Bio 113 Midterm #4 Chapters 13-15

Lakukan tugas rumah & ujian kamu dengan baik sekarang menggunakan Quizwiz!

What is the biological significance of the pH shift in oxygen saturation of hemoglobin?

. Hemoglobin moves from a pH of 7.6 in your lungs to pH 7.2 in your muscles. The lower pH is caused by a number of factors including the fact that CO2 is a weak acid, and muscle cells consume O2 and produce a lot of CO2 as a waste product. Changing the pH from 7.6 to 7.2 is a change of 100.4 or 2.5-fold increase in H+ ion concentration. When a protein is surrounded by more H+ ions, its side chains move in response to the increased acidity. Amino acids with a positive charge are repelled by the H+ ions, whereas negatively charged amino acids may move closer to the periphery of the protein surface. When a protein changes its shape, it also changes its function. In this case, hemoglobin alters the way it binds and releases oxygen. The pH-sensitive switch of hemoglobin is another example of an emergent property

Summarize the molecular process that converts UV light into a signal for a lytic lifestyle. Why is responding to UV light an adaptive strategy for λ phage?

After UV treatment, the cells that used to be lysogenic quickly convert to lytic, as indicated by the reciprocal curves in Figure 13.8B. It takes about 30 minutes for all the lysogenic λ to make the genetic switch, produce new viruses, and rupture their host cells. Converting from lysogenic to lytic requires new protein production to kill the host cells. Therefore, the conversion to lytic viruses produced a smooth curve and not a digital one-step change because the process of switching lifestyles involved random diffusion of proteins and DNA. {Connections: Diffusion of proteins inside cells was explored in Section 8.2.} Random diffusion of molecules means it takes an average of 25 minutes after UV exposure for the infected cells to lyse.

Return to AnAge database, and search for the opossum Didelphis virginiana. Are Austad's data described in this database? Austad only measured females; speculate about male senescence and reproductive rates on Sapelo Island and then refine Austad's original scientific question and associated experimental design.

AnAge database contains information about Didelphis virginiana, including Austad's data from Sapelo Island and Figure 15.30. Male opossums mature later than females, and you would expect island males to mature later and reproduce longer than mainland opossums if the disposable soma hypothesis were true.

Distinguish the two methods used in Figure 14.3 to detect bioluminescence. Use the data to show how the methods detect two different cellular components that affect bioluminescence production.

As shown in Figure 14.3, the scientists used two different methods to measure induction of bioluminescence over time. In the first method, they measured the amount of light emitted directly from the bacterial culture using a light detector in a dark box. In the second method, they extracted bacterial cytoplasm at different times to determine when cells produced the bioluminescent enzyme called luciferase, the same enzyme used by fireflies. After extracting the cytoplasm, the investigators mixed the cytoplasm with a known luciferase substrate to allow the enzyme to produce light. If luciferase enzyme had been produced by the bacteria, the cytoplasm would have emitted light in proportion to the amount of enzyme present. The biologists recorded the light intensity immediately after mixing substrate with the extracted cytoplasm (Figure 14.3). As you can see from the data, the two different methods produced different results which provide additional information about the regulation of light production by V. fischeri.

Compare and contrast first-set and second-set immune rejection of tissue transplants.

As shown in the last panel of Figure 15.9, second-set rejection is caused by T cells, a subtype of white blood cells. First-set rejection in half of the mice takes about 10 days, whereas second-set rejection in half the mice only takes about 6 days. The experiment in Figure 15.9 was a critical breakthrough in understanding the normal immune system. Second set rejection happens much faster than first set rejection due to previous exposure.

When a naïve mouse is given a first-set allograft in Figure 15.9, how many days does it take for 50% of the recipients to completely reject the graft? How many days does it take for a second-set rejection to occur in 50% of the recipients? What subtype of immune cells are responsible for the "memory" of previous allograft exposure?

As shown in the last panel of Figure 15.9, second-set rejection is caused by T cells, a subtype of white blood cells. First-set rejection in half of the mice takes about 10 days, whereas second-set rejection in half the mice only takes about 6 days. The experiment in Figure 15.9 was a critical breakthrough in understanding the normal immune system. Immunologists would need to understand a normal immune system before they could determine how a fetus avoids being rejected.

Do you detect substantial weight differences among the three genotypes of free-feeding mice in Table 15.3? What about among the pair-fed mice? What similarities do you detect among the ob and db mice compared to the wt mice when you look at the pair-fed mice?

