Biology Midterm 2

(Ch. 15) SCIENTIFIC INQUIRY DRAW IT Assume you are mapping genes A, B, C, and D in Drosophila. You know that these genes are linked on the same chromosome, and you determine the recombination frequencies between each pair of genes to be as follows: A and B, 8%; A and C, 28%; A and D, 25%; B and C, 20%; B and D, 33%. (a) Describe how you determined the recombination frequency for each pair of genes. (b) Draw a chromosome map based on your data.

(a) For each pair of genes, you had to generate an F1 dihybrid fly; let's use the A and B genes as an example. You obtained homozygous parental flies, either the first with dominant alleles of the two genes (AABB) and the second with recessive alleles (aabb), or the first with dominant alleles of gene A and recessive alleles of gene B (AAbb) and the second with recessive alleles of gene A and dominant alleles of gene B (aaBB). Breeding either of these pairs of P generation flies gave you an F1 dihybrid, which you then testcrossed with a doubly homozygous recessive fly (aabb). You classed the offspring as parental or recombinant, based on the genotypes of the P generation parents (either of the two pairs described above). You added up the number of recombinant types and then divided by the total number of offspring. This gave you the recombination percentage (in this case, 8%), which you can translate into map units (8 map units) to construct your map B) (https://docs.google.com/document/d/1BcOkTc3gujeGjZ9SDbhOhYKefn_9vUk7GA1b01JNOvc/edit#heading=h.avoi5zlegcpj Figure 15)

7(Ch. 8) DRAW IT Using a series of arrows, draw the branched metabolic reaction pathway described by the following statements, and then answer the question at the end. Use red arrows and minus signs to indicate inhibition. L can form either M or N. M can form O. O can form either P or R. P can form Q. R can form S. O inhibits the reaction of L to form M. Q inhibits the reaction of O to form P. S inhibits the reaction of O to form R Which reaction would prevail if both Q and S were present in the cell in high concentrations? (A) L to M (B) M to O (C) L to N (D) O to P

(https://docs.google.com/document/d/1BcOkTc3gujeGjZ9SDbhOhYKefn_9vUk7GA1b01JNOvc/edit Figure 1) C) L to N

(Ch. 9) VISUAL SKILLS This computer model shows the four parts of ATP synthase, each part consisting of a number of polypeptide subunits (the solid gray part is still an area of active research). Using Figure 9.13 as a guide, label the rotor, stator, internal rod, and catalytic knob of this molecular motor.

(https://docs.google.com/document/d/1BcOkTc3gujeGjZ9SDbhOhYKefn_9vUk7GA1b01JNOvc/edit# Figure 3)

(Ch. 10) SCIENTIFIC INQUIRY • DRAW IT The following diagram represents an experiment with isolated thylakoids. The thylakoids were first made acidic by soaking them in a solution at pH 4. After the thylakoid space reached pH 4, the thylakoids were transferred to a basic solution at pH 8. The thylakoids then made ATP in the dark. (See Concept 3.3 to review pH.) Draw an enlargement of part of the thylakoid membrane in the beaker with the solution at pH 8. Draw ATP synthase. Label the areas of high H+ concentration and low H+ concentration. Show the direction protons flow through the enzyme, and show the reaction where ATP is synthesized. Would ATP end up in the thylakoid or outside of it? Explain why the thylakoids in the experiment were able to make ATP in the dark.

(https://docs.google.com/document/d/1BcOkTc3gujeGjZ9SDbhOhYKefn_9vUk7GA1b01JNOvc/edit# Figure 7) The ATP would end up outside the thylakoid. The thylakoids were able to make ATP in the dark because the researchers set up an artificial proton concentration gradient across the thylakoid membrane; thus, the light reactions were not necessary to establish the H+ gradient required for ATP synthesis by ATP synthase.

(Ch. 8) A researcher has developed an assay to measure the activity of an important enzyme present in pancreatic cells growing in culture. She adds the enzyme's substrate to a dish of cells and then measures the appearance of reaction products. The results are graphed as the amount of product on the y-axis versus time on the x-axis. The researcher notes four sections of the graph. For a short period of time, no products appear (section A). Then (section B) the reaction rate is quite high (the slope of the line is steep). Next, the reaction gradually slows down (section C). Finally, the graph line becomes flat (section D). Draw and label the graph, and propose a model to explain the molecular events occurring at each stage of this reaction profile.

(https://docs.google.com/document/d/1BcOkTc3gujeGjZ9SDbhOhYKefn_9vUk7GA1b01JNOvc/edit#heading=h.2fe63bw9rab3 Figure 2) A. The substrate molecules are entering the pancreatic cells, so no product is made yet. B. There is sufficient substrate, so the reaction is proceeding at a maximum rate. C. As the substrate is used up, the rate decreases (the slope is less steep). D. The line is flat because no new substrate remains and thus no new product appears.

(Ch. 12) DRAW IT Draw one eukaryotic chromosome as it would appear during interphase, during each of the stages of mitosis, and during cytokinesis. Also draw and label the nuclear envelope and any microtubules attached to the chromosome(s).

(https://docs.google.com/document/d/1BcOkTc3gujeGjZ9SDbhOhYKefn_9vUk7GA1b01JNOvc/edit#heading=h.6t1i4rnma0um Figure 10)

(10.4) DRAW IT Draw a simple version of the Calvin cycle that shows only the number of carbon atoms at each step, using numerals rather than gray spheres (for example, "3 * 1C = 3C" at the beginning of the cycle). Explain how the total number of carbon atoms remains constant for every three turns of the cycle, noting the forms in which the carbon atoms enter and leave the cycle.

(https://docs.google.com/document/d/1BcOkTc3gujeGjZ9SDbhOhYKefn_9vUk7GA1b01JNOvc/edit#heading=h.83n6miuy2nop Figure 6) For every three turns of the Calvin cycle, the total number of carbon atoms remains constant because three carbons enter as part of CO2 molecules, replacing the three that left as part of a G3P molecule.

(Ch. 9) DRAW IT The graph (Ref Chart Fig. 5) here shows the pH difference across the inner mitochondrial membrane over time in an actively respiring cell. At the time indicated by the vertical arrow, a metabolic poison is added that specifically and completely inhibits all function of mitochondrial ATP synthase. Draw what you would expect to see for the rest of the graphed line over the next short period of time. Explain.

(https://docs.google.com/document/d/1BcOkTc3gujeGjZ9SDbhOhYKefn_9vUk7GA1b01JNOvc/edit#heading=h.c2hc2v1d25z5 Figure 4) H+ would continue to be pumped across the membrane into the intermembrane space, increasing the difference between the matrix pH and the intermembrane space pH. H+ would not be able to flow back through ATP synthase, since the enzyme is inhibited by the poison, so rather than maintaining a constant difference across the membrane, the difference would continue to increase. (Over a longer period of time, the H+ concentration in the intermembrane space would be so high that no more H+ would be able to be pumped against the gradient. This wasn't asked for in the question, but if your graph levels off at the right end and this is your reasoning, your answer is correct.)

(Ch. 12) VISUAL SKILLS The light micrograph shows dividing cells near the tip of an onion root. Identify a cell in each of the following stages: prophase, prometaphase, metaphase, anaphase, and telophase. Describe the major events occurring at each stage.

(https://docs.google.com/document/d/1BcOkTc3gujeGjZ9SDbhOhYKefn_9vUk7GA1b01JNOvc/edit#heading=h.kvpybj5xibxj Figure 9)

(14.1) . DRAW IT Pea plants heterozygous for flower position and stem length (AaTt) are allowed to self-pollinate, and 400 of the resulting seeds are planted. Draw a Punnett square for this cross. How many offspring would be predicted to have terminal flowers and be dwarf? (See Table 14.1.)

(https://docs.google.com/document/d/1BcOkTc3gujeGjZ9SDbhOhYKefn_9vUk7GA1b01JNOvc/edit#heading=h.zcqm4hinubwl Figure 12) According to the law of independent assortment, 25 plants (1⁄ 16 of the offspring) are predicted to be aatt, or recessive for both characters. The actual result is likely to differ slightly from this value.

(15.4) MAKE CONNECTIONS The gene that is activated on the Philadelphia chromosome codes for an intracellular tyrosine kinase. Review the discussion of cell cycle control in Concept 12.3, and explain how the activation of this gene could contribute to the development of cancer.

. Activation of this gene could lead to the production of too much of this kinase. If the kinase is involved in a signaling pathway that triggers cell division, too much of it could trigger unrestricted cell division, which in turn could contribute to the development of a cancer (in this case, a cancer of one type of white blood cell).

(13.3) WHAT IF? After the synaptonemal complex disappears, how would any pair of homologous chromosomes be associated if crossing over did not occur? What effect might this have on gamete formation?

