Chapter 1. QUICK CHECK ANSWERS

Chapter 1 Quick Check Answers

  1. One hypothesis is that their tan fur protects the mice from predators by allowing them to blend in with their surroundings. To test this idea, you could devise an experiment in which dark and tan mice (or, to be more humane, artificial mice) are exposed to a predator, such as a hawk, on a variety of backgrounds. Do hawks preferentially attack dark mice on light sands and light mice on dark backgrounds?
  2. One test would be to take two large populations of laboratory mice and introduce cigarette smoke into the cages of one population but not the other. Do mice subjected to smoke develop cancer at rates that are significantly higher than those of the control group? We can also make observations of human populations: Do smokers develop lung cancer at rates significantly higher than nonsmokers?
  3. DNA directs the formation of proteins that do the cell’s work. Mutations can result in proteins with altered structure and, therefore, different functions. Changes in protein function can cause the cell to work improperly, or fail altogether, resulting in some cases in disease.
  4. The use of antibiotics selects for antibiotic resistance among bacteria. Bacteria do not become antibiotic resistant out of “need” or because resistance would be advantageous for them. Instead, before the application of antibiotics, bacteria with antibiotic resistance exist in low numbers. The application of antibiotics allows these bacteria to grow and reproduce more successfully than those that are susceptible to antibiotics. Mutations, such as those responsible for antibiotic resistance, are not influenced by whether or not the organism is in an environment in which that mutation would be advantageous.

Chapter 2 Quick Check Answers

  1. One conclusion is that atoms consist mainly of empty space, and hence most positively charged particles passing through the gold foil do not come close enough to any positive charge to be deflected. Another conclusion is that the positively charged protons in the nucleus must be small and densely packed.
  2. Hydrogen and lithium are in the same column, or group, in the periodic table. They both have one valence electron in their outer orbital. As a result, one atom of lithium combines with one atom of hydrogen to make lithium hydride, with a full complement of two electrons in the single molecular orbital.
  3. Ice is less dense than liquid water. As a result, when water freezes it expands in volume and can burst closed containers, such as cans of soda or water pipes in houses. This property is unusual. For most substances, the solid phase is more dense than the liquid phase.
  4. Glucose and galactose differ only in the orientation of the –OH and –H groups attached to carbon number 4.

Chapter 3 Quick Check Answers

  1. Because R = A or G, then %R = %A + %G; and because %A = %T and %G = %C, we can write %R = %A + %G = %T + %C. But T or C = Y, and so %T + %C = %Y. It follows that %R = %Y.
  2. The RNA transcript has the sequence 5'–AUCGCUGAAAGU–3'.
  3. The incorporation of a nucleotide with a 3' H rather than a 3' OH will stop subsequent elongation because the 3' OH is necessary to attack the high energy phosphate bond of the incoming nucleoside triphosphate. The incorporation of a nucleotide with a 2' H rather than a 2' OH will have no effect on elongation, as this group is not involved in the polymerization reaction.
  4. The eukaryotic DNA sequence contains introns, which the bacterial cell cannot splice out properly, and so the correct protein is not produced from the information in the bacterial RNA transcript.

Chapter 4 Quick Check Answers

  1. The sequence of amino acids in a protein determines how a protein folds, so a change in even a single amino acid can affect the way the protein folds and can disrupt its function.
  2. The three reading frames are
    UUU GGG UUU GGG…, which codes for repeating
    Phe–Gly–Phe–Gly…
    UUG GGU UUG GGU…, which codes for repeating
    Leu–Gly–Leu–Gly…
    UGG GUU UGG GUU…, which codes for repeating
    Trp–Val–Trp–Val…
  3. With proper eukaryotic processing, the RNA transcript from the bacterial DNA will be capped at the 5' end. The initiation complex will form at the 5' cap and move along the mRNA until the first AUG codon is encountered, and then translation begins. When one of the termination codons is encountered, the polypeptide is released. Translation of the downstream polypeptides cannot take place because the Shine–Dalgarno sequences preceding them are not recognized by the eukaryotic translational machinery.
  4. A mutation that decreases survival or reproduction decreases in number in each generation because nonmutant forms leave more offspring. Eventually the harmful mutation disappears from the population because its final carriers failed to survive or reproduce. In the extreme case when the harmful mutation causes death or sterility, it disappears in only one generation.

Chapter 5 Quick Check Answers

  1. Saturated fatty acids have reduced membrane mobility compared to that of unsaturated fatty acids. As a result, fatty acids tend to be solid at room temperature, while unsaturated fatty acids tend to be liquid. Margarine and animal fats contain saturated fatty acids and are solid, while many plant and fish oils, containing unsaturated fatty acids, are liquid at room temperature.
  2. Water molecules move in both directions, but the net movement of water molecules is from side A to side B. This process is called osmosis: Water moves from regions of higher water concentration to regions of lower water concentration. Likewise, sodium and chloride ions move in both directions, but the net movement of sodium and chloride ions is from side B to side A. Movement of water and ions results from diffusion, the random motion of substances. Even when the concentration of all molecules is the same on the two sides, diffusion still occurs, but there is no net movement of water molecules or ions.
  3. If the sodium-potassium pump is poisoned, cells tend to swell and even burst, as the extracellular fluid becomes hypotonic relative to the inside of the cell and water moves into the cell by osmosis.

Chapter 6 Quick Check Answers

  1. No, because the second law of thermodynamics applies to the universe as a whole. This means that we have to consider not just the air in the room but the heat released to the outdoors as well. An air conditioner produces more hot air than cold air, and therefore total entropy increases, as expected from the second law of thermodynamics.
  2. Increasing the temperature increases the value of T∆S, which decreases ∆G, since ∆G = ∆H - T∆S. As a result, an increase in temperature makes it more likely that a reaction will proceed without a net input of energy.
  3. Enzymes increase the reaction rate and decrease the activation energy. The other parameters are not changed by enzymes.

Chapter 7 Quick Check Answers

  1. The higher-energy forms are ATP, NADH, FADH2, C6H12O6, and reduced molecules; the lower-energy forms are ADP, NAD+, FAD, CO2, and corresponding oxidized molecules.
  2. At the end of glycolysis, the energy in the original glucose molecule is contained in pyruvate, ATP, and NADH.
  3. At the end of the citric acid cycle, the energy in the original glucose molecule is held in ATP, NADH, and FADH2.
  4. Oxygen is consumed in cellular respiration. Oxygen is the final electron acceptor and is converted to water in the electron transport chain.
  5. Uncoupling agents decrease the proton gradient and therefore decrease the levels of ATP. The energy of the proton gradient is not used for oxidative phosphorylation, but instead is dissipated as heat. Uncoupling agents are found naturally in certain tissues, such as fat, for heat generation. They can also act as poisons.
  6. Yeast cells are eukaryotes. In breadmaking, yeast use sugar as a food source for ethanol fermentation. The carbon dioxide produced in the process causes the bread to rise. The ethanol is removed in the baking process.