As you would expect, the wt mice in Table 15.3 gained weight over time and their percent body fat was similar regardless of feeding regime. By comparison, free-feeding ob and db mice gained about twice as much weight as wt, but pair-fed ob and db mice gained no more weight than the control wt mice. However, the ob and db pair-fed mice had much larger percentages of fat than the wt mice despite their equivalent overall weights. Once again, you can see that although ob and db mice would eat more if given the opportunity, the amount of food is not the reason they accumulated more fat. Something about their mutations caused the pair-fed mutant mice to accumulate fat even when their diets were restricted to wt amounts of food and their overall weight was no greater than wt mice. The rat data suggest that human obesity might be influenced by diet as well as genetics.

In Figure 13.10D, the cI dimer sitting on operator 2 physically interacts with the RNA polymerase. What sort of modulation occurs to stimulate RNA polymerase to begin transcribing mRNA? Would it be possible for a second RNA polymerase to initiate transcription of cro from PR when cI is being transcribed? Explain your rationale.

At this point in the book, you should be recognizing some familiar lessons learned earlier. {Connections: Chapter 7 addressed protein structure/function and allosteric modulation.} In the case of cI interacting with RNA polymerase, cI dimer bound to operator 2 uses allosteric modulation to activate RNA polymerase. "The structure of a protein determines its function" is a very important rule in biology, and one you have seen repeatedly. A second RNA polymerase cannot transcribe cro when cI is bound to operators 1 and 2, because cI dimers function as repressors for PR. In the absence of cI, cro is produced because RNA polymerase has a higher affinity for the DNA of PR than PRM. PR is active unless RNA polymerase is blocked by one of two conditions. cI binds operator 2 very quickly after binding to operator 1 due to the cooperativity of two cI dimers interacting. However, cro lacks cooperativity because cro dimers are composed of a single domain that binds DNA. An infected cell either contain cI dimers blocking access to PR or the cell contains high concentrations of cro dimers that bind to operator 3.

Which portion of the preimplantation embryo on the right side of Figure 15.14B displayed the greatest concentration of MHC IG protein on its surface? Why does this expression pattern of MHC IG make sense given the anatomy of a fetus inside the uterus?

Based on the data in Figure 15.14B, trophoblast cells produce more MHC IG than the small circular embryo, as indicated by the brighter green cytotrophoblasts. Given that cytotrophoblasts directly interact with maternal NK and T cells, it makes sense that the trophoblast would benefit from a higher density of MHC IG than the fetus, which is physically separated from the mother's immune system.

Did reproduction rates of island or mainland opossums vary more between the 2 years of breeding in Figure 15.30? Are these results consistent or contradictory to disposable soma hypothesis? If disposable soma hypothesis is correct, what reproductive outcome would you predict for island opossums that lived 3 to 4 years?

Based on the data in Figures 15.30C and 15.30D, island opossums do not vary in total mass of their offspring between year one and two, whereas mainland opossums produce a greater mass in year one. The mass of mainland opossums' second year litter is approximately the same as the mass for both litters for island opossums. If island opossums reproduce during year three and into year four, you would expect the island opossums eventually to produce more offspring than their mainland counterparts, but the data do not extend beyond two years. Disposable soma hypothesis correctly predicted that short-lived individuals would produce more offspring sooner because these mainland opossums allocated less energy to longevity than island opossums. Island opossums allocated more energy to long-term survival because the reduced island predation means they enjoy the possibility of living four years instead of only two.

Explain the logical contradiction between a pregnant woman carrying a fetus and allograft rejection. Given that tissue near allograft transplants swells and turns red prior to dying, what physiological process do you think causes the rejection of the foreign tissue?

Based on the redness and swelling, Medawar correctly hypothesized that allograft rejection was caused by an immune response. Autografts are "self" tissue meaning the tissue is genetically identical to the recipient, whereas allografts are "non-self" tissues and rejected the same way virally infected host cells are destroyed. Immune systems attack and reject non-self tissue, which makes the contradiction of a pregnant mammal more apparent. By definition, a fetus is 50% non-self and therefore should be rejected like an allograft, but fetuses are not rejected. This contradiction stimulated a very compelling set of experiments to understand why pregnant women fail to reject the non-self fetal tissue inside them.

If you were a T cell trying to distinguish self from non-self by touch, which part or parts of MHC-I plus peptide do you think you would feel in Figure 15.10? Because MHC I molecules only display peptides translated within the cell on which they are presented, do you expect the peptide to bind weakly or tightly to the MHC I molecule displaying it?