. If crossing over did not occur, the two homologs would not be associated in any way. This is because each sister chromatid would be either all maternal or all paternal DNA, and the single DNA molecule would therefore not have been joined to the DNA of a nonsister chromatid, holding the complex together. The lack of association of homologs could easily result in the incorrect arrangement of homologs during metaphase I (both might go toward the same pole, for instance) and ultimately in formation of gametes with an abnormal number of chromosomes.

(Ch. 15) A man with hemophilia (a recessive, sex-linked condition) has a daughter without hemophilia who marries a man without hemophilia. What is the probability of their daughter having hemophilia? Their son? If they have four sons, that all will be affected?

0; 1⁄ 2; 1⁄ 16

(Ch. 15) A wild-type fruit fly (heterozygous for gray body color and normal wings) is mated with a black fly with vestigial wings. The offspring have the following phenotypic distribution: wildtype, 778; black vestigial, 785; black normal, 158; gray vestigial, 162. What is the recombination frequency between these genes for body color and wing size? Is this consistent with the results of the experiment in Figure 15.9? Draw the chromosomes in the wild-type and black parents.

17%; yes, it is consistent. In Figure 15.9, the recombination frequency was also 17%. (You'd expect this to be the case since these are the very same two genes, and their distance from each other wouldn't change from one experiment to another.) (https://docs.google.com/document/d/1BcOkTc3gujeGjZ9SDbhOhYKefn_9vUk7GA1b01JNOvc/edit#heading=h.4fbk92qg29ms Figure 14)

(12.1) How many chromosomes are drawn in each part of Figure 12.5? (Ignore the micrograph in step 2.)

1; 1; 2

(14.2) For any gene with a dominant allele A and recessive allele a, what proportions of the offspring from an AA * Aa cross are expected to be homozygous dominant, homozygous recessive, and heterozygous?

1⁄ 2 homozygous dominant (AA), 0 homozygous recessive (aa), and 1⁄ 2 heterozygous (Aa)

(15.2) Neither Tim nor Shonda has Duchenne muscular dystrophy, but their firstborn son does. What is the probability that a second child will have the disease? What is the probability if the second child is a boy? A girl?

1⁄ 4 (1⁄ 2 chance that the child will inherit a Y chromosome from the father and be male * 1⁄ 2 chance that he will inherit the X carrying the disease allele from his mother). If the child is a boy, there is a 1⁄ 2 chance he will have the disease; a female would have zero chance (but 1⁄ 2 chance of being a carrier).

(14.2) Two organisms, with genotypes BbDD and BBDd, are mated. Assuming independent assortment of the B>b and D>d genes, write the genotypes of all possible offspring from this cross and use the rules of probability to calculate the chance of each genotype occurring.

1⁄ 4 BBDD; 1⁄ 4 BbDD; 1⁄ 4 BBDd; 1⁄ 4 BbDd

(14.4) Lucia and Jared each have a sibling with cystic fibrosis, but neither Lucia nor Jared nor any of their parents have the disease. Calculate the probability that if this couple has a child, the child will have cystic fibrosis. What would be the probability if a test revealed that Jared is a carrier but Lucia is not? Explain your answers.

1⁄ 9 (Since cystic fibrosis is caused by a recessive allele, Lucia and Jared's siblings who have CF must be homozygous recessive. Therefore, each parent must be a carrier of the recessive allele. Since neither Lucia nor Jared has CF and are thus not cc (using C/c as the alleles for the CF gene), this means they each have a 2⁄ 3 chance of being a carrier. If they are both carriers, there is a 1⁄ 4 chance that they will have a child with CF; 2⁄ 3 * 2⁄ 3 * 1⁄ 4 = 1⁄ 9); virtually 0 (Both Lucia and Jared would have to be carriers to produce a child with the disease, unless a very rare mutation (change) occurred in the DNA of cells making eggs or sperm in a noncarrier that resulted in the CF allele.)

(12.1) WHAT IF? A chicken has 78 chromosomes in its somatic cells. How many chromosomes did the chicken inherit from each parent? How many chromosomes are in each of the chicken's gametes? How many chromosomes will be in each somatic cell of the chicken's offspring?

39; 39; 78

(Ch. 15) Two genes of a flower, one controlling blue (B) versus white (b) petals and the other controlling round (R) versus oval (r) stamens, are linked and are 10 map units apart. You cross a homozygous blue/oval plant with a homozygous white/ round plant. The resulting F1 progeny are crossed with homozygous white/oval plants, and 1,000 offspring plants are obtained. How many F2 plants of each of the four phenotypes do you expect?

450 each of blue/oval and white/round (parentals) and 50 each of blue/round and white/oval (recombinants)

(12.2) How many chromosomes are shown in the illustration in Figure 12.8? Are they duplicated? How many chromatids are shown?

6 chromosomes; they are duplicated; 12 chromatids

(Ch. 15) A wild-type fruit fly (heterozygous for gray body color and red eyes) is mated with a black fruit fly with purple eyes. The offspring are wild-type, 721; black/purple, 751; gray/purple, 49; black/red, 45. What is the recombination frequency between these genes for body color and eye color? Using information from problem 3, what fruit flies (genotypes and phenotypes) would you mate to determine the order of the body color, wing size, and eye color genes on the chromosome?

6%; wild-type heterozygous for normal wings and red eyes * recessive homozygous for vestigial wings and purple eyes

(10.4) To synthesize one glucose molecule, the Calvin cycle uses ________ molecules of CO2, ________ molecules of ATP, and ________ molecules of NADPH.

6, 18, 12

(8.5) . WHAT IF? Regulation of isoleucine synthesis is an example of feedback inhibition of an anabolic pathway. With that in mind, explain how ATP might be involved in feedback inhibition of a catabolic pathway.

A catabolic pathway breaks down organic molecules, generating energy that is stored in ATP molecules. In feedback inhibition of such a pathway, ATP (one product) would act as an allosteric inhibitor of an enzyme catalyzing an early step in the catabolic process. When ATP is plentiful, the pathway would be turned off and no more would be made.

(9.5) Consider the NADH formed during glycolysis. What is the final acceptor for its electrons during fermentation? During aerobic respiration? During anaerobic respiration?

A derivative of pyruvate, such as acetaldehyde during alcohol fermentation, or pyruvate itself during lactic acid fermentation; O2; another electron acceptor at the end of an electron transport chain, such as sulfate (SO4 2 - )

(13.2) Using shoes as an analogy for chromosomes, how would you describe the collection of "shoes" in human diploid and haploid cells?

A human diploid cell would have 23 "pairs of shoes." A haploid cell would have 23 "shoes," one of each pair.

(12.2) A kinetochore has been compared to a coupling device that connects a motor to the cargo that it moves. Explain.

A kinetochore connects the spindle (a motor; note that it has motor proteins) to a chromosome (the cargo it will move).

(11.4) WHAT IF? Some human diseases are associated with malfunctioning protein phosphatases. How would such proteins affect signaling pathways? (Review the discussion of protein phosphatases in Concept 11.3 and see Figure 11.10.)

A malfunctioning protein phosphatase would not be able to dephosphorylate a particular receptor or relay protein. As a result, the signaling pathway, once activated, would not be able to be terminated. (In fact, one study found altered protein phosphatases in cells from 25% of colorectal tumors.)

(11.3) What is a protein kinase, and what is its role in a signal transduction pathway?

A protein kinase is an enzyme that transfers a phosphate group from ATP to a protein, usually activating that protein (often a second type of protein kinase). Many signal transduction pathways include a series of such interactions, in which each phosphorylated protein kinase in turn phosphorylates the next protein kinase in the series. Such phosphorylation cascades carry a signal from outside the cell to the cellular protein(s) that will carry out the response.

(8.3) Which combination has more free energy: glutamic acid + ammonia + ATP or glutamine + ADP + ○P i ? Explain.

A set of coupled reactions can transform the first combination into the second. Since this is an exergonic process overall, ΔG is negative and the first combination must have more free energy (see Figure 8.10).

(8.4) Many spontaneous reactions occur very slowly. Why don't all spontaneous reactions occur instantly?

A spontaneous reaction is a reaction that is exergonic. However, if it has a high activation energy that is rarely attained, the rate of the reaction may be low.

(12.3) How does MPF allow a cell to pass the G2 phase checkpoint and enter mitosis? (See Figure 12.16.)

A sufficient amount of MPF has to exist for a cell to pass the G2 checkpoint; this occurs through the accumulation of cyclin proteins, which combine with Cdk to form (active) MPF. MPF then phosphorylates other proteins, initiating mitosis.