Chapter 8 Quick Check Answers

  1. You should label the oxygen in CO2 (that is, inject C18O2) because the entire CO2 molecule is combined with RuBP by rubisco, whereas H2O is the electron donor needed for the reduction step of the Calvin cycle. The extraction of electrons from water releases O2 as a by-product.
  2. NADPH supplies the major input of energy that is used to synthesize carbohydrates in the Calvin cycle.
  3. Antenna chlorophylls transfer absorbed light energy from one antenna chlorophyll molecule to another, and ultimately to the reaction center. Reaction center chlorophylls transfer absorbed light energy and a high-energy electron to an electron acceptor, resulting in the oxidation of the reaction center chlorophyll molecule.
  4. The energy difference between the oxidation of water and the reduction of NADP+ is greater than the amount of energy that can be harvested when a single photosystem absorbs light energy.
  5. Rubisco faces a fundamental trade-off between selectivity and speed because it can use both CO2 and O2 as substrates. High selectivity of CO2 over O2 requires that the reaction have a high energy barrier, leading to a lower catalytic rate.
  6. Like cellular respiration, photorespiration releases CO2. Unlike photorespiration, it consumes rather than produces ATP.

Chapter 9 Quick Check Answers

  1. Only cells that have receptors for the hormone respond to the signal. Therefore, signaling can be specific for particular cells.
  2. No. A G protein-coupled receptor is a transmembrane receptor protein, which interacts with another protein called a G protein located inside the cell on the cytoplasmic side of the plasma membrane.
  3. The signal to increase heart rate carried by adrenaline can be reversed in at least four ways: (1) Decreasing the amounts of the signaling molecule available to bind and activate the G protein-coupled receptor; (2) inactivating the G protein; (3) decreasing the amount of second messenger cAMP; and (4) dephosphorylating the target proteins that caused the increased rate of contraction of the muscle cells.

Chapter 10 Quick Check Answers

  1. A defect in dynein would cause the melanin granules to remain dispersed since dynein transports the granules back toward the minus end of the microtubule during granule aggregation.
  2. Adherens junctions form a belt around cells, whereas desmosomes are button-like attachments. In addition, adherens junctions connect to actin microfilaments, whereas desmosomes connect to intermediate filaments.
  3. Tight junctions prevent the passage of materials through the space between cells. Gap junctions in animal cells and plasmodesmata in plant cells permit the exchange of materials between the cytoplasm of adjacent cells.
  4. Integrins are responsible for the production of milk proteins by mammary cells in response to the extracellular matrix because integrins bind extracellular matrix proteins. By contrast, cadherins bind to other cadherins when cells adhere to other cells.

Chapter 11 Quick Check Answers

  1. A mutation that disrupts the function of the FtsZ protein will block cell division.
  2. Sister chromatids are the result of DNA replication during S phase, and so they have identical DNA sequences (with the exception of a few changes due to rare mutations). The two homologous chromosomes are inherited from two different parents. The DNA sequences of these chromosomes are therefore similar, but not identical.
  3. A cell that undergoes mitosis but not cytokinesis will become a single cell with two nuclei (and therefore twice the normal amount of DNA); this type of cell is called a multinucleate cell.
  4. In human cells at the end of prophase I, there are 92 chromatids, 46 centromeres, and 23 bivalents.
  5. In meiosis I, homologous chromosomes pair, undergo crossing over, and segregate from each other. These events do not occur in mitosis. In mitosis, centromeres divide and sister chromatids separate, events that do not take place in meiosis I.
  6. The products of meiosis are different from each other as a result of two key processes: (1) crossing over, which occurs at essentially random positions along the chromosomes and creates unique combinations of genetic differences that may be present in the maternal and paternal chromosomes, and (2) random orientation of the homologous chromosomes on the spindle in metaphase I, so each nucleus receives a random combination of maternal and paternal homologs.
  7. The function of the p53 protein can be disrupted by a mutation in the p53 gene. Alternatively, certain viral proteins, such as the E6 protein of HPV discussed in Case 2: Cancer, can interfere with the function of p53.
  8. An oncogene causes cancer by producing an excess of protein activity that pushes the cell to divide. A tumor suppressor like p53 functions oppositely. Its normal function is to prevent cell division and its absence is what allows the cell to divide uncontrollably.

Chapter 12 Quick Check Answers

  1. After one round of replication, you would predict only 14N/15N hybrid DNA, which has a density of 1.715 gm/cm3. After two rounds of replication, you would predict half the molecules to be 14N/15N hybrid (density 1.715 gm/cm3) and half to be 15N/15N heavy (density 1.722 gm/cm3).
  2. There are many possible reasons that might explain the absence of a band following PCR and gel electrophoresis. The DNA polymerase might be defective or have low activity, or you may have inadvertently left out one of the components of the reaction, or the primer sequences might have been incorrect so that there was not a complementary sequence for them to anneal to, or you may have connected the positive and negative leads to the wrong ends of the gel box so that the products ran off the top of the gel.
  3. The sequence of the template strand is antiparallel to the synthesized strand and inferred from complementary base pairing as 5'-ACTCGGTAGT-3'.
  4. The value of sticky ends is that they give the researcher greater control over which restriction fragments can come together and be attached. Sticky ends can pair only with other sticky ends that have complementary 3' and 5' overhangs. Hence, restriction fragments produced by Bam HI can combine only with other fragments produced by Bam HI, and not with fragments produced by Hin dIII, for example. However, any blunt end can be attached to any other blunt end, for example, a Hpa I end to a Sma I end.

Chapter 13 Quick Check Answers

  1. DNA sequencing technology is limited to DNA molecules that are less than about 1000 nucleotides, so the challenge of genome sequencing is the size of many genomes, which can far exceed 1000 nucleotides. Sequencing whole genomes required a method of piecing together smaller sequenced DNA fragments.
  2. Even when short repeated sequences can be sequenced, there is an assembly problem similar to that for longer repeats in which the researcher has no way of knowing where in the repeat any particular sequenced fragment should be assigned. The result is that the total number of repeats remains unresolved. It could be in the hundreds, thousands, tens of thousands, hundreds of thousands, or even millions.
  3. The observation that closely related lentiviruses have closely related host organisms suggests that each type of lentivirus can infect only one or a small number of closely related organisms.
  4. You can’t tell. The C-value paradox means that you cannot predict genome size from the complexity of the organism. In fact, the genome of an amoeba is much larger than the genome of a human.