Because MHC I molecules display self protein fragments, these peptides must bind tightly to prevent peptides produced in other cells from binding to an empty space and incorrectly appearing as a self peptide. Go to this online Jsmol tutorial of MHC I that allows you to understand better the relationship between an MHC I molecule and its bound peptide. T cells use receptor proteins in their plasma membranes to "feel" MHC I + peptide on the surfaces of all cells to distinguish self from non-self. In their efforts to recognize cells, T cell receptors physically interact with the peptide and MHC I molecules, as you might imagine when you study their structure in Figure 15.12A. The peptide and the "antlers" of MHC I molecules are about the same height, which gives a T cell receptor equal access to both at the same time.

Describe the difference between ob and ob2J in Figure 15.21A. What type of mutation is present in the ob allele (Figure 15.21B)? Hypothesize how both ob- and ob2J alleles could cause the same ob phenotype despite their transcriptional differences.

Before looking at the data in Figure 15.21, it might have seemed difficult to understand how it could be possible to have two different mutant alleles of leptin. From Figure 15.21A, you can see that the recessive ob allele produced much more leptin mRNA than the wt allele, which is counterintuitive. Given that ob is a recessive mutation, the mutant ob leptin mRNA cannot encode functional leptin protein. The mRNA over abundance appears to be the body's homeostatic mechanism attempting to balance food intake and energy storage in a broken system. Leptin protein is produced in fat cells of wt mice, and an ob mouse has excessive fat deposits. As ob fat cells accumulated, the fat cells transcribed more and more leptin mRNA. Figure 15.21B shows that the ob allele contains a nonsense mutation, a premature stop codon. Producing a truncated, nonfunctional leptin protein means ob mice lack the wildtype negative feedback loop that would normally slow eating and fat storage. The ob mouse continued to produce more fat cells, and these fat cells transcribed more defective leptin mRNA and more nonfunctional truncated leptin protein in a futile feedback loop of failed regulation. However, the more recently discovered ob2J allele does not produce any detectable levels of leptin mRNA, indicating that the mutation may be in the ob2J leptin promoter. Therefore, the two leptin alleles contain two different mutations in the same gene, and the alleles produce two different errors that result in identical phenotypes caused by a lack of functional leptin protein.

Distinguish the two ob mutant alleles presented in Figure 15.21.

Differentiate ob and db mutant strains of mice using the data in Figure 15.19, Figure 15.22, and Table 15.3.

Both ob and db are autosomal recessive mutations and their phenotypes are only present in homozygous individuals. Parabiotic mice co-circulate blood due to the surgery used to connect the two individuals. Figure 15.22 demonstrated for the first time that the leptin protein could functionally complement the ob phenotype. When human or mouse leptin protein was produced in E. coli and injected into ob mice, the mice ate significantly less food and lost significantly more weight, predominantly in the form of less fat. The two negative controls of no injection or injecting buffer did not affect the mice. The db mice, however, did not change their eating behavior, nor did they lose any weight when injected with leptin. How can a functional leptin protein have no effect on db mice? You may have speculated a mechanism based on what you have already learned about signal transduction—ligands require receptors. {Connections: You learned about signal transduction in Section 7.2.} Leptin is a protein ligand secreted by fat cells that circulates in the blood. Circulating leptin explains the parabiotic results of Figure 15.19 because functional leptin protein from wt mice was able to complement ob mice and cause the obese mice to lose weight. To affect change, a ligand must bind to a functional receptor to initiate signal transduction. Understanding signal transduction in general led to the prediction that the db mutation might produce a nonfunctional leptin receptor. This hypothesis is consistent with the data, including the inability of leptin injections to rescue db mice in Figure 15.22 and the inability of wt mice to rescue conjoined db mice in Figure 15.19. The parabiotic data indicated the leptin receptor did not circulate in the blood. You would expect a receptor to be embedded in the plasma membrane of the appropriate cells, whatever those are. The receptor hypothesis for db led many investigators to search for the leptin receptor gene and determine the cause of the db phenotype. Table 15.3 presents the averaged weights and percent body fat for each of the six groups of mice. As you would expect, every group showed an increase in weight as the young mice grew. The investigators presented the variance in their data, which is helpful, but they did not indicate whether any of the differences were significant or not. You are left to estimate significant differences based on your prior biology experience.

Compose a defense of quorum sensing as evolutionarily beneficial to the squid and the bacteria compared to constitutive glowing by symbiotic V. fischeri.