(Ch. 10) Which of the following statements is a correct distinction between autotrophs and heterotrophs? (A) Autotrophs, but not heterotrophs, can nourish themselves beginning with CO2 and other nutrients that are inorganic. (B) Only heterotrophs require chemical compounds from the environment. (C) Cellular respiration is unique to heterotrophs. (D) Only heterotrophs have mitochondria.

A) Autotrophs, but not heterotrophs, can nourish themselves beginning with CO2 and other nutrients that are inorganic.

(Ch. 12) Cell A has half as much DNA as cells B, C, and D in a mitotically active tissue. Cell A is most likely in (A) G1. (B) G2. (C) prophase. (D) metaphase.

A) G1

(Ch. 9) The final electron acceptor of the electron transport chain that functions in aerobic oxidative phosphorylation is (A) O2. (B) water. (C) NAD+. (D) pyruvate.

A) O2

(Ch. 9) When electrons flow along the electron transport chains of mitochondria, which of the following changes occurs? (A) The pH of the matrix increases. (B) ATP synthase pumps protons by active transport. (C) The electrons gain free energy. (D) NAD+ is oxidized

A) The pH of the matrix increases.

(Ch. 13) A human cell containing 22 autosomes and a Y chromosome is (A) a sperm. (B) an egg. (C) a zygote. (D) a somatic cell of a male.

A) a sperm.

4(Ch. 8) If an enzyme in solution is saturated with substrate, the most effective way to obtain a faster yield of products is to (A) add more of the enzyme. (B) heat the solution to 90°C. (C) add more substrate. (D) add a noncompetitive inhibitor

A) add more of the enyme

(Ch. 11) Consider this pathway: epinephrine S G protein-coupled receptor S G protein S adenylyl cyclase S cAMP. Identify the second messenger. (A) cAMP (B) G protein (C) GTP (D) adenylyl cyclase

A) cAMP

(Ch. 10) Which of the following occurs during the Calvin cycle? (A) carbon fixation (B) reduction of NADP+ (C) release of oxygen (D) generation of CO2

A) carbon fixation

(Ch. 12) In the cells of some organisms, mitosis occurs without cytokinesis. This will result in (A) cells with more than one nucleus. (B) cells that are unusually small. (C) cells lacking nuclei. (D) cell cycles lacking an S phase.

A) cells with more than one nucleus

(Ch. 11) The activation of receptor tyrosine kinases is characterized by (A) dimerization and phosphorylation. (B) dimerization and IP3 binding. (C) a phosphorylation cascade. (D) GTP hydrolysis.

A) dimerization and phosphorylation.

(Ch. 12) Vinblastine is a standard chemotherapeutic drug used to treat cancer. Because it interferes with the assembly of microtubules, its effectiveness must be related to (A) disruption of mitotic spindle formation. (B) suppression of cyclin production. (C) myosin denaturation and inhibition of cleavage furrow formation. (D) inhibition of DNA synthesis.

A) disruption of miotic spindle formation.

(Ch. 13) Meiosis II is similar to mitosis in that (A) sister chromatids separate during anaphase. (B) DNA replicates before the division. (C) the daughter cells are diploid. (D) homologous chromosomes synapse.

A) sister chromatids separate during anaphase.

(9.6) VISUAL SKILLS What will happen in a muscle cell that has used up its supply of O2 and ATP? (Review Figures 9.17 and 9.19.)

AMP will accumulate, stimulating phosphofructokinase, and thus increasing the rate of glycolysis. Since oxygen is not present, the cell will convert more pyruvate to lactate, providing a supply of ATP.

(8.3) How does ATP typically transfer energy from an exergonic to an endergonic reaction in the cell?

ATP usually transfers energy to an endergonic process by phosphorylating (adding a phosphate group to) another molecule. (Exergonic processes, in turn, phosphorylate ADP to regenerate ATP.)

(8.3) . MAKE CONNECTIONS Does Figure 8.11a show passive or active transport? Explain. (See Concepts 7.3 and 7.4.)

Active transport: The solute is being transported against its concentration gradient, which requires energy, provided by ATP hydrolysis.

(11.4) How can a target cell's response to a single hormone molecule result in a response that affects a million other molecules?

At each step in a cascade of sequential activations, one molecule or ion may activate numerous molecules functioning in the next step. This causes the response to be amplified at each such step and overall results in a large amplification of the original signal.

3(Ch. 8) Which of the following metabolic processes can occur without a net influx of energy from some other process? (A) ADP + ○P i + S ATP + H2O (B) C6H12O6 + 6 O2 S 6 CO2 + 6 H2O (C) 6 CO2 + 6 H2O S C6H12O6 + 6 O2 (D) Amino acids S Protein

B) C6H12O6 + 6 O2 S 6 CO2 + 6 H2O

(Ch. 10) Which of the following sequences correctly represents the flow of electrons during photosynthesis? (A) NADPH to O2 to CO2 (B) H2O to NADPH to Calvin cycle (C) H2O to photosystem I to photosystem II (D) NADPH to electron transport chain to O2

B) H2O to NADPH to Calvin cycle

(Ch. 12) Through a microscope, you can see a cell plate beginning to develop across the middle of a cell and nuclei forming on either side of the cell plate. This cell is most likely (A) an animal cell in the process of cytokinesis. (B) a plant cell in the process of cytokinesis. (C) a bacterial cell dividing. (D) a plant cell in metaphase.

B) a plant cell in the process of cytokinesis.

1(Ch. 8) . Choose the pair of terms that correctly completes this sentence: Catabolism is to anabolism as ____________ is to ____________. (A) exergonic; spontaneous (B) exergonic; endergonic (C) free energy; entropy (D) work; energy

B) exergonic; endergonic

(Ch. 11) Lipid-soluble signaling molecules, such as aldosterone, cross the membranes of all cells but affect only target cells because (A) only target cells retain the appropriate DNA segments. (B) intracellular receptors are present only in target cells. (C) only target cells have enzymes that break down aldosterone. (D) only in target cells is aldosterone able to initiate the phosphorylation cascade that turns genes on.

B) intracellular receptors are present only in target cells.

(Ch. 13) The two homologs of a pair move toward opposite poles of dividing cell during (A) mitosis. (B) meiosis I. (C) meiosis II. (D) fertilization.

B) meiosis I.

(Ch. 10) In mechanism, photophosphorylation is most similar to (A) substrate-level phosphorylation in glycolysis. (B) oxidative phosphorylation in cellular respiration. (C) carbon fixation. (D) reduction of NADP+

B) oxidative phosphorylation

(Ch. 9) In mitochondria, exergonic redox reactions (A) are the source of energy driving prokaryotic ATP synthesis. (B) provide the energy that establishes the proton gradient. (C) reduce carbon atoms to carbon dioxide. (D) are coupled via phosphorylated intermediates to endergonic processes.

B) provide the energy that establishes the proton gradient.

(Ch. 12) Which of the following occurs during S phase? (A) condensation of the chromosomes (B) replication of the DNA (C) separation of sister chromatids (D) spindle formation

B) replication of the DNA

(Ch. 9) Most CO2 from catabolism is released during (A) glycolysis. (B) the citric acid cycle. (C) lactate fermentation. (D) electron transport.

B) the citric acid cycle.

(Ch. 15) Banana plants, which are triploid, are seedless and therefore sterile. Thinking about meiosis, propose a possible explanation.

Because bananas are triploid, homologous pairs cannot line up during meiosis. Therefore, it is not possible to generate gametes that can fuse to produce a zygote with the triploid number of chromosomes.

(10.1) Why are heterotrophs dependent on autotrophs?

Because heterotrophs cannot photosynthesize, they cannot capture the energy of sunlight and produce energy-rich compounds like sugars, as autotrophs can. Sugars are oxidized by cellular respiration, providing energy (in the form of ATP) for cellular processes. Without this ability, heterotrophs depend on autotrophs to provide sugars as the food molecules that fuel their vital processes.

(15.2) A white-eyed female Drosophila is mated with a red-eyed (wild-type) male, the reciprocal cross of the one shown in Figure 15.3. What phenotypes and genotypes do you predict for the offspring from this cross?

Because the gene for this eye color character is located on the X chromosome, all female offspring will be red-eyed and heterozygous (Xw+ Xw); all male offspring will inherit a Y chromosome from the father and be white-eyed (XwY). (Another way to say this is that 1⁄ 2 the offspring will be red-eyed heterozygous [carrier] females, and 1⁄ 2 will be white-eyed males.)