Chapter 14 Quick Check Answers

  1. Mutant genes that are harmful or neutral are much less likely to persist in a population than ones that result in increased survival and reproduction because of greater fitness.
  2. The number of nucleotides that are inserted or deleted is almost always an exact multiple of 3 because each codon in the genetic code consists of three nucleotides. Any insertion or deletion that includes a number of nucleotides that is not an exact multiple of 3 shifts the reading frame, and the resulting sequencing will most likely code for a nonfunctional protein.
  3. Mutations are random; they are not directed by the environment. This does not mean that the environment cannot affect the rate of mutation. Mutagens increase the rate of the mutation, but they cannot induce specific mutations that would be beneficial to the organism in response to the environment.

Chapter 15 Quick Check Answers

  1. Mutations that cause antibiotic resistance are clearly beneficial to bacteria when antibiotic is present, but are neutral or even harmful in the absence of antibiotic. The effect of a mutation on phenotype often depends on the environment.
  2. Only a small fraction of the human genome codes for proteins or other functional elements (Chapter 13), so most mutations are neutral.
  3. If the suspect is the actual source of a sample, then the DNA “fingerprints” must match exactly, and any mismatch rules out the suspect as the source of the sample. When there is a match, however, there is always a small chance (typically very small) that the sample came from another person with the same DNA “fingerprint.”
  4. In a VNTR, the restriction fragments are different lengths because of a variable number of repeats between restriction sites (one, two, three, four, or more). In a RFLP, the restriction fragments are different lengths because a restriction site is removed in one DNA sequence and not in another, making the distances between flanking sites different.
  5. A point mutation is a change in a single nucleotide in an individual cell, such as C to G. A SNP results from a point mutation that occurred at some time in the past, so that now in the population there are two or more different single nucleotides at a given position. For example, some people might have C at a certain position and others might have G at that position.
  6. Both Y chromosomes in the XYY baby must come from the father, so nondisjunction took place in the father. Consider now normal meiosis, shown in Fig. 15.12a, and suppose we let dark blue represent the X chromosome and light blue the Y chromosome. The first meiotic division separates the X chromosome from the Y chromosome, and the second meiotic division separates the sister chromatids of the X chromosome and the sister chromatids of the Y chromosome. For the Y chromosomes to remain together and be included in the same sperm, nondisjunction must take place in the second meiotic division (Fig. 15.12c).

Chapter 16 Quick Check Answers

  1. For the true-breeding plants with yellow seeds, the phenotype is yellow seeds, and the genotype is AA. For the true-breeding plants with green seeds, the phenotype is green seeds, and the genotype is aa.
  2. Yes in both cases. When an allele is recessive, the homozygous dominant and heterozygous individuals have the same phenotype but different genotypes, as in AA and Aa plants. When a trait is influenced by the environment, individuals with the same genotype can have different phenotypes because of the environment. An example discussed in Chapter 15 is how smoking tobacco increases the severity of emphysema resulting from the PiZ allele of alpha-1 antitrypsin.
  3. Half the progeny will have the AA genotype and half will have the Aa genotype. All the plants will have yellow seeds.
  4. The probability that the first pea is green is ¼; the probability that the second pea is green is also ¼; the probability that the third pea is yellow is ¾; and the probability that the fourth pea is yellow is ¾. We use the multiplication rule (because these are mutually exclusive events) to figure out that the probability of this configuration is ¼ × ¼ × ¾ × ¾ = 9/256. There are six different ways that a pod of four seeds can have two green seeds and two yellow seeds (the green seeds can be in positions 1 and 2; 1 and 3; 1 and 4; 2 and 3; 2 and 4; or 3 and 4). Each of these configurations occurs with a probability of 9/256 (as previously described). The probability of any of these occurring (either one configuration or another or another) is given by the addition rule, or 9/256 + 9/256 + 9/256 + 9/256 + 9/256 + 9/256 = (9/256) × 6 = 54/256, or about 21%.
  5. The F1 chickens have white feathers because the dominant inhibitor allele I inhibits expression of the pigment allele C.
  6. In the population as a whole, there are many copies of each chromosome, so any gene can have multiple alleles present in the different copies. Any one individual can have only two copies of any chromosome, and so any individual can have no more than two different alleles.

Chapter 17 Quick Check Answers

  1. A woman whose father is color blind must be heterozygous for the mutant allele. If she has children with a man who is color blind, then half of the female off spring are expected to be homozygous mutant and therefore color blind.
  2. Even when one (or more) crossovers occurs between the genes, only two of the four products of meiosis are recombinant because crossing over takes place at the four-strand stage of meiosis (Fig. 17.10a). With two strands that are recombinant and two that are nonrecombinant, the frequency of recombination is 2/4 = 50%, and so this is the maximum.
  3. Independent assortment means that a doubly heterozygous genotype like AB ab produces gametes in the ratio 1/4 AB, 1/4 Ab, 1/4 aB, 1/4 ab. The first and last are nonrecombinant gametes, and the second and third are recombinant gametes. The frequency of recombination is therefore 1/4 + 1/4 = 1/2, or 50%. This means that independent assortment is observed for genes that are far apart in the same chromosome as well as for genes in different chromosomes.

Chapter 18 Quick Check Answers

  1. No. Regression toward the mean simply indicates that, in any population, parents with extreme phenotypes (very tall or very short, for example) will tend to have offspring that are closer to the average height of the population.
  2. Let’s consider some possibilities, some true, some false.
    • In the development of fingerprint ridges in the embryo, genes play a predominant role.
    • In the development of fingerprint ridges in the embryo, the environment plays only a minor role.
    • Differences in the number of fingerprint ridges among individuals are determined largely by differences in genotype.
    • Differences in the number of fingerprint ridges among individuals are almost unaffected by differences in the environment.
    • The average number of fingerprint ridges in the offspring is not much different from the average number of fingerprint ridges i the parents.
    • The average number of fingerprint ridges in the offspring is almost unrelated to the average number of fingerprint ridges in the parents.
    Statements a and b are false no matter what the heritability because heritability refers to differences (variation) among individuals and not to the phenotype of any one individual. High heritability implies that statements c, d, and e are all correct. Statement f is false because such a trait would have a low heritability.
  3. The easier genes to detect are the few genes with large effects. The magnitude of gene effects that can be detected depends on the number of individuals studied. The larger the number, the smaller the effects that can be detected. Whatever the size of the study, however, the genes with the largest effects are always easier to detect, even though there may be few of them.