Both the squid and the bacteria benefit from quorum sensing to regulate light production in V. fischeri. The bacteria don't want to waste energy and produce light when it does not help the squid. There is no point in producing light when the cells are too sparse to produce a useful amount of light. Each morning, the squid purges most of the bacteria into the ocean, and thus turns off light production. The cells offer no benefit to the squid during the day when it is trying to hide in the sand. The squid benefits from nighttime production of light, which is activated once sufficient density of V. fischeri cells accumulate in early evening. The emergent property of quorum sensing benefits both the squid and the bacteria. The squid provides shelter and nutrients for the bacteria in exchange for producing the squid's nightlight. You can include single-celled organisms in your list of organisms that transfer information to communicate with individuals of a population. Individual V. fischeri cells use communication for species recognition, determination of population density, and cooperation. The bacteria do not generate light for their own benefit; they do so for the squid. Without bioluminescence, the nocturnal squid would be easier to see by predators below the squid. In addition, a glowing squid does not cast a moon shadow, which means it can sneak up on its prey better. The squid benefits from its bacterial symbiont and in exchange provides the bacteria with shelter and nutrients. Therefore, V. fischeri benefits indirectly from its ability to produce light inside squid. From

View interactive Jsmol tutorials of cro dimer and cI dimer bound to the λ DNA bistable toggle switch. Compare their structures to see why cI can bind with cooperativity but cro cannot. In which DNA groove do these two proteins bind the double helix, major or minor groove?

By viewing the two structures for Integrating Question #21, you may have noticed cro and cI proteins bind to DNA using domains that contain only alpha helices. Both DNA-binding proteins fit into the major groove and interact with the paired bases but not the phosphodiester backbones. Both proteins physically touch the DNA and bind with affinities which are determined by their amino acid sequences. Therefore, if you wanted to search for potential common evolutionary origins, you could align the amino acids as you did for Integrating Question #22. Your BLASTp results showed the two proteins have very little amino acid sequence similarity (Figure 13.16). The lack of similarity indicates these proteins probably have different evolutionary histories rather than evolving from a common ancestral gene. Perhaps the lack of amino acid conservation should not be surprising given that the proteins have different affinities for the three operators. However, you could not have known until you aligned the sequences. Proposing hypotheses and testing them is how scientists work and contribute to the collective knowledge. Now you are doing science too.

Connect the allosteric modulation of hemoglobin subunits with the oxygen saturation curve in Figure 13.1. Explain why the slope of the hemoglobin curve is not linear like the water line.

Cooperativity is an emergent property caused by allosteric interactions initiated by the subtle movement of one heme group. Hemoglobin exhibits a steep Hill plot slope because of the rapid change in O2 binding affinity from low to high after the first oxygen is bound. When O2 gas molecules dissolve in plain water, there is no cooperativity and O2's solubility is linear with the amount of O2. Natural selection has favored two globin genes instead of only one, presumably because it takes two different subunits working as dimers within the four subunit hemoglobin tetramer to produce allosteric interactions. You have learned much of what is known about hemoglobin function, and now you have hit the limits of the field's understanding. Biologists are still trying to determine the exact steps required for cooperativity and, specifically, why two different subunits work but α alone or β alone is insufficient for the emergent property of wild-type hemoglobin.

Describe how E. coli's use of quorum sensing differs from V. fischeri's based on the data in Figure 14.5. Speculate what E. coli seems to be "saying" about the amount of glucose present.

E. coli does not use quorum sensing to produce light. It uses quorum sensing to indicate the amount of available food in the surrounding environment. When cells detected five times more glucose, they made more autoinducer than cells at the same density in 0.1% glucose. Unlike V. fischeri, E. coli appears to destroy its own autoinducer when the availability of food goes below a critical threshold, as demonstrated by the reduced signaling activity when the glucose concentration approached zero in Figure 14.5. Once again, the ability of bacterial cells to communicate generated an emergent property of group behavior that increases the population's efficiency for energy acquisition and growth.

Hypothesize how the fetus avoids destruction by T cells and NK cells. Do you think fetal cells lack MHC I?

Fetal cells must have MHC I molecules or the NK cells would attack and destroy the fetus.

Based on the time zero data in Figure 13.8B, do all lysogenic (colony forming) bacteria remain lysogenic in the absence of UV light? What do the data at time zero tell you about the amount of noise (random variation) in the choice between lytic and lysogenic?