(Ch. 15) A planet is inhabited by creatures that reproduce with the same hereditary patterns seen in humans. Three phenotypic characters are height (T = tall, t = dwarf), head appendages (A = antennae, a = no antennae), and nose shape (S = upturned snout, s = downturned snout). Since the creatures are not "intelligent," Earth scientists are able to do some controlled breeding experiments using various heterozygotes in testcrosses. For tall heterozygotes with antennae, the offspring are tall/antennae, 46; dwarf/antennae, 7; dwarf/no antennae, 42; tall/no antennae, 5. For heterozygotes with antennae and an upturned snout, the offspring are antennae/upturned snout, 47; antennae/downturned snout, 2; no antennae/downturned snout, 48; no antennae/upturned snout, 3. Calculate the recombination frequencies for both experiments.

Between T and A, 12%; between A and S, 5%

(Ch. 15) Scientists do a testcross of the creatures from problem 4 using heterozygotes for height and nose shape. The offspring are tall/ upturned snout, 41; dwarf/upturned snout, 8; dwarf/downturned snout, 43; tall/downturned snout, 8. Calculate the recombination frequency from these data, and then use your answer from problem 4 to determine the correct order of the three linked genes.

Between T and S, 16%; sequence of genes is T-A-S

(11.2) . MAKE CONNECTIONS How is ligand binding similar to the process of allosteric regulation of enzymes? (See Figure 8.20.)

Binding of a ligand to a receptor changes the shape of the receptor, altering the ability of the receptor to transmit a signal. Binding of an allosteric regulator to an enzyme changes the shape of the enzyme, either promoting or inhibiting enzyme activity.

(10.5) MAKE CONNECTIONS Refer to the discussion of ocean acidification in Concept 3.3. Ocean acidification and changes in the distribution of C3 and C4 plants may seem to be two very different problems, but what do they have in common? Explain.

Both problems are caused by a drastic change in Earth's atmosphere due to burning of fossil fuels. The increase in CO2 concentration affects ocean chemistry by decreasing pH, thus affecting calcification by marine organisms. On land, CO2 concentration and air temperature are conditions that plants have become adapted to, and changes in these characteristics have a strong effect on photosynthesis by plants. Thus, alteration of these two fundamental factors could have critical effects on organisms all around the planet, in all different habitats.

(9.1) Compare and contrast aerobic and anaerobic respiration, including the processes involved.

Both processes include glycolysis, the citric acid cycle, and oxidative phosphorylation. In aerobic respiration, the final electron acceptor is molecular oxygen (O2); in anaerobic respiration, the final electron acceptor is a different substance.

(Ch. 11) Which observation suggested to Sutherland the involvement of a second messenger in epinephrine's effect on liver cells? (A) Enzymatic activity was proportional to the amount of calcium added to a cell-free extract. (B) Receptor studies indicated that epinephrine was a ligand. (C) Glycogen breakdown was observed only when epinephrine was administered to intact cells. (D) Glycogen breakdown was observed only when epinephrine and glycogen phosphorylase were mixed.

C) Glycogen breakdown was observed only when epinephrine was administered to intact cells.

(Ch. 9) The immediate energy source that drives ATP synthesis by ATP synthase during oxidative phosphorylation is the (A) oxidation of glucose and other organic compounds. (B) flow of electrons down the electron transport chain. (C) H+ concentration gradient across the membrane holding ATP synthase. (D) transfer of phosphate to ADP.

C) H+ concentration gradient across the membrane holding ATP synthase.

(Ch. 10) How is photosynthesis similar in C4 plants and CAM plants? (A) In both cases, only photosystem I is used. (B) Both types of plants make sugar without the Calvin cycle. (C) In both cases, rubisco is not used to fix carbon initially. (D) Both types of plants make most of their sugar in the dark.

C) In both cases, rubisco is not used to fix carbon initially.

(Ch. 11) Protein phosphorylation is commonly involved with which of the following? (A) ligand binding by receptor tyrosine kinases. (B) activation of G protein-coupled receptors. (C) activation of protein kinase molecules. (D) release of Ca2+ from the ER lumen.

C) activation of protein kinase molecules.

(Ch. 12) One difference between cancer cells and normal cells is that cancer cells (A) are unable to synthesize DNA. (B) are arrested at the S phase of the cell cycle. (C) continue to divide even when they are tightly packed together. (D) cannot function properly because they are affected by density-dependent inhibition.

C) continue to divide even when they are tightly packed together.

(Ch. 9) Which metabolic pathway is common to both fermentation and cellular respiration of a glucose molecule? (A) the citric acid cycle (B) the electron transport chain (C) glycolysis (D) reduction of pyruvate to lactate

C) glycolysis

2(Ch. 8) Most cells cannot harness heat to perform work because (A) heat does not involve a transfer of energy. (B) cells do not have much thermal energy; they are relatively cool. (C) temperature is usually uniform throughout a cell. (D) heat can never be used to do work.

C) temperature is usually uniform throughout a cell

(Ch. 12) The decline of MPF activity at the end of mitosis is due to (A) the destruction of the protein kinase Cdk. (B) decreased synthesis of Cdk. (C) the degradation of cyclin. (D) the accumulation of cyclin.

C) the degradation of cyclin.

5(Ch. 8) Some bacteria are metabolically active in hot springs because (A) they are able to maintain a lower internal temperature. (B) high temperatures make catalysis unnecessary. (C) their enzymes have high optimal temperatures. (D) their enzymes are completely insensitive to temperature.

C) their enzymes have high optimal temperatures

(Ch. 10) Which process is most directly driven by light energy? (A) creation of a pH gradient by pumping protons across the thylakoid membrane (B) reduction of NADP + molecules (C) transfer of energy from pigment molecule to pigment molecule (D) ATP synthesis

C) transfer of energy from pigment molecule to pigment molecule

(Ch. 13) . If we continue to follow the cell lineage from question 4, then the DNA content of a single cell at metaphase of meiosis II will be (A) 0.25x. (B) 0.5x. (C) x. (D) 2x.

C) x

(9.1) WHAT IF? If the following redox reaction occurred, which compounds would be oxidized? Reduced? C4H6O5 + NAD+ S C4H4O5 + NADH + H+

C4H6O5 would be oxidized and NAD+ would be reduced.

(10.2) MAKE CONNECTIONS How do the CO2 molecules used in photosynthesis reach and enter the chloroplasts inside leaf cells? (See Concept 7.2.)

CO2 enters the leaves via stomata, and, being a nonpolar molecule, can cross the leaf cell membrane and the chloroplast membranes to reach the stroma of the chloroplast.

(8.2) VISUAL SKILLS How would the processes of catabolism and anabolism relate to Figure 8.5c?

Catabolism breaks down organic molecules, releasing their chemical energy and resulting in smaller products with more entropy, as when moving from the top to the bottom of Figure 8.5c. Anabolism consumes energy to synthesize larger molecules from simpler ones, as when moving from the bottom to the top of part (c).

(8.2) Cellular respiration uses glucose and O2, which have high levels of free energy, and releases CO2 and H2O, which have low levels of free energy. Is cellular respiration spontaneous or not? Is it exergonic or endergonic? What happens to the energy released from glucose?

Cellular respiration is a spontaneous and exergonic process. The energy released from glucose is used to do work in the cell or is lost as heat.

(15.3) When two genes are located on the same chromosome, what is the physical basis for the production of recombinant offspring in a testcross between a dihybrid parent and a double-mutant (recessive) parent?

Crossing over during meiosis I in the heterozygous parent produces some gametes with recombinant genotypes for the two genes. Offspring with a recombinant phenotype arise from fertilization of the recombinant gametes by homozygous recessive gametes from the double-mutant parent.

(Ch. 13) . If the DNA content of a diploid cell in the G1 phase of the cell cycle is x, then the DNA content of the same cell at metaphase of meiosis I will be (A) 0.25x. (B) 0.5x. (C) x. (D) 2x.

D) 2x

(Ch. 10) The light reactions supply the Calvin cycle with (A) light energy. (B) CO2 and ATP. (C) H2O and NADPH. (D) ATP and NADPH.

D) ATP and NADPH

6(Ch. 8) If an enzyme is added to a solution where its substrate and product are in equilibrium, what will occur? (A) Additional substrate will be formed. (B) The reaction will change from endergonic to exergonic. (C) The free energy of the system will change. (D) Nothing; the reaction will stay at equilibrium.

D) Nothing; the reaction will stay at equilibrium

(Ch. 12) The drug cytochalasin B blocks the function of actin. Which of the following aspects of the animal cell cycle would be most disrupted by cytochalasin B? (A) spindle formation (B) spindle attachment to kinetochores (C) cell elongation during anaphase (D) cleavage furrow formation and cytokinesis

D) cleavage furrow formation and cytokinesis

(Ch. 11) Binding of a signaling molecule to which type of receptor leads directly to a change in the distribution of substances on opposite sides of the membrane? (A) intracellular receptor (B) G protein-coupled receptor (C) phosphorylated receptor tyrosine kinase dimer (D) ligand-gated ion channel

D) ligand-gated ion channel

(Ch. 9) What is the oxidizing agent in the following reaction? Pyruvate + NADH + H+ to Lactate + NAD+ (A) oxygen (B) NADH (C) lactate (D) pyruvate

D) pyruvate

(9.4) . WHAT IF? In the absence of O2, as in question 1, what do you think would happen if you decreased the pH of the intermembrane space of the mitochondrion? Explain your answer.