Chapter 19 Quick Check Answers

  1. Combinatorial control requires far fewer genes because there is not a one-to-one correspondence between the genes that are regulated and those that do the regulating. The products of a relatively small number of regulatory genes can be used in combination to control the expression of a far larger number of target genes. As an analogy, consider the virtually limitless combinations of sounds that can be played on a standard piano that has only 88 keys (36 black and 52 white).
  2. Small regulatory RNAs are not translated into proteins (Chapter 3). As a result, they are often called noncoding RNAs since they do not encode for proteins. Their function is often to regulate the expression of other genes. By contrast, mRNAs are translated into proteins.
  3. An activator turns on the expression of a gene directly by binding to DNA. An inducer binds a repressor, preventing the repressor from turning off the expression of a gene.
  4. A mutation in the repressor gene that does not allow the repressor protein to bind allactose will lead to a cell that is not able to produce (ß-galactosidase, in the presence or absence of lactose. In other words, the lactose operon would not be inducible.

Chapter 20 Quick Check Answers

  1. A cell from the inner cell mass of a blastocyst has more developmental potential than a cell from the ectoderm, which is one of the three germ layers, because the blastocyst occurs earlier in development than the formation of the three germ layers.
  2. In Chapter 19, we saw that many common traits result from an interaction of genes and the environment, and so even clones can look different from each other because of environmental influences. Furthermore, if the clone is produced by nuclear transplantation, as in the cases of Dolly and CopyCat, the nuclear genomes of clone and parent are identical, but their mitochondrial genomes are different.
  3. Yes. The embryo would develop normally, even though the embryo does not have normal bicoid function, because bicoid is a maternal-effect gene, in which the mother’s genotype affects the phenotype of the offspring.
  4. The gap genes control the expression of the pair-rule genes, which in turn control the expression of segment-polarity genes. Therefore, if gap-gene expression is not normal, you would predict that segment-polarity gene expression would also not be normal.
  5. Sexually reproducing organisms carry two copies of most genes, one from the mother and the other from the father. If either of these is knocked out by a loss-of-function mutation, the other is still present and compensates for the mutant. Hence, a loss-of-function mutation is expected to be recessive. In a gain-of-function mutation, one of the gene copies is expressed in the wrong amount, or the wrong tissue, or at the wrong time, and if such a gene turns on a developmental pathway, expression of only one copy of the gene is sufficient to turn the pathway on. Hence, a gain-of-function mutation in a gene that controls a developmental pathway is expected to be dominant.
  6. Because A and B together result in petals, and B and C together result in stamens, a flower expressing B activity in all four whorls would have, from outside to inside, petals, petals, stamens, and stamens.

Chapter 21 Quick Check Answers

  1. The allele frequency of a was calculated as follows:
    frequency(a) = [2 × (number aa) + 1 × (number Aa)]/[2 × (total number)]
    This equation can be rewritten as
    frequency(a) = [(number aa) + ½ × (number Aa)]/(total number)
    Note that
    number aa/total number = freqency(aa)
    and
    ½ × (number Aa)/total number = ½ frequency(Aa)
    Therefore,
    frequency(a) = frequency(aa) + ½ frequency(Aa)
    Stated in words, the frequency of allele a equals the frequency of aa homozygotes plus half the frequency of Aa heterozygotes. Because we were given genotype frequencies (50% aa, 25% Aa, and 25% AA), we can simply substitute into the preceding equation to solve for the frequency of a:
    0.50 + ½(0.25) = 0.625, or 62.5%
    By similar logic,
    frequency(A) = frequency(AA) + ½ frequency(Aa)
    which equals
    0.25 + ½(0.25) = 0.375, or 37.5%
    These equations are very useful for determining allele frequencies directly from genotype frequencies.
  2. There is no next step. DNA sequence provides full genetic resolution.
  3. We can conclude that the population is evolving. What we cannot tell is what mechanism—selection, migration, mutation, genetic drift, or non-random mating—is causing it to evolve. To determine what mechanism(s) is (are) driving the process requires more detailed population genetics analysis.
  4. Adaptation is the fit between an organism and its environment. Of all the evolutionary mechanisms, only selection causes allele frequencies to change based on how they contribute to the success of an individual in terms of survival and reproduction. This means that allele frequencies in the next generation are ultimately governed by the environment. Because phenotype is in part determined by genotype, organisms become adapted to their environment under the influence of selection over time.

Q-6

Chapter 22 Quick Check Answers

  1. Species change over time, making a single definition that can be applied in all cases difficult. The biological species concept (BSC) works well in many cases and is useful in that it gives us a specific phenomenon—reproductive isolation—to study. In addition, it focuses our attention on the flow of genes between individuals of a single species. However, there are plenty of cases, including asexual organisms and organisms known only from fossils, in which it is not applicable.
  2. Volcanic island chains offer many examples of adaptive radiation because they are dependent on the serendipitous process of colonization. This means that only some plants and animals are able to get there. In addition, because of the absence of competitors on these islands, there are often many available ecological opportunities for the colonizers. For example, there may be no insect-eating birds on the islands, providing an ecological niche to be filled.
  3. If populations of fish became separated from other populations of fish in different ponds during dry periods, this scenario suggests that speciation was more likely to be allopatric rather than sympatric. The allopatrically evolved species became sympatric only when the lake flooded again, combining all the separate ponds into a single body of water.

Chapter 23 Quick Check Answers

  1. No. The two trees are equivalent. The node shared by humans and mice is simply rotated in one tree compared to the other tree. In both trees, the closest relative of humans is mice.
  2. A group called “fish” is paraphyletic because it includes some, but not all, of the descendants of a common ancestor. The descendants of the common ancestor of fish also include amphibians, sauropsids, and mammals.
  3. The traits are analogous, the result of convergent evolution. Both animals are adapted for swimming in water and converged independently on similar traits, including a streamlined body and fins. However, they are only distantly related, as one is a fish and the other is a mammal.
  4. Character traits can be measured for fossils as well as living organisms, so you can examine the features of the skeleton and then construct a phylogenetic tree that takes into account the skeletal traits they share with known vertebrates.