Figure 13.8B shows that lysogenic phage do not remain 100% lysogenic because the switch that determines their lifestyle is imperfect and therefore "noisy." If all lysogenic cells remained lysogenic, the percent of colony-forming cells at zero minutes would have been 100%, and the percentage of plaque-forming cells would have been 0%. By evolving a noisy or leaky switch, λ produces new phage at a low rate all the time, which is probably adaptive because lysis results in more copies of the virus floating around in search of more host cells to infect. After UV treatment, the cells that used to be lysogenic quickly convert to lytic, as indicated by the reciprocal curves in Figure 13.8B. It takes about 30 minutes for all the lysogenic λ to make the genetic switch, produce new viruses, and rupture their host cells. Converting from lysogenic to lytic requires new protein production to kill the host cells. Therefore, the conversion to lytic viruses produced a smooth curve and not a digital one-step change because the process of switching lifestyles involved random diffusion of proteins and DNA. {Connections: Diffusion of proteins inside cells was explored in Section 8.2.} Random diffusion of molecules means it takes an average of 25 minutes after UV exposure for the infected cells to lyse.

Diagram the paired promoters that govern the genetic choice made by λ phage. Incorporate into your diagram how the two promoters are mutually exclusive.

Figure 13.9 Promoter PR initiates transcription by RNA polymerase to the right, whereas promoter PRM initiates transcription to the left. Two RNA polymerases would never be bound to both promoters at the same time in an infected E. coli.

As indicated in Figure 13.9, RNA polymerases transcribe only in one direction. If an RNA polymerase binds to a subset of the operators, hypothesize whether the transcription factors bind to operators covered by RNA polymerase or to operators not covered by RNA polymerase. Explain the logic you used to deduce this answer given the limited information you have so far.

Figure 13.9 depicts two RNA polymerases bound to the λ genetic switch, but this would not happen in a living cell. Only one RNA polymerase binds to a single promoter. Once bound, the RNA polymerase could move only in one direction to transcribe one of the two critical genes adjacent to the PRM or PR promoters. The choice between lytic and lysogenic is determined by which promoter binds the RNA polymerase and whether cro or cI is transcribed. Transcription factors bind to DNA and stimulate the adjacent RNA polymerase to produce mRNA. If a transcription factor were bound to operator 1, then RNA polymerase could not bind to PR, and it would have to bind to PRM and transcribe cI. Conversely, transcription factors bound to operator 3 would block access to PRM and result in the RNA polymerase binding to PR and initiating transcription of cro mRNA.

Use the data in Figures 14.2 and 14.5 to explain how quorum sensing can be used to communicate different types of information.

Figure 14.2: Once marine biologists realized that the light produced in squid was caused by bacteria, many biologists wanted to understand this unexpected mutualistic symbiosis. Margaret McFall-Ngai and Edward Ruby wondered how baby squid are born without bacteria but later obtain their mutualistic bacteria. Squid continuously circulate seawater through their bodies to obtain oxygen. {Connections: The use of oxygen in cellular respiration was examined in Section 10.4.} The water enters the body through paired openings near the animal's eyes. The bacteria normally accumulate and grow in a specialized compartment called the light organ. The investigators exposed baby squid to seawater from two distant locations—Hawaii and California. As part of the experiment, they sterilized some water samples to kill the V. fischeri living in the water. Some water samples were spiked with known sources of V. fischeri as positive controls. They placed the baby squid in the water and later measured the amount of light produced in the squid (Figure 14.2). From the data in Figure 14.2, you can see that squid are not born with the V. fischeri needed for bioluminescence. When grown in sterile sea water or sea water from California, the Hawaiian squid never developed the ability to produce their own light. When exposed to non-sterile water from Hawaii or water spiked with bacteria from Hawaiian squid, the babies were able to generate light. The water from California did contain a strain of V. fischeri, but these bacteria were not able to take up residence in the Hawaiian squid. These results indicate the symbiosis is more complicated than first expected. It appears the animals and bacteria coevolved with regional strain differences because the Hawaiian squid could not host the Californian strain of bacteria. Figure 14.5: The Bassler lab not only revealed the quorum sensing mechanism in V. fischeri, they also studied quorum sensing in E. coli (Figure 14.5). In these experiments, Bassler's lab grew E. coli in two different concentrations of glucose, which was used as an energy source for the bacteria. The biologists measured the amount of autoinducer secreted and plotted it as a percentage of maximum signaling activity while simultaneously measuring the amount of glucose and the number of E. coli cells in the media. Notice that the amount of autoinducer is affected by the amount of glucose in the media. E. coli's autoinducer must be saying something different than V. fischeri, which produces autoinducer proportional to cell density. E. coli does not use quorum sensing to produce light. It uses quorum sensing to indicate the amount of available food in the surrounding environment. When cells detected five times more glucose, they made more autoinducer than cells at the same density in 0.1% glucose. Unlike V. fischeri, E. coli appears to destroy its own autoinducer when the availability of food goes below a critical threshold, as demonstrated by the reduced signaling activity when the glucose concentration approached zero in Figure 14.5. Once again, the ability of bacterial cells to communicate generated an emergent property of group behavior that increases the population's efficiency for energy acquisition and growth.