Decreasing the pH means addition of H+. This would establish a proton gradient even without the function of the electron transport chain, and we would expect ATP synthase to function and synthesize ATP. (In fact, it was experiments like this that provided support for chemiosmosis as an energy-coupling mechanism.)

(12.2) Compare the roles of tubulin and actin during eukaryotic cell division with the roles of tubulin-like and actin-like proteins during bacterial binary fission.

During eukaryotic cell division, tubulin is involved in spindle formation and chromosome movement, while actin functions during cytokinesis. In bacterial binary fission, it's the opposite: Actin-like molecules are thought to move the daughter bacterial chromosomes to opposite ends of the cell, and tubulin-like molecules are thought to act in daughter cell separation.

(15.5) WHAT IF? Mitochondrial genes are critical to the energy metabolism of cells, but mitochondrial disorders caused by mutations in these genes are generally not lethal. Why not?

Each cell contains numerous mitochondria, and in affected individuals, most cells contain a variable mixture of normal and mutant mitochondria. The normal mitochondria carry out enough cellular respiration for survival. (The situation is similar for chloroplasts.)

(13.2) MAKE CONNECTIONS In Figure 13.4, how many DNA molecules (double helices) are present (see Figure 12.5)? What is the haploid number of this cell? Is a set of chromosomes haploid or diploid?

Each of the six chromosomes is duplicated, so each contains two DNA molecules (double helices), so there are 12 DNA molecules in the cell. The haploid number, n, is 3. One set is always haploid.

(Ch. 15) Assume that genes A and B are 50 map units apart on the same chromosome. An animal heterozygous at both loci is crossed with one that is homozygous recessive at both loci. What percentage of the offspring will show recombinant phenotypes resulting from crossovers? Without knowing these genes are on the same chromosome, how would you interpret the results of this cross?

Fifty percent of the offspring will show phenotypes resulting from crossovers. These results would be the same as those from a cross where A and B were not on the same chromosome, and you would interpret the results to mean that the genes are unlinked. (Further crosses involving other genes between A and B on the same chromosome would reveal the genetic linkage and map distances.)

(12.2) Compare cytokinesis in animal cells and plant cells.

Following mitosis, cytokinesis results in two genetically identical daughter cells in both plant cells and animal cells. However, the mechanism of dividing the cytoplasm is different in animals and plants. In an animal cell, cytokinesis occurs by cleavage, which divides the parent cell in two with a contractile ring of actin filaments. In a plant cell, a cell plate forms in the middle of the cell and grows until its membrane fuses with the plasma membrane of the parent cell. A new cell wall grows inside the cell plate, thus eventually between the two new cells.

(12.2) During which stages of the cell cycle does a chromosome consist of two identical chromatids?

From the end of S phase in interphase through the end of metaphase in mitosis

(11.1) WHAT IF? If epinephrine were mixed with glycogen phosphorylase and glycogen in a cell-free mixture in a test tube, would glucose 1-phosphate be generated? Why or why not?

Glucose 1-phosphate would not be generated because the activation of the enzyme requires an intact cell, with an intact receptor in the membrane and an intact signal transduction pathway. The enzyme cannot be activated directly by interaction with the signaling molecule in the cell-free mixture.

(11.1) In liver cells, glycogen phosphorylase acts in which of the three stages of the signaling pathway associated with an epinephrine-initiated signal?

Glycogen phosphorylase acts in the third stage, the cellular response to epinephrine signaling.

(10.3) What color of light is least effective in driving photosynthesis? Explain.

Green, because green light is mostly transmitted and reflected—not absorbed—by photosynthetic pigments

(14.3) If a man with type AB blood marries a woman with type O, what blood types would you expect in their children? What fraction would you expect of each type?

Half of the children would be expected to have type A blood and half type B blood.

(13.4) WHAT IF? If maternal and paternal chromatids have the same two alleles for every gene, will crossing over lead to genetic variation?

If the segments of the maternal and paternal chromatids that undergo crossing over are genetically identical and thus have the same two alleles for every gene, then the recombinant chromosomes will be genetically equivalent to the parental chromosomes. Crossing over contributes to genetic variation only when it involves the rearrangement of different alleles.

(15.2) MAKE CONNECTIONS Consider what you learned about dominant and recessive alleles in Concept 14.1. If a disorder were caused by a dominant X-linked allele, how would the inheritance pattern differ from what we see for recessive X-linked disorders?

In a disorder caused by a dominant allele, there is no such thing as a "carrier," since those with the allele have the disorder. Because the allele is dominant, the females lose any "advantage" in having two X chromosomes, since one disorder-associated allele is sufficient to result in the disorder. All fathers who have the dominant allele will pass it along to all their daughters, who will also have the disorder. A mother who has the allele (and thus the disorder) will pass it to half of her sons and half of her daughters.

(9.3) . VISUAL SKILLS The conversions shown in Figure 9.9 and step 4 of Figure 9.11 are each catalyzed by a large multienzyme complex. What similarities are there in the reactions that occur in these two cases?

In both cases, the precursor molecule loses a CO2 molecule and then donates electrons to an electron carrier in an oxidation step. Also, the product has been activated due to the attachment of a CoA group by its S atom.

(15.3) VISUAL SKILLS For each type of offspring of the testcross in Figure 15.9, explain the relationship between its phenotype and the alleles contributed by the female parent. (It will be useful to draw out the chromosomes of each fly and follow the alleles throughout the cross.)

In each case, the alleles contributed by the female parent (in the egg) determine the phenotype of the offspring because the male in this cross contributes only recessive alleles. Thus, identifying the phenotype of the offspring tells you what alleles were in the mother's (the dihybrid female's) egg.

(10.4) MAKE CONNECTIONS Review Figures 9.8 and 10.18, and then discuss the roles of intermediate and product played by glyceraldehyde 3-phosphate (G3P) in the two processes shown in these figures.

In glycolysis, G3P acts as an intermediate. The six-carbon sugar fructose 1,6-bisphosphate is cleaved into two three-carbon sugars, one of which is G3P. The other is an isomer called dihydroxyacetone phosphate (DHAP), which can be converted to G3P by an isomerase. Because G3P is the substrate for the next enzyme, it is constantly removed, and the reaction equilibrium is pulled in the direction of conversion of DHAP to more G3P. In the Calvin cycle, G3P acts as both an intermediate and a product. For every three CO2 molecules that enter the cycle, six G3P molecules are formed, five of which must remain in the cycle and become rearranged to regenerate three five-carbon RuBP molecules. The one remaining G3P is a product, which can be thought of as the result of "reducing" the three CO2 molecules that entered the cycle into a three-carbon sugar that can later be used to generate energy.

(15.4) About 5% of individuals with Down syndrome have a chromosomal translocation in which a third copy of chromosome 21 is attached to chromosome 14. If this translocation occurred in a parent's gonad, how could it lead to Down syndrome in a child?

In meiosis, a combined 14-21 chromosome will behave as one chromosome. If a gamete receives the combined 14-21 chromosome and a normal copy of chromosome 21, trisomy 21 will result when this gamete combines with a normal gamete (with its own chromosome 21) during fertilization.

(14.4) MAKE CONNECTIONS In Table 14.1, note the phenotypic ratio of the dominant to recessive trait in the F2 generation for the monohybrid cross involving flower color. Then determine the phenotypic ratio for the offspring of the secondgeneration couple in Figure 14.15b. What accounts for the difference in the two ratios?

In the monohybrid cross involving flower color, the ratio is 3.15 purple : 1 white, while in the human family in the pedigree, the ratio in the third generation is 1 taster of PTC : 1 nontaster of PTC. The difference is due to the small sample size (two offspring) in the human family. If the second-generation couple in this pedigree were able to have 929 offspring as in the pea plant cross, the ratio would likely be closer to 3:1. (Note that none of the pea plant crosses in Table 14.1 yielded exactly a 3:1 ratio.)

(8.4) DRAW IT A mature lysosome has an internal pH of around 4.5. Using Figure 8.17b as a guide, draw a graph showing what you would predict for the rate of reaction for a lysosomal enzyme. Label its optimal pH, assuming its optimal pH matches its environment.

In the presence of malonate, increase the concentration of the normal substrate (succinate) and see whether the rate of reaction increases. If it does, malonate is a competitive inhibitor.