Chapter 24 Quick Check Answers

  1. No. Modern humans and modern chimpanzees share a common ancestor. We are not descended from chimpanzees because chimpanzees are living today.
  2. The multiregional hypothesis suggests that the most recent common ancestor of all living humans lived about 2 million years ago, when H. ergaster first migrated from Africa to colonize Europe and Asia. The out-of-Africa hypothesis suggests that H. sapiens evolved much more recently in Africa and then moved out of Africa, replacing the remnants of populations originally established by H. ergaster when it left Africa. The out-of-Africa hypothesis suggests a much more recent common ancestor for all present-day humans, one that lived about 200,000 years ago.
  3. Studies of mtDNA, which is maternally transmitted, indicated that there was no input of Neanderthal mtDNA into the modern human population. In contrast, studies of Neanderthal genomic DNA showed that 1% to 4% of non-Africans’ genomes are composed of Neanderthal DNA, implying that there was interbreeding between the ancestors of non-Africans and Neanderthals, presumably when the H. sapiens population first left Africa. If there is evidence of genetic input but no evidence of a female-based genetic contribution, we can hypothesize that male Neanderthals were the key contributors.
  4. It is inaccurate to label FOXP2 the “language gene” for several reasons. First, the gene is expressed in many tissues, so its effects are not limited to speech and language. Second, many genes are required for language, not just FOXP2. Nevertheless, its association with vocal communication in multiple species is striking.

Chapter 25 Quick Check Answers

  1. The coincidence in timing between the Industrial Revolution and an increase in CO2 levels—following a millennium of little change—suggests that human activities played a role in recent changes in atmospheric composition. Alternative hypotheses would focus on other processes that add CO2 to the atmosphere. Perhaps, for example, increased volcanic activity added more CO2 to the atmosphere, or perhaps warming induced by changes in solar radiation caused thawing of permafrost, facilitating increased respiration of soil organic matter at high latitudes. These hypotheses can be tested against the historical record of volcanic activity and measurements of solar radiation over the past century.
  2. All else being equal, increasing the elevation of mountains should increase rates of chemical weathering and erosion. As chemical weathering of continental rocks consumes CO2, atmospheric CO2 levels should decline.
  3. Antelopes are primary consumers—that is, animals that eat vegetation. Lions are secondary consumers that eat antelopes. Because the transfer of energy from one level in the food web to the next higher level is inefficient, a given abundance of antelopes can support only a much smaller biomass of lions.

Q-7

Chapter 26 Quick Check Answers

  1. In prokaryotic organisms, horizontal gene transfer transports genes from one organism to another, facilitating the generation of new gene combinations.
  2. Archaeal cell and genome organization are similar to those of Bacteria. It was only through the use of molecular-sequence comparisons that Archaea were shown to occupy a distinct branch on the tree of life.
  3. Primary production could be accomplished by anoxygenic photosynthesis, which does not generate oxygen. Oxidation of organic matter to carbon dioxide could be accomplished by anaerobic respiration (and fermentation), which does not use oxygen.
  4. It wouldn’t. Bacteria and archaeons cycle nitrogen between the atmosphere (nitrogen gas) and biologically useful forms. In the absence of these prokaryotes, life as we know it would not be possible.
  5. Not necessarily. Today, hyperthermophiles live in hot springs and hydrothermal ridges on the ocean floor. These might have been the environment in which the last common ancestors of these organisms thrived.

Chapter 27 Quick Check Answers

  1. In general, eukaryotes employ a subset of the metabolic pathways used by bacteria. Most eukaryotes are capable of aerobic respiration; most can also gain at least some energy by fermentation; and some can also photosynthesize. In many ways, then, carbon cycling by eukaryotes is much like carbon cycling by aerobic prokaryotes. The novel contribution of eukaryotes is the ability to capture and ingest other cells, thus introducing predation into the carbon cycle.
  2. No. Plants form one branch of the green algal tree, whose early branches are all occupied by unicellular forms. Similarly, the branch containing animals has choanoflagellates and other unicellular forms in its lower branches. Thus, animals and plants evolved multicellularity independently of each other. (Note that genetic evidence further supports this phylogenetic hypothesis. Plants and animals use different sets of genes to regulate multicellular development; see Chapter 28.)
  3. Molecular sequences for genes in red and green algae show the close evolutionary relationship of these organelles to free-living cyanobacteria. This supports the hypothesis that eukaryotes initially gained the ability to photosynthesize by the symbiotic incorporation of cyanobacterial cells. Gene sequences in the chloroplasts of other photosynthetic eukaryotes (brown algae, diatoms, euglenids, and other algae) are very similar to those of red and green algal chloroplasts, indicating that photosynthesis spread throughout the eukaryotic tree by the symbiotic incorporation of green or red algal cells into other eukaryotes.

Chapter 28 Quick Check Answers

  1. In simple multicellular organisms, all or nearly all cells are in direct contact with the environment. In complex multicellular organisms, most cells are completely surrounded by other cells.
  2. Without mechanisms for bulk transport, the movement of oxygen, nutrients, and molecular signals through organisms is limited by diffusion. Rates of diffusion, in turn, limit the size and shape that an organism can achieve. Bulk transport allows key molecules to be transported over distances much greater than those possible by diffusion alone, making larger organisms possible.
  3. All multicellular organisms require molecules that promote adhesion between cells, communication between cells, and differentiation into different cell types within the body. Plants and animals have largely distinct sets of molecules for these functions. Many of the differences between plants and animals reflect the fact that plant cells have cell walls, whereas animal cells do not.

