of your body provide the first line of defense. This includes skin, sweat, saliva, and tears.
6. The humoral response targets extracellular pathogens and relies on B cells to produce antibodies to bind to and help remove the invader.
7. Passive immunity is when antibodies are given to the patient (and are not produced by the patient’s own B cells, as they are in active immunity). Passive immunity is an important treatment option when the patient does not have the luxury of time to produce their own antibodies (venom from a snake bite, or babies receiving antibodies from their mother).
8. The flu virus mutates every year, and the previous year’s vaccine will no longer be close enough of a “match.”
28
Populations
1. 1-f; 2-d; 3-e; 4-a; 5-c; 6-b
2. A species with a Type III survivorship curve would need to produce a large number of offspring because most will die off very early in life. Plants, insects, and heavily preyed-upon mammals exhibit a Type III survivorship curve.
3.
The logistic curve is more realistic because the initial exponential growth is unsustainable. Resources will eventually become limited and population growth will slow and reach the carrying capacity.
4. Mutualism, because both the leafhopper and the bacteria benefit from the arrangement.
5. A population with a low death rate and a high birthrate would experience an overall increase in the population numbers. This population would most likely have an age structure with mostly young individuals.
6.
Large mammals exhibit the Type I survivorship curve. The probability of survival remains high throughout their lives (which is good, considering the relatively few offspring produced), and most die after they reach old age.
7.
a. A viral disease (DD)
b. Competition for mates (DD)
c. Volcanic eruption (DI)
d. Predation (DD)
e. Wildfire (DI)
f. Oil tanker spill in ocean (DI)
g. Clearcutting forest (DI)
8. To be commensalism, the louse must benefit, and the fish must be neither harmed nor helped. If the louse perfectly replaces the function of the fish’s tongue (and no harm comes to the fish), it would be a commensal relationship.
29
Ecosystems and Their Interconnectivity
1. A habitat only refers to where it lives. A niche, however, also takes into consideration its role in the environment: what it eats, how it reproduces, if its prey for other organisms, when it sleeps, and so on.
2. The graph depicts the competitive exclusion principle. Both species are initially growing well, but at day 4, resources become limiting and species A outcompetes species B. Species B eventually dies off, leaving only species A to occupy this particular niche.
3. Each step in a food chain depicts the transference of energy as one organism eats another. The second law of thermodynamics state that every time energy is converted from one form to another (such as an herbivore eating plants and then transforming the energy of the plants’ tissues into ATP), the process is not 100% efficient. In fact, only a measly 10% of the plants’ stored energy is able to be used by the herbivore. That means that, eventually, the available energy will become too small to support another step in the food chain (another trophic level).
4. Photosynthesis is the only process through which carbon dioxide is removed from the atmosphere. Once it is locked into tissues (thanks to food chains), carbon dioxide is released back into the atmosphere through cellular respiration and decomposition. Burning of fossil fuels also releases a massive amount of carbon dioxide.
5. The competitive exclusion principle states that two different species cannot occupy the same ecological niche. The survivor is the one who was able to outcompete the other.
6. The first trophic level must be a producer (autotroph) because every chain is built upon the energy provided by the sun. The only type of organism able to harness photons of sunlight are autotrophs, either photosynthetic plants, protists, or bacteria.
7. Earth is a closed system in regards to matter, but an open system in regards to energy.
8. The first step of the nitrogen cycle is nitrogen fixation, when nitrogen gas is fixed into organic molecules. The two means of nitrogen fixation are lightning and nitrogen-fixing bacteria.
30
Human Impact on Ecosystems
1. Both primary and secondary succession are processes by which ecosystems reestablish their populations after a disturbance. The difference is that primary succession must start from scratch, re-forming soil from the rocks that remain. It takes much longer than secondary succession (which has intact soil).
2. A renewable resource can be replaced at the same rate by which it is consumed. Give an example of this type of resource: solar, wind, hydroelectric, or geothermal energy.
3. Secondary pollutants are created when primary pollutants undergo chemical reactions in the atmosphere. Primary pollutants are released directly into the atmosphere from burning fossil fuels.
4. When excess fertilizer enters a body of water, the phosphorus causes an algal bloom (phosphorus is normally a limiting factor and keeps algal growth in check). The overgrowth of algae soon begins to die, and decomposers begin to consume the dead algae. The decomposers are undergoing massive amounts of cellular respiration and using up tons of oxygen in the process, leading to hypoxic (low oxygen) conditions in the water. This causes a die-off of aquatic animals and plants.
5. a. Chernobyl nuclear contamination (SS), b. California wildfires (SS), c. Mount St. Helens volcanic eruption (either PS or SS, depending on severity of damage. If soil remains intact, it is SS.), d. Deforestation of Amazon rainforest (SS), e. Retreating glaciers in Alaska (PS).
6. Lichen is a pioneer species that is able to live on the newly exposed rocks and help break them down into soil (the first step of primary succession).
7. Burning fossil fuels releases carbon dioxide into the atmosphere. The CO2 in the atmosphere (along with other greenhouse gases) absorbs the infrared radiation released from the heated earth, warming the air.