Analyze the first 7.5 hours of data in Figure 14.3. What was the cell density of the bacteria when light was first detected? Did the two methods of measuring bioluminescence induction give different results? What was happening to cell density during this time?

Figure 14.3 indicates that bioluminescence of V. fischeri only occurs when the bacteria use quorum sensing to determine cell density of the population (about 10% maximum density). When the culture media was not crowded with cells, there was no bioluminescence. When a bacterial culture reaches a threshold number of bacteria, a quorum all of the cells in the culture turn on their bioluminescence. Both bioluminescence detection methods produced identical data for the first 7.5 hours, indicating induction of bioluminescence is regulated by production of the luciferase enzyme.

What effect did the injected leptin have on ob and db mice in Figure 15.22? What was the function of the buffer and no injection treatments? Speculate about a mechanism to explain why leptin injections affected the ob mice but not the db mice.

Figure 15.22 demonstrated for the first time that the leptin protein could functionally complement the ob phenotype. When human or mouse leptin protein was produced in E. coli and injected into ob mice, the mice ate significantly less food and lost significantly more weight, predominantly in the form of less fat. The two negative controls of no injection or injecting buffer did not affect the mice. The db mice, however, did not change their eating behavior, nor did they lose any weight when injected with leptin. How can a functional leptin protein have no effect on db mice? You may have speculated a mechanism based on what you have already learned about signal transduction—ligands require receptors. {Connections: You learned about signal transduction in Section 7.2.} Leptin is a protein ligand secreted by fat cells that circulates in the blood. Circulating leptin explains the parabiotic results of Figure 15.19 because functional leptin protein from wt mice was able to complement ob mice and cause the obese mice to lose weight. To affect change, a ligand must bind to a functional receptor to initiate signal transduction. Understanding signal transduction in general led to the prediction that the db mutation might produce a nonfunctional leptin receptor. This hypothesis is consistent with the data, including the inability of leptin injections to rescue db mice in Figure 15.22 and the inability of wt mice to rescue conjoined db mice in Figure 15.19. The parabiotic data indicated the leptin receptor did not circulate in the blood. You would expect a receptor to be embedded in the plasma membrane of the appropriate cells, whatever those are. The receptor hypothesis for db led many investigators to search for the leptin receptor gene and determine the cause of the db phenotype.

Look at the amino acid sequence alignments in Figure 13.5. Find the seventh amino acid in human β hemoglobin that causes sickle cell disease when mutated from glutamic acid (E) to valine (V). Do any species have V in this position? Another common blood disease is β thalassemia. Go to the NCBI OMIM database ID# 613985. What is the molecular cause of β-zero-thalassemia? What does β-zero-thalassemia tell you about the ability of α hemoglobin to function without β hemoglobin?

From Figure 13.5, you can see more evidence of vertebrate evolution when you study the α and β hemoglobin protein sequences. In Section 6.2, you saw the evolutionary conservation of histone protein sequences, but their termini were more variable than their central regions (see histone amino acid alignment). Hemoglobin subunits have less diversity at their termini, which indicates that their termini are critical to the overall structure of the four-subunit protein. None of the β hemoglobins in Figure 13.5 have a valine (V) for amino acid seven that causes sickle cell disease. {Connection: Sickle cell disease is in Section 22.1.} You would not expect other animals to use V at position seven given the severe consequences of mutating that amino acid. Beta-zero-thalassemia alleles produce no functional β hemoglobin protein, which tells you that α hemoglobin is necessary, but not sufficient, for normal hemoglobin function. From the β-zero disease phenotype, you know that hemoglobin must be composed of two α and two β subunits, a combination which is also supported by the high degree of conservation for the animals shown in Figure 13.5. By analyzing the hemoglobin data, you have constructed a model of the molecule's emergent property of cooperativity that allows hemoglobin to switch from binding to releasing O2. What you need to learn next is how hemoglobin's structure facilitates the cooperative binding.

Use the data in Figure 15.25A to explain why you get hungry at lunch time when you ate breakfast 4 hours earlier, but you can sleep without hunger from 10 PM to 8 AM. Interpret the negative correlation with leptin and body fat in Figure 15.25B. How are the data in Figure 15.25B connected to the data in Figure 15.25A?