Chapter 29 Quick Check Answers

  1. Plants transpire to obtain CO2 from the atmosphere. Plants cannot completely seal themselves off against water loss because that would prevent them from obtaining the CO2 needed for photosynthesis. As CO2 diffuses into a leaf to reach the photosynthetic cells, water vapor evaporates from surfaces within the leaf and diffuses out of the leaf. Thus, transpiration is an unavoidable byproduct of acquiring CO2 by diffusion from the atmosphere.
  2. CAM plants open their stomata at night when rates of evaporation are low and close them during the day to conserve water. During the night, CO2 that diffuses into the leaf becomes incorporated into 4-carbon organic acids that are stored in the vacuole. During the day, these organic acids are retrieved from the vacuole and broken apart, and CO2 is released into the cytoplasm. Because the stomata are closed, this CO2 does not diffuse out of the leaf but instead can be used in the Calvin cycle to synthesize carbohydrates. C4 plants produce 4-carbon organic acids in mesophyll cells, which then diffuse through plasmodesmata into bundle-sheath cells. In the bundle-sheath cells, each 4-carbon acid is broken up into a 3-carbon compound and CO2. The CO2 is used in the Calvin cycle to produce carbohydrates, while the 3-carbon compound diffuses back to the mesophyll cells, completing the C4 cycle. Because the C4 cycle operates much faster than the Calvin cycle, the concentration of CO2 in the bundle-sheath cells is greatly increased. It is this ability to concentrate CO2 that allows C4 plants to suppress photorespiration.
  3. The statement is correct in that the force that pulls water from the soil results from the evaporation of water from leaf cells, the energy for which comes from the heating of the leaf by the sun. It is incorrect in that there are structures within the plant that make this possible, notably the xylem conduits, and also the roots and the stem. The plant must expend energy to produce all these structures.
  4. The movement of carbohydrates in phloem is from source to sink. Sources can be actively photosynthesizing leaves, or any organ with a large amount of stored carbohydrates. Carrots are examples of roots with large amounts of stored carbohydrates. In contrast, sinks are organs that require an import of carbohydrates to meet their growth and respiratory needs. Examples of sinks include young leaves, developing fruits, and most roots. Typically, the shoot is the source and the roots are a sink. At the end of winter or following a severe drought, however, a plant may regrow its shoot by drawing upon carbohydrates that had been stored in the roots. In this case, the movement of carbohydrates is from roots to shoot.

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Chapter 30 Quick Check Answers

  1. Spores can be dispersed by the wind. In addition, their walls contain sporopollenin, a tough, resistant covering that allows spores to withstand environmental stresses such as ultraviolet radiation and desiccation.
  2. Both mosses and ferns, like all plants, exhibit an alternation of a haploid (gametophyte) generation and a diploid (sporophyte) generation. However, in mosses, the sporophyte is dependent on the gametophyte, whereas in ferns the gametophyte and sporophyte are independent of each other.
  3. Pollen is the male gametophyte of seed plants that develops within the spore wall. A seed develops from an ovule, which is the female gametophyte enclosed in layers of the sporophyte.
  4. Compared to spores, seeds can store more resources, slow down their metabolism, and exhibit dormancy, all of which aid their dispersal.
  5. A flower consists of four whorls of organs: From outside in, they are the sepals, petals, stamens, and carpels. The sepals generally protect developing flowers; the petals commonly serve to attract pollinators; the stamens produce pollen; and the carpel protects the ovules inside that develop into seeds following fertilization.
  6. A fruit forms from the ovary that encloses the seeds and, in some plants, other parts of the flower, including the petals and sepals.

Chapter 31 Quick Check Answers

  1. The shoot apical meristem is the site of cell division and therefore of the growth of stems. It controls the pattern of nodes and internodes and therefore determines the patterns of leaves. Finally, it produces new apical meristems, allowing stems to branch.
  2. One of a plant’s two lateral meristems produces new vascular tissues, and the other maintains an outer protective layer.
  3. Roots and stems both grow from apical meristems, have regions of cell division, elongation, and differentiation, and respond to light and gravity. However, root apical meristems are covered by a root cap, have a single vascular bundle, and initiate branching from the pericycle; all these features are different from what is seen in stems.

Chapter 32 Quick Check Answers

  1. No. The specific branch of the plant immune system can defend only against biotrophic pathogens, not necrotrophic ones, because this branch of the immune system is based on receptor molecules within living cells. Necrotrophic pathogens kill cells before colonizing them.
  2. In the hypersensitive response, plants block off an infected area. In SAR, plants send a signal to uninfected tissue, enabling it to mount a defense and prevent infection.
  3. Preventing jasmonic acid synthesis results in plants that are unable to mount a defense throughout the plant body and as a result suffer more damage from herbivores.
  4. The trade-off between growth and defense creates a diversifying force that favors different species that are specialized for each habitat, rather than a single species that dominates across all soil types.

Chapter 33 Quick Check Answers

  1. The green algae are paraphyletic because the last common ancestor of all green algae also gave rise to land plants; the bryophytes are paraphyletic because the last common ancestor of mosses, liverworts, and hornworts also gave rise to vascular plants. Spore-bearing vascular plants are paraphyletic because the last common ancestor of lycophytes, ferns, and horsetails also gave rise to seed plants. You can’t tell by the figure, but when fossils are considered, gymnosperms are also paraphyletic because the last common ancestor of all gymnosperms also gave rise to the angiopserms.
  2. An alternative explanation is that stomata evolved once, in the common ancestor of bryophytes and vascular plants.
  3. Xylem vessels evolved convergently in gnetophytes and angiosperms; insect pollination evolved convergently in cycads and angiosperms; fleshy tissues to facilitate seed dispersal evolved convergently in early seed plants (reflected today by ginkgo and cycad seeds), conifers such as juniper and yew, and flowering plants.
  4. Monocots produce only a single cotyledon; monocot roots are produced directly from the stem; many monocots produce creeping stems; and the strap-shaped leaves of many monocots elongate from the base and the many parallel veins connect to the stem. Because the vascular bundles of monocot stems are distributed throughout the cross section, as opposed to arranged in a ring, they do not form a vascular cambium.

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Chapter 34 Quick Check Answers

  1. Hyphae enable fungi to seek out new food resources; their mechanical strength allows them to penetrate large food bodies such as rotting logs or animal carcasses. In addition, hyphae transport nutrients from one part of the fungus to another.
  2. A diploid cell has a single nucleus with two complete sets of chromosomes; a dikaryotic cell has two genetically distinct haploid nuclei.
  3. Complex multicellular fruiting bodies evolved independently on two branches, the Basidiomycetes and the Ascomycetes.
  4. You are eating the multicellular fruiting body built from dikaryotic hyphae.

Chapter 35 Quick Check Answers

  1. Nerve signals are transmitted electrically (in the form of an action potential) from one end of a neuron to the other. They are transmitted chemically (by neurotransmitters) across a synapse from one neuron to another.
  2. Nerve impulses are slowed or completely stopped in response to loss of myelin surrounding nerve cells.
  3. Nerve gases that block acetylcholinesterase, the enzyme that inactivates acetylcholine, cause violent muscle contractions, leading to paralysis of the diaphragm (the main muscle involved in breathing) and death by asphyxiation.