(8.4) WHAT IF? Malonate is an inhibitor of the enzyme succinate dehydrogenase. How would you determine whether malonate is a competitive or noncompetitive inhibitor?

In the presence of malonate, increase the concentration of the normal substrate (succinate) and see whether the rate of reaction increases. If it does, malonate is a competitive inhibitor.

(10.3) WHAT IF? In an experiment, isolated chloroplasts placed in an illuminated solution with the appropriate chemicals can carry out ATP synthesis. Predict what would happen to the rate of synthesis if a compound is added to the solution that makes membranes freely permeable to hydrogen ions.

In this experiment, the rate of ATP synthesis would slow and eventually stop. Because the added compound would not allow a proton gradient to build up across the membrane, ATP synthase could not catalyze ATP production.

(15.5) Gene dosage—the number of copies of a gene that are actively being expressed—is important to proper development. Identify and describe two processes that establish the proper dosage of certain genes.

Inactivation of an X chromosome in females and genomic imprinting. Because of X inactivation, the effective dose of genes on the X chromosome is the same in males and females. As a result of genomic imprinting, only one allele of certain genes is phenotypically expressed.

(14.3) Incomplete dominance and epistasis are both terms that define genetic relationships. What is the most basic distinction between these terms?

Incomplete dominance describes the relationship between two alleles of a single gene, whereas epistasis relates to the genetic relationship between two genes (and the respective alleles of each).

(14.4) Juanita was born with six toes on each foot, a dominant trait called polydactyly. Two of her five siblings and her mother, but not her father, also have extra digits. What is Juanita's genotype for the number-of-digits character? Explain your answer. Use D and d to symbolize the alleles for this character.

Juanita's genotype is Dd. Because the allele for polydactyly (D) is dominant to the allele for five digits per appendage (d), the trait is expressed in people with either the DD or Dd genotype. But because Juanita's father does not have polydactyly, his genotype must be dd, which means that Juanita inherited a d allele from him. Therefore, Juanita, who does have the trait, must be heterozygous.

(12.2) MAKE CONNECTIONS What other functions do actin and tubulin carry out? Name the proteins they interact with to do so. (Review Figures 6.21a and 6.26a.)

Microtubules made up of tubulin in the cell provide "rails" along which vesicles and other organelles can travel, based on interactions of motor proteins with tubulin in the microtubules. In muscle cells, actin in microfilaments interacts with myosin filaments to cause muscle contraction.

(13.4) . What is the original source of variation among the different alleles of a gene?

Mutations in a gene lead to the different versions (alleles) of that gene.

(9.2) VISUAL SKILLS During the redox reaction in glycolysis (see step 6 in Figure 9.8), which one of the molecules acts as the oxidizing agent? The reducing agent?

NAD+ acts as the oxidizing agent in step 6, accepting electrons from glyceraldehyde 3-phosphate (G3P), which thus acts as the reducing agent.

(9.3) . VISUAL SKILLS In the citric acid cycle shown in Figure 9.11, what molecules capture energy from the redox reactions? How is ATP produced?

NADH and FADH2; one ATP is produced during substrate-level phosphorylation in step 5.

(11.2) . Nerve growth factor (NGF) is a water-soluble signaling molecule. Would you expect the receptor for NGF to be intracellular or in the plasma membrane? Explain

NGF is water-soluble (hydrophilic), so it cannot pass through the lipid membrane to reach intracellular receptors, as steroid hormones can. Therefore, you'd expect the NGF receptor to be in the plasma membrane—which is, in fact, the case

(15.4) MAKE CONNECTIONS The ABO blood type locus has been mapped on chromosome 9. A father who has type AB blood and a mother who has type O blood have a child with trisomy 9 and type A blood. Using this information, can you tell in which parent the nondisjunction occurred? Explain your answer. (See Figures 14.11 and 15.13.)

No. The child can be either I AI Ai or I Aii. A sperm of genotype I AI A could result from nondisjunction in the father during meiosis II, while an egg with the genotype ii could result from nondisjunction in the mother during either meiosis I or meiosis II.

(8.4) Say you are using a Bunsen burner in the lab. Why doesn't the flame creep back into the gas line and set the gas supply on fire?

O2 is required as a substrate to react with the gas. There is O2 in the lab air, which is why the Bunsen burner has a flame above the gas delivery opening, but there is no O2 in the rubber tubing or the gas supply.

(9.4) MAKE CONNECTIONS Membranes must be fluid to function properly (see Concept 7.1). How does the operation of the electron transport chain support that assertion?

One of the components of the electron transport chain, ubiquinone (Q), must be able to diffuse within the membrane. It could not do so if the membrane components were locked rigidly into place.

(9.4) . WHAT IF? What effect would an absence of O2 have on the process shown in Figure 9.14?

Oxidative phosphorylation would eventually stop entirely, resulting in no ATP production by this process. Without oxygen to "pull" electrons down the electron transport chain, H+ would not be pumped into the mitochondrion's intermembrane space and chemiosmosis would not occur.

(13.1) . MAKE CONNECTIONS Using your knowledge of gene expression in a cell, explain what causes the traits of parents (such as hair color) to show up in their offspring. (See Concept 5.5.)

Parents pass genes to their offspring; by dictating the production of messenger RNAs (mRNAs), the genes program cells to make specific enzymes and other proteins, whose cumulative action produces an individual's inherited traits.

(10.5) Describe how photorespiration lowers photosynthetic output for plants.

Photorespiration decreases photosynthetic output by adding O2, instead of CO2, to the Calvin cycle. As a result, no sugar is generated (no carbon is fixed), and O2 is used rather than generated.

(10.6) MAKE CONNECTIONS How do plants use the sugar they produce during photosynthesis to directly power the work of the cell? Provide some examples of cellular work. (See Figures 8.10, 8.11, and 9.5.)

Plants can break down the sugar they make (in the form of glucose) by cellular respiration, producing ATPs for various cellular processes such as endergonic chemical reactions, transport of substances across membranes, and movement of molecules in the cell. ATPs are also used for the movement of chloroplasts during cellular streaming in some plant cells (see Figure 6.26).

(11.3) When a signal transduction pathway involves a phosphorylation cascade, how does the cell's response get turned off?

Protein phosphatases reverse the effects of the kinases by dephosphorylation, and unless the signaling molecule is at a high enough concentration that it is continuously rebinding the receptor, the kinase molecules will all be returned to their inactive states by phosphatases.

(Ch. 15) Pseudohypertrophic muscular dystrophy is an inherited disorder that causes gradual deterioration of the muscles. It is seen almost exclusively in boys born to unaffected parents and usually results in death in the early teens. Is this disorder caused by a dominant or a recessive allele? Is its inheritance sex-linked or autosomal? How do you know? Explain why this disorder is almost never seen in girls.

Recessive; if the disorder were dominant, it would affect at least one parent of a child born with the disorder. The disorder's inheritance is sexlinked because it is seen only in boys. For a girl to have the disorder, she would have to inherit recessive alleles from both parents. This would be very rare, since males with the recessive allele on their X chromosome die in their early teens.

(11.4) WHAT IF? If two cells have different scaffolding proteins, explain how they might behave differently in response to the same signaling molecule.

Scaffolding proteins hold molecular components of signaling pathways in a complex with each other. Different scaffolding proteins would assemble different collections of proteins, facilitating different molecular interactions and leading to different cellular responses in the two cells.

(14.1) MAKE CONNECTIONS In some pea plant crosses, the plants are self-pollinated. Is self-pollination considered asexual or sexual reproduction? Explain. (See Concept 13.1.)

Self-pollination is sexual reproduction because meiosis is involved in forming gametes, which unite during fertilization. As a result, the offspring in self-pollination are genetically different from the parent. (As mentioned in the footnote near the beginning of Concept 14.1, we have simplified the explanation in referring to the single pea plant as a parent. Technically, the gametophytes in the flower are the two "parents.")

(13.1) WHAT IF? A horticulturalist breeds orchids, trying to obtain a plant with a unique combination of desirable traits. After many years, she finally succeeds, and wants to produce more plants like this one. Discuss whether she should crossbreed it with another plant or cause it to undergo asexual reproduction (forming a clone), and why.

She should clone (generate a clone of) it. Crossbreeding it with another plant would generate offspring that have additional variation, which she no longer desires now that she has obtained her ideal orchid.

(Ch. 9) MAKE CONNECTIONS Step 3 in Figure 9.8 is a major point of regulation of glycolysis. The enzyme phosphofructokinase is allosterically regulated by ATP and related molecules (see Concept 8.5). Considering the overall result of glycolysis, would you expect ATP to inhibit or stimulate activity of this enzyme? Explain. (Hint: Make sure you consider the role of ATP as an allosteric regulator, not as a substrate of the enzyme.)