From Figure 15.25A, you can see that leptin increases at night when most people sleep and therefore you can go longer without feeling hungry at night than you can during the day when your leptin levels are lower. These data also illustrate why it is a bad idea to have a midnight snack if you are trying to reduce your fat accumulation. Snacking at night overrides your natural tendency not to eat at night, and this could hasten the onset of obesity. For the girls in Figure 15.25B with lower percent changes in levels of nocturnal leptin, you would predict an increase in fat accumulation. As expected, the girls with lower percent changes in nocturnal leptin levels gained more body fat over the 6 months, whereas those with higher percent changes in levels of nocturnal leptin lost body fat.

Which tissues express the leptin receptor mRNA and presumably the protein too? What tissues are not included in Figure 15.23A that you would like to see in this experiment? What controls are lacking from this experiment? What do you learn in Figure 15.23C that is especially informative?

From the RNA Northern blot data in Figure 15.23A, it appears every tissue produces some leptin receptor mRNA, but some tissues produce more than others. However, the investigators did not include an actin control that would allow you to evaluate the amount and quality of mRNA loaded in each lane. Although lung cells appear to produce the most leptin receptor mRNA, you cannot be sure because perhaps more mRNA was loaded in the lung lane than any of the other lanes. Scroll down some on the NCBI website to see a bar graph showing how much leptin receptor mRNA is transcribed in 20 human tissues. From the NCBI information, you can see that human lung cells produce a surprising amount of leptin receptor mRNA but no more than the human ovaries and less than human liver. You might have expected the investigators to include fat cells in their RNA Northern blot. From NCBI, you can see that adipose, or fat cells, produces about the same amount as lung and ovary, but less than liver. It seems odd that the tissue with the highest levels of leptin receptor mRNA is the liver, meaning more research needs to be conducted to understand all of leptin's emergent properties. The NCBI data reveal the importance of a loading control for Northern blots that was absent in Figure 15.23A. In the brain tissue of Figure 15.23C, you might have expected to see the hypothalamus labeled brightly with radioactive leptin, but instead you see the choroid plexus contains the highest concentration of leptin receptor protein. The choroid plexus is composed of capillary blood vessels within the spaces, called ventricles, of the brain. The cerebrospinal fluid is produced by the choroid plexus and provides a physical barrier to exclude many blood-borne molecules from the brain and spinal cord. The presence of leptin in cerebrospinal fluid indicates some cells of the brain and spinal cord probably express leptin receptors on their plasma membranes, too. The current understanding of why the choroid plexus has a very high concentration of leptin receptors is that this capillary bed transports leptin from the blood into the cerebrospinal fluid. Leptin is produced in fat cells, circulates through the blood, and is transported to the cerebrospinal fluid, which supplies the leptin ligand to cells of the hypothalamus. It would be reasonable to predict at least some hypothalamus neurons produce leptin receptors on their plasma membranes, but you are not able to see the hypothalamus in Figure 15.23C.

Are squid born with the ability to generate bioluminescence? If they are not born with the bacteria, how do the squid acquire their symbiotic bacteria? Support your answer with data from Figure 14.2.

From the data in Figure 14.2, you can see that squid are not born with the V. fischeri needed for bioluminescence. When grown in sterile sea water or sea water from California, the Hawaiian squid never developed the ability to produce their own light. When exposed to non-sterile water from Hawaii or water spiked with bacteria from Hawaiian squid, the babies were able to generate light. The water from California did contain a strain of V. fischeri, but these bacteria were not able to take up residence in the Hawaiian squid. These results indicate the symbiosis is more complicated than first expected. It appears the animals and bacteria coevolved with regional strain differences because the Hawaiian squid could not host the Californian strain of bacteria.

Review the mechanism that permits non-self embryos to avoid immune rejection by their mothers.

Given its high blood content, the placenta is where you would expect the maternal immune system to attack fetal cells and thus kill the entire embryo. The key to an embryo's survival is in the MHC I alleles that it presents on every cell derived from the fertilized egg. Given that cytotrophoblasts directly interact with maternal NK and T cells, it makes sense that the trophoblast would benefit from a higher density of MHC IG than the fetus, which is physically separated from the mother's immune system. MHC IG protects trophoblast cells from lysis by NK cells in Figure 15.15, as indicated by the increased cell lysis when incubated with the anti-MHC IG antibody. The two irrelevant antibodies in Figure 15.15 did not prevent NK cells from ignoring non-self cytotrophoblasts. However, the anti-MHC IG antibody masked the protective MHC IG proteins and allowed the NK cells to attack the non-self cells. Only about 20% of the cytotrophoblasts were lysed when incubated with anti-MHC IG because in vitro experiments are never 100% efficient.