Chapter 36 Quick Check Answers

  1. Prostaglandins depolarize nociceptors and increase their firing rate.
  2. Excess mucus secretion blocks airborne odor molecules from reaching the chemosensory hairs of olfactory neurons, reducing their ability to bind odorants and therefore reducing the sense of smell. Taste is also affected because much of the sense of taste depends on smell.
  3. Spinning causes the hair cells in the semicircular canals to accommodate to the new motion. When you stop spinning, the hair cells detect a change in angular motion of the head (relative to the motion when your head was spinning), which leads to a sense of imbalance when you stand or try to walk.
  4. Rod cells are extremely sensitive to light, allowing an animal to see in dimly lit conditions. Cone cells provide color vision through photopigments with different light wavelength absorption peaks, and they provide high spatial acuity, and thus sharp vision.
  5. The three areas of the brain that are used to remember and recognize a face are the occipital cortex for processing visual information, the hippocampus for forming long-term memory of a face, and the temporal lobe for facial recognition.

Chapter 37 Quick Check Answers

  1. Without newly synthesized ATP, myosin cross-bridges formed with actin cannot detach from their actin binding sites, so they remain in the bound state and make the muscle stiff.
  2. Because acetylcholine is blocked and cannot depolarize the muscle cell membrane, calcium is not released from the sarcoplasmic reticulum, and the muscle is unable to contract.
  3. By contracting at an intermediate length, actin and myosin filament overlap of the biceps fibers is maximized, allowing the greatest number of cross-bridges to form and produce force to lift a weight.
  4. Glass is composed of a single material (silica), which is rigid but can crack easily. Bone is a composite material, built of crystalline mineral reinforced by the protein collagen, enabling bone to absorb much more energy before breaking.

Chapter 38 Quick Check Answers

  1. Diabetes can result either from decreased insulin production by the pancreas (type 1 diabetes) or decreased effect of insulin on target cells (type 2 diabetes). Type 1 diabetes is an autoimmune disease in which the insulin-producing cells in the pancreas are attacked by the immune system. The result is inadequate insulin secretion. Daily injections of insulin can maintain stable normal circulating levels of blood glucose. In individuals with type 2 diabetes, insulin production is unaffected, but cells are not able to respond to normal circulating levels of insulin. Type 2 diabetes is commonly linked to obesity and is most effectively treated by proper diet and exercise.
  2. Peptide hormones bind cell membrane receptors because they are hydrophilic and cannot diffuse across the cell membrane, activating intracellular second messengers that stimulate signaling cascades within the target cell. Steroid hormones diffuse across the cell membrane and bind intracellular cytoplasmic or nuclear receptors, leading to changes in gene expression and protein synthesis within the target cell.
  3. Hormones are secreted by a cell or gland and transported in the bloodstream to distant sites where they exert their effect. Hormones achieve specificity by binding to receptors, which are present only on their target cells.
  4. Paracrine signals are chemicals secreted by a cell that influence the activity of neighboring cells without entering the bloodstream. Pheromones are chemicals secreted into the environment and transmitted to other species members, influencing their behavior. Hormones are released into the bloodstream and have effects on distant cells.

Chapter 39 Quick Check Answers

  1. Diffusion is increased by a large surface area for exchange and a short diffusion distance.
  2. In individuals with asthma, resistance to airflow increases, thereby decreasing the flow of air. Resistance increases because the diameter of the airways decreases as a result of smooth muscle contraction in the airway walls, as well as mucus secretion.
  3. Air has a low density and viscosity, and a high O2 content relative to water.
  4. Sickle-cell anemia is life threatening because transport of the deformed sickle-shaped red blood cells is limited in small blood vessels. Therefore, the supply of O2 to tissues is reduced and cellular metabolism in the tissues supplied by the blood vessels is disrupted, possibly leading to organ failure.
  5. Change in a vessel’s diameter (or radius) affects blood flow most. For example, a twofold reduction in radius increases resistance 16 times.
  6. The pressure in the ventricle exceeds the pressure in the atrium.

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Chapter 40 Quick Check Answers

  1. Glycogen stores are used first, then fat, and finally muscle protein.
  2. No. Metabolic rate increases with increasing mass, but the relationship between the two is not linear. Instead, metabolic rate increases with mass raised to the ¾ power. This means that a dog that is twice as heavy as a cat, for example, has a metabolic rate that is less than twice as much as the cat.
  3. “Essential,” when used to describe dietary amino acids, minerals, or vitamins, does not mean that other nutrients are less important or useful. “Essential” in this context means that these nutrients must be obtained in the diet—the organism cannot synthesize them on its own. In other words, it is “essential” that these nutrients are consumed.
  4. Rocks and sediments in gizzards break apart food, as teeth do in vertebrates.
  5. By eating the products of the cecum, rabbits can pass the partially digested food by the small intestine a second time, where nutrients are then absorbed.

Chapter 41 Quick Check Answers

  1. When a cell is placed in a hypotonic solution, water moves into the cell by osmosis and the cell swells or bursts. When a cell is placed in a hypertonic solution, water leaves the cell by osmosis and the cell shrinks.
  2. Osmosis depends on the difference in total solute concentration on the two sides of a membrane. There is no net movement of water molecules across a semipermeable membrane with a high concentration of a particular solute on one side of a membrane if this concentration is matched by a similar concentration of some other solute on the other side of the membrane. For example, the concentration of sodium ions may be high on one side of a semipermeable membrane and low on the other, whereas the distribution of potassium ions may be just the reverse, so the total solute concentration is the same on the two sides of the membrane. The result is no net water movement.
  3. Ammonia excreted by fish is rapidly diluted to nontoxic levels in the surrounding water. Mammals convert ammonia to urea, which is much less toxic than ammonia and therefore can be concentrated and stored before being eliminated.
  4. Although the filtrate is dilute at the start and end of the loop of Henle, the loop creates a concentration gradient from the cortex to the medulla, which is important for subsequent water reabsorption from the collecting ducts. The loops also leave urea as the main solute.

Chapter 42 Quick Check Answers

  1. Sexual reproduction generates genetically unique offspring by (1) chance mutation; (2) recombination between homologous chromosomes during meiosis I; (3) random segregation of homologous chromosomes during meiosis I; and (4) new combinations of chromosomes in the process of fertilization. Asexual reproduction, like sexual reproduction, is also subject to chance mutations.
  2. The vas deferens carries sperm from the epididymis to the ejaculatory duct. Cutting the vas deferens on both sides prevents sperm from entering the ejaculatory duct and leads to male sterility. This procedure, called a vasectomy, is a method of male contraception.
  3. Estrogens or progesterones (or a combination of these) in the pill suppress GnRH, FSH, and LH and therefore block oocyte development and ovulation.
  4. In females, oogonia differentiate into primary oocytes before birth and therefore a female is born with all the primary oocytes she will have. In males, by contrast, spermatogonia do not differentiate into primary spermatocytes until puberty, but sperm can be produced throughout a male’s lifetime. In addition, only one of the four products of meiosis becomes a gamete in females, whereas all four products of meiosis become sperm in males.
  5. In the process of development, cells differentiate into different cell types. The process of differentiation involves the turning on and turning off of specific genes. So, for example, genes that are expressed in a skin cell are different from those that are expressed in a liver or muscle cell. These changes in gene expression, once established, are for the most part stable through mitotic cell divisions.