Since the overall process of glycolysis results in net production of ATP, it would make sense for the process to slow down when ATP levels have increased substantially. Thus, we would expect ATP to allosterically inhibit phosphofructokinase.

(13.1) Describe how an asexually reproducing eukaryotic organism produces offspring that are genetically identical to each other and to their parents.

Such organisms reproduce by mitosis, which generates offspring whose genomes are exact copies of the parent's genome (in the absence of mutation).

(9.3) What processes in your cells produce the CO2 that you exhale?

The CO2 that we exhale is produced by pyruvate oxidation and the citric acid cycle.

(11.3) WHAT IF? If you exposed a cell to a ligand that binds to a receptor and activates phospholipase C, predict the effect the IP3-gated calcium channel would have on Ca2 + concentration in the cytosol.

The IP3-gated channel would open, allowing calcium ions to flow out of the ER and into the cytoplasm, which would raise the cytosolic Ca2 + concentration.

(8.5) 1. How do an activator and an inhibitor have different effects on an allosterically regulated enzyme?

The activator binds in such a way that it stabilizes the active form of an enzyme, whereas the inhibitor stabilizes the inactive form.

(8.1) Describe the forms of energy found in an apple as it grows on a tree, then falls, then is digested by someone who eats it.

The apple has potential energy in its position hanging on the tree, and the sugars and other nutrients it contains have chemical energy. The apple has kinetic energy as it falls from the tree to the ground. Finally, when the apple is digested and its molecules broken down, some of the chemical energy is used to do work, and the rest is lost as thermal energy.

(14.3) WHAT IF? A rooster with gray feathers and a hen of the same phenotype produce 15 gray, 6 black, and 8 white chicks. What is the simplest explanation for the inheritance of these colors in chickens? What phenotypes would you expect in the offspring of a cross between a gray rooster and a black hen?

The black and white alleles are incompletely dominant, with heterozygotes being gray in color. A cross between a gray rooster and a black hen should yield approximately equal numbers of gray and black offspring.

(11.2) WHAT IF? What would the effect be if a cell made defective receptor tyrosine kinase proteins that were unable to dimerize?

The cell with the faulty receptor would not be able to respond appropriately to the signaling molecule when it was present. This would most likely have dire consequences for the cell since regulation of the cell's activities by this receptor would not occur appropriately.

(9.5) WHAT IF? A glucose-fed yeast cell is moved from an aerobic environment to an anaerobic one. How would its rate of glucose consumption change if ATP were to be generated at the same rate?

The cell would need to consume glucose at a rate about 16 times the consumption rate in the aerobic environment (2 ATP are generated by fermentation versus up to 32 ATP by cellular respiration).

(13.3) MAKE CONNECTIONS Compare the chromosomes in a cell at metaphase of mitosis with those in a cell at metaphase II. (See Figures 12.7 and 13.8.)

The chromosomes are similar in that each is composed of two sister chromatids, and the individual chromosomes are positioned similarly at the metaphase plate. The chromosomes differ in that in a mitotically dividing cell, sister chromatids of each chromosome are genetically identical, but in a meiotically dividing cell, sister chromatids are genetically distinct because of crossing over in meiosis I. Moreover, the chromosomes in metaphase of mitosis can be a diploid set or a haploid set, but the chromosomes in metaphase of meiosis II always consist of a haploid set.

(9.6) MAKE CONNECTIONS Compare the structures of a carbohydrate and a fat (see Figures 5.3 and 5.9). What features make fat a much better fuel?

The fat is much more reduced; it has many —CH2— units, and in all these bonds the electrons are equally shared. The electrons present in a carbohydrate molecule are already somewhat oxidized (shared unequally in bonds; there are more C—O and O—H bonds), as quite a few of them are bound to oxygen. Electrons that are equally shared, as in fat, have a higher energy level than electrons that are unequally shared, as in carbohydrates. Thus, fat is a much better fuel than carbohydrate.

(15.5) Reciprocal crosses between two primrose varieties, A and B, produced the following results: A female * B male S offspring with all green (nonvariegated) leaves; B female * A male S offspring with patterned (variegated) leaves. Explain these results.

The genes for leaf coloration are located in plastids within the cytoplasm. Normally, only the maternal parent transmits plastid genes to offspring. Since variegated offspring are produced only when the female parent is of the B variety, we can conclude that variety B contains both the wild-type and mutant alleles of pigment genes, producing variegated leaves. (Variety A must contain only the wild-type allele of pigment genes.)

(14.2) WHAT IF? Three characters (flower color, seed color, and pod shape) are considered in a cross between two pea plants: PpYyIi * ppYyIi. Using the rules of probability, determine the fraction of offspring predicted to be homozygous recessive for at least two of the three characters.

The genotypes that fulfill this condition are ppyyII, ppyyIi, ppYYii, ppYyii, Ppyyii, and ppyyii. Use the multiplication rule to find the probability of getting each genotype, and then use the addition rule to find the overall probability of meeting the conditions of this problem: (https://docs.google.com/document/d/1BcOkTc3gujeGjZ9SDbhOhYKefn_9vUk7GA1b01JNOvc/edit#heading=h.ow0dhwugjov3 Figure 13)

(12.3) . MAKE CONNECTIONS Explain how receptor tyrosine kinases and intracellular receptors might function in triggering cell division. (Review Figures 11.8 and 11.9 and Concept 11.2.)

The intracellular receptor (for example, an estrogen receptor), once activated, would be able to act as a transcription factor in the nucleus, turning on genes that may cause the cell to pass a checkpoint and divide. The RTK receptor, when activated by a ligand, would form a dimer, and each subunit of the dimer would phosphorylate the other. This would lead to a series of signal transduction steps, ultimately turning on genes in the nucleus. As in the case of the estrogen receptor, the genes would code for proteins necessary to cause the cell to pass a checkpoint and divide.

(15.1) Which one of Mendel's laws describes the inheritance of alleles for a single character? Which law relates to the inheritance of alleles for two characters in a dihybrid cross?

The law of segregation describes the inheritance of alleles for a single character. The law of independent assortment of alleles describes the inheritance of alleles for two characters.

(10.2) WHAT IF? The Calvin cycle requires ATP and NADPH, products of the light reactions. If a classmate asserted that the light reactions don't depend on the Calvin cycle and, with continual light, could just keep on producing ATP and NADPH, how would you respond?

The light reactions could not keep producing NADPH and ATP without the NADP+ , ADP, and ~ P i that the Calvin cycle generates. The two cycles are interdependent.

(10.1) WHAT IF? Fossil fuels are being depleted faster than they are replenished. Therefore, researchers are developing ways to produce "biodiesel" from the products of photosynthetic algae. They have proposed placing containers of these algae near industrial plants or highly congested city streets. Considering the process of photosynthesis, how does this arrangement make sense?

The main product of fossil fuel combustion (for example, by cars) is CO2. Placing containers of algae near sources of CO2 emission makes sense because the algae need CO2 to carry out photosynthesis. The higher the CO2 concentration, the higher will be the rate of algal photosynthesis. (At the same time, algae would be reducing the CO2 concentration in those areas, which would otherwise contribute to climate change; see Concept 1.1.)

(10.4) How are the large numbers of ATP and NADPH molecules used during the Calvin cycle consistent with the high value of glucose as an energy source?

The more potential energy and reducing power a molecule stores, the more energy and reducing power are required for the formation of that molecule. Glucose is a valuable energy source because it is highly reduced (has lots of C—H bonds), storing lots of potential energy in its electrons. To reduce CO2 to glucose, a large amount of energy and a lot of reducing power are required in the form of large numbers of ATP and NADPH molecules, respectively.

(12.3) . In Figure 12.14, why do the nuclei resulting from experiment 2 contain different amounts of DNA?

The nucleus on the right was originally in the G1 phase; therefore, it had not yet duplicated its chromosomes. The nucleus on the left was in the M phase, so it had already duplicated its chromosomes.

(15.1) MAKE CONNECTIONS Review the description of meiosis (see Figure 13.8) and Mendel's laws of segregation and independent assortment (see Concept 14.1). What is the physical basis for each of Mendel's laws?

The physical basis for the law of segregation is the separation of homologs in anaphase I. The physical basis for the law of independent assortment is the alternative arrangements of all the different homologous chromosome pairs in metaphase I.

(14.1) WHAT IF? List all gametes that could be made by a pea plant heterozygous for seed color, seed shape, and pod shape (YyRrIi; see Table 14.1). How large a Punnett square would you need to draw to predict the offspring of a self-pollination of this "trihybrid"?