Chapter 43 Quick Check Answers

  1. Damage to cilia in the respiratory tract increases the probability of lung infection by impairing the ability of the respiratory tract to sweep out microorganisms.
  2. A second novel antigen injected on day 40 will produce a primary response, not a secondary response, because it is a novel antigen that has not been encountered before by the immune system.
  3. Although a given B cell produces only one antibody, the entire population of B cells produces many different antibodies. Each B cell produces a different antibody as the result of a unique pattern of genomic rearrangement.
  4. T cells are activated when their surface T cell receptors bind to an antigen in association with an MHC molecule. By contrast, B cells are activated when their surface antibodies bind to a free antigen.

Chapter 44 Quick Check Answers

  1. Neither Cnidaria nor Bilateria is more closely related to sponges. They share a common ancestor that is the closest relative of sponges.
  2. A jellyfish would use more available nutrients than a sponge of the same size because it can move toward nutrients and capture and digest larger food items.
  3. The shared features among snails, squids, and clams suggest that all mollusks descended from a common ancestor that had a head, radula, foot, mantle, and well-differentiated organ systems.
  4. Young humans and other animals often have features that they lose as they age. Our first “baby” teeth are an example. If you could look at your own photograph when you were a one-month-old, inch-long embryo, you would see a notochord, gill slits, and a tail.
  5. Scales are found on all of these animals due to their presence in a common ancestor. Even feathers are modified scales.

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Chapter 45 Quick Check Answers

  1. In fruit flies, there are two alleles of the for gene that are expressed in larvae, leading to two different phenotypes in the population. As a result, larvae are either sitters or rovers. By contrast, in honeybees, the for gene is up-regulated during development, so a single individual begins by staying in the hive and then becomes a forager.
  2. In classical conditioning, one stimulus is exchanged for another (the smell of food is exchanged for the sound of a bell in the case of Pavlov’s dogs). As a result, a novel stimulus evokes a behavior. In operant conditioning, a behavior is associated with a reward or punishment, so the behavior becomes more likely (with a reward) or less likely (with a punishment). Therefore, in the case of classical conditioning, an association is made between a stimulus and a behavior, and in operant conditioning, an association is made between a behavior and a response.
  3. Photoperiod is a good marker for time of year, and therefore for season. Many species have season-specific behaviors—hibernation in bears, migration in birds—that must be timed appropriately.
  4. Organisms do not act “for the good of the species” because natural selection operates on individuals: It is the individual that lives or dies, reproduces or fails to reproduce. Traits that are disadvantageous to the individual are therefore selected against by natural selection even if they are beneficial to the species as a whole.
  5. With reference to Fig. 45.17, the probability of each offspring inheriting an A allele is 0.5. The probability of a cousin sharing an A allele is 0.25, so the probability of each of the animal’s cousin’s offspring having an A is 0.25 × 0.5 = 0.125. In this case, the animal would have to substitute four of its cousin’s offspring for each of its own.

Chapter 46 Quick Check Answers

  1. The per capita growth rate, r, equals r = (∆N/∆t)/N1. In this case, the change in time ∆t is 1 year, so r = ∆N/N1. If the starting population is x, at triple the size it is 3x. Therefore, ∆N is 3x - x, or 2x, and r = 2x/x, or 2. In general, when a population triples in size in a year, it grows by 200%, so the per capita growth rate is 2.
  2. One cent per day, doubled each day (a rate of increase of 100%) for 30 days exceeds $10,000,000.
  3. Logistic curves are S shaped because exponential increase during early stages of population growth causes the first half of the curve to rise steeply, but then growth becomes limited by resource availability and so slows as population size approaches the carrying capacity of the environment.
  4. Every organism invests all available energy in the ways that make most sense for the current environment. Increase in one area means decrease in another. To increase investments in all these areas, more resources in terms of food or energy would have to be acquired.
  5. The closer island will initially receive more colonists than the farther one, but the larger island can eventually support more kinds of species.

Chapter 47 Quick Check Answers

  1. The fundamental niche is larger than the realized niche because the fundamental niche represents all the habitats where a species could possibly live, but competition among different species usually restricts a species to a smaller area, which is the realized niche.
  2. The cost to the apple tree is resources in the form of nectar, which takes energy to produce, and the benefit is pollination. The cost to the bee is the energy required to search for food and transfer pollen, and the benefit is food.
  3. Although they can affect many other species in the Arctic, lemmings are not a keystone species because their effects are a function of their great abundance. Keystone species have effects on communities that are disproportionate to their numbers because of the particular roles they play, often as predators.
  4. A community is a group of organisms that share a given place. A community together with the physical environment in which the organisms live constitutes an ecosystem.
  5. A diversity gradient can be seen with altitude that is similar to the one seen with latitude: Typically, the higher the altitude, the fewer the number of species.

Chapter 48 Quick Check Answers

  1. In the 130 years from 1800 to 1930, our numbers doubled from 1 to 2 billion, and they doubled again in just the 45 years from 1930 to 1975. This pattern indicates that the rate of human population growth increased in this time.
  2. Global warming is the measured increase in Earth surface temperatures over the past 50 years. The greenhouse effect describes a process by which global warming can occur. The greenhouse effect is the result of the capacity of some molecules—especially carbon dioxide, methane, and water vapor—to absorb heat energy and then emit it in all directions. As a result, the energy incoming from solar radiation exceeds the amount radiated back into space as heat. Without the greenhouse effect, Earth would not be habitable. However, in recent decades, increasing amounts of greenhouse gases in the atmosphere have resulted in global warming.
  3. As ecosystems warm, plants able to open flowers earlier appear to have higher fitness, enabling them to maintain or increase their geographic ranges.
  4. Invasive species may outcompete native species for available resources, diminishing the population size and, in time, the diversity of native species. Introduced animals may prey on native species, reducing their numbers.
  5. The Red Queen hypothesis suggests that organisms need to keep evolving (running) just to stay in the same ecological niche (place). As new antibiotics are developed, pathogens like malaria frequently evolve resistance, so researchers must continually strive to develop new medicines as natural selection erodes the effectiveness of the old.

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