The plant could make eight different gametes (YRI, YRi, YrI, Yri, yRI, yRi, yrI, and yri). To fit all the possible gametes in a self-pollination, a Punnett square would need 8 rows and 8 columns. It would have spaces for the 64 possible unions of gametes in the offspring.

(11.4) Epinephrine affects heart muscle cells by causing them to mobilize glucose, contract faster, and increase heart rate. The muscle cells around lungs and airways, on the other hand, have the opposite response to epinephrine: They relax, allowing more air to be breathed in. What might explain why respiratory (breathing-related) muscle cells can respond so differently from heart muscle cells?

The proteins in the two cells are different, so the cellular response is different. In heart muscle cells, the pathway shown in Figure 11.16 allows glucose to fuel faster muscle contractions and heart rate. In respiratory muscles, the relay proteins must be different, so that the effect is to block muscle contraction. (In fact, the steps are the same through protein kinase A (PKA), but in respiratory muscle cells, PKA phosphorylates a protein that is required for muscle contraction—and in this case, phosphorylation inactivates that protein. So muscle contraction cannot occur.)

(Ch. 9) MAKE CONNECTIONS The proton pump shown in Figures 7.18 and 7.19 is a type of ATP synthase like that in Figure 9.13. Compare the processes shown in the three figures, and say whether they are involved in active or passive transport (see also Concepts 7.3 and 7.4).

The proton pump in Figures 7.18 and 7.19 is carrying out active transport, using ATP hydrolysis to pump protons against their concentration gradient. Because ATP is required, this is active transport of protons. The ATP synthase in Figure 9.13 is using the flow of protons down their concentration gradient to power ATP synthesis. Because the protons are moving down their concentration gradient, no energy is required, and this is passive transport.

(8.2) WHAT IF? Some partygoers wear glow-in-the-dark necklaces that start glowing once they are "activated" by snapping the necklace. This allows two chemicals to react and emit light in the form of chemiluminescence. Is the reaction exergonic or endergonic? Explain.

The reaction is exergonic because it releases energy—in this case, in the form of light. (This is a nonbiological version of the bioluminescence seen in Figure 8.1.)

(8.1) MAKE CONNECTIONS How does the second law of thermodynamics help explain the diffusion of a substance across a membrane? (See Figure 7.11.)

The second law is the trend toward randomization, or increasing entropy. When the concentrations of a substance on both sides of a membrane are equal, the distribution is more random than when they are unequal. Diffusion of a substance to a region where it is initially less concentrated increases entropy, making it an energetically favorable (spontaneous) process as described by the second law. This explains the process seen in Figure 7.11.

(11.3) What is the actual "signal" that is being transduced in any signal transduction pathway, such as those shown in Figures 11.6 and 11.10? In what way is this information being passed from the exterior to the interior of the cell?

The signal that is being transduced is the information that a signaling molecule is bound to the cell-surface receptor. Information is transduced by way of sequential protein-protein interactions that change protein shapes, causing them to function in a way that passes the signal (the information) along.

(8.1) WHAT IF? If you place a teaspoon of sugar in the bottom of a glass of water, it will dissolve completely over time. Left longer, eventually the water will disappear and the sugar crystals will reappear. Explain these observations in terms of entropy.

The sugar crystals become less ordered (entropy increases) as they dissolve and become randomly spread out in the water. Over time, the water evaporates, and the crystals form again because the water volume is insufficient to keep them in solution. While the reappearance of sugar crystals may represent a "spontaneous" increase in order (decrease in entropy), it is balanced by the decrease in order (increase in entropy) of the water molecules, which changed from a relatively compact arrangement as liquid water to a much more dispersed and disordered form as water vapor.

(11.1) Explain how signaling is involved in ensuring that yeast cells fuse only with cells of the opposite mating type.

The two cells of opposite mating type (a and a) each secrete a unique signaling molecule, which can only be bound by receptors carried on cells of the opposite mating type. Thus, the a mating factor cannot bind to another a cell and cause it to grow toward the first a cell. Only an a cell can "receive" the signaling molecule and respond by directed growth.

(13.2) VISUAL SKILLS In the karyotype in Figure 13.3, how many pairs of chromosomes are present? How many sets?

There are 23 pairs of chromosomes and two sets.

(Ch. 13) Explain how you can tell that the cell in question 6 is undergoing meiosis, not mitosis.

This cell must be undergoing meiosis because the two homologs of a homologous pair are associated with each other at the metaphase plate; this does not occur in mitosis. Also, chiasmata are clearly present, meaning that crossing over has occurred, another process unique to meiosis.

(13.2) WHAT IF? A certain eukaryote lives as a unicellular organism, but during environmental stress, it produces gametes. The gametes fuse, and the resulting zygote undergoes meiosis, generating new single cells. What type of organism could this be?

This organism has the life cycle shown in Figure 13.6c, since the zygote doesn't undergo mitosis but immediately undergoes meiosis. Therefore, it must be a fungus or a protist, perhaps an alga.

(15.1) WHAT IF? Propose a possible reason that the first naturally occurring mutant fruit fly Morgan saw involved a gene on a sex chromosome and was found in a male.

To show the mutant phenotype, a male needs to possess only one mutant allele. If this gene had been on a pair of autosomes, an individual would show the recessive phenotype only if both alleles were mutant, a much less probable situation.

(10.2) Explain how the use of an oxygen isotope helped elucidate the chemistry of photosynthesis.

Using 18O, a heavy isotope of oxygen, as a label, researchers were able to confirm van Niel's hypothesis that the oxygen atoms in O2 produced during photosynthesis comes from H2O, not from CO2.

(10.3) . In the light reactions, what is the initial electron donor? Where do the electrons finally end up?

Water (H2O) is the initial electron donor; NADP+ accepts electrons at the end of the electron transport chain, becoming reduced to NADPH.

(9.6) VISUAL SKILLS During intense exercise, can a muscle cell use fat as a concentrated source of chemical energy? Explain. (Review Figures 9.17 and 9.18.)

When O2 is present, the fatty acid chains containing most of the energy of a fat are oxidized and fed into the citric acid cycle and the electron transport chain. During intense exercise, however, O2 is scarce in muscle cells, so ATP must be generated by glycolysis alone. A very small part of the fat molecule, the glycerol backbone, can be oxidized via glycolysis, but the amount of energy released by this portion is insignificant compared to that released by the fatty acid chains. (This is why moderate exercise, staying below 70% maximum heart rate, is better for burning fat—because enough O2 remains available to the muscles.)

(9.6) Under what circumstances might your body synthesize fat molecules?

When you consume more food than necessary for metabolic processes, your body synthesizes fat as a way of storing energy for later use.

(10.5) The presence of only PS I, not PS II, in the bundle-sheath cells of C4 plants has an effect on O2 concentration. What is that effect, and how might that benefit the plant?

Without PS II, no O2 is generated in bundle-sheath cells. This avoids the problem of O2 competing with CO2 for binding to rubisco in these cells.

(13.4) . The diploid number for fruit flies is 8, and the diploid number for grasshoppers is 46. If no crossing over took place, would the genetic variation among offspring from a given pair of parents be greater in fruit flies or grasshoppers? Explain.

Without crossing over, independent assortment of chromosomes during meiosis I theoretically can generate 2n possible haploid gametes, and random fertilization can produce 2n * 2n possible diploid zygotes. Because the haploid number (n) of grasshoppers is 23 and that of fruit flies is 4, two grasshoppers would be expected to produce a greater variety of zygotes than would two fruit flies.

(10.4) WHAT IF? Consider a poison that inhibits an enzyme of the Calvin cycle. Do you think such a poison will also inhibit the light reactions? Explain.

Yes, it would inhibit the dark reactions. The light reactions require ADP and NADP+ , which would not be formed in sufficient quantities from ATP and NADPH if the Calvin cycle stopped.

(Ch. 13) DRAW IT The diagram shows a cell in meiosis. (a) Label the appropriate structures with these terms: chromosome (label as duplicated or unduplicated), centromere, kinetochore, sister chromatids, nonsister chromatids, homologous pair (use a bracket when labeling), homolog (label each one), chiasma, sister chromatid cohesion, and gene loci. Circle and label the alleles of the F gene. (b) Precisely describe the makeup of a haploid set and a diploid set in this cell. (c) Identify the stage of meiosis shown.

a) (https://docs.google.com/document/d/1BcOkTc3gujeGjZ9SDbhOhYKefn_9vUk7GA1b01JNOvc/edit# Figure 11) b) A haploid set is made up of one long, one medium, and one short chromosome, no matter what combination of colors. For example, one red long, one blue medium, and one red short chromosome make up a haploid set. (In cases where crossovers have occurred, a haploid set of one color may include segments of chromatids of the other color.) All red and blue chromosomes together make up a diploid set. (c) Metaphase I