How many oxygen molecules bind to hemoglobin composed of two α subunits and two β subunits? How many O2 molecules bind as a result of cooperativity? Use the data in Figure 13.3 to explain why myoglobin has a slope of 1 in the Hill plot of Figure 13.4. Given that myoglobin is found in muscle cells, speculate on its function.

Hemoglobin binds four oxygen molecules—one for each subunit, because each subunit carries one iron-containing heme group. Myoglobin only binds a single O2 molecule, so it does not provide a large volume of O2 buffer. However, myoglobin does provide an emergency supply of oxygen in extreme cases of low O2 supply inside muscle cells. Based on the Hill plot in Figure 13.4, you can see that myoglobin exhibits no cooperativity and thus no emergent properties related to O2 binding. Myoglobin is similar to the high affinity binding protein in Figure 13.3 and lacks a sigmoidal O2 saturation curve because it lacks cooperativity. Hemoglobin has a sigmoidal O2 saturation curve and a Hill plot slope close to three, indicating that once the first O2 binds, the other three bind very rapidly, producing the emergent property of cooperativity.

Summarize the significance of the saturation curve of hemoglobin in Figure 13.1. What emergent property is revealed by this figure? How can you quantify this emergent property?

Hemoglobin is your first example of an emergent property at the molecular level. A very small movement by one iron ion leads to massive behavioral changes for one of the most abundant proteins in your body. It still may be difficult for you to visualize the impact of hemoglobin's emergent property, but you can feel its effects every time you exercise vigorously and need to "catch your breath." In a hemoglobin solution, the amount of dissolved oxygen increases in an unexpected way, revealing an emergent property of hemoglobin. The hemoglobin binds to oxygen very slowly until its concentration reaches about 1 kilopascal (kPa) of pressure, which is 1% of standard atmospheric pressure (100 kPa). From Table 13.1, you can see that the partial pressure of oxygen in air would be 21% of the standard atmospheric pressure, or approximately 20 kPa. When exposed to air, a solution of hemoglobin is fully saturated with oxygen. The emergent property of hemoglobin is revealed between 1 and 9 kPa of oxygen concentration. With very little increase in oxygen concentration, the amount of oxygen dissolved in a hemoglobin solution increases very rapidly. The shape of this curve is often referred to as sigmoidal, which is derived from the Greek word sigmoid, which means curved in two directions like the letter S. {Connections: The logistic model for population growth discussed in BME 24.2.} The sigmoid-shaped oxygen solubility curve is what allows you to breathe in air and deliver oxygen to all of your cells.

AnAge database documents the maximum age for a wide range of species. Sort by longevity and determine what species has the longest known lifespan. Is the oldest species a plant or an animal? You might have to search the Internet if you don't recognize the Latin name.

Hexactinellid sponge is the surprising longevity winner with an estimated lifespan of 15,000 years, which is about 100 longer than the oldest human! Sponges are colonial animals that feed on small plankton in the water. Perhaps you were also surprised to realize that 14 of the oldest species live in the water and that 14 species are known to live longer than humans. The oldest silver-haired gorilla was 55.4 years old, lived in the Dallas, Texas zoo, died of cancer, and was diagnosed with cataracts in her eyes. Humans live a long time compared to other primates, but we cannot compete with sponges for longevity.

Look at the data between 7.5 and 12 hours. Did the two methods of measuring bioluminescence induction give different results? What was happening to cell density during this time?

However, between 7.5 and 12 hours, the amount of enzyme remains at its maximum level but direct luminescence from the culture of live cells is reduced by about 70%. Because enzyme accumulation was sustained for 4.5 hours, the cells must have glowed less because they lacked available substrate. Remember the enzyme accumulation was determined after the investigators added additional substrate to the cytoplasmic extract, whereas direct bacterial bioluminescence relied on substrate produced within live bacterial cells. The Hawaiian bobtail squid allows V. fischeri to grow inside its light organ so that the cells can reach the quorum density and induce bioluminescence. From the data in Figure 14.3, it is clear that cells only turn on bioluminescence when the bacteria reach a certain density. V. fischeri do not obtain luciferase enzyme or its substrate from the squid.

Search PubMed using the three author names, Hylenius, Melbye, and Hviid, to locate a 2004 paper that they and others published in Molecular Human Reproduction. What mutation is responsible for some cases of preeclampsia?

Hylenius and colleagues documented a correlation between particular MHC IG alleles and preeclampsia (Figure 15.16). The human MHC IG gene is composed of eight exons with the last two exons being part of the 3' untranslated region of the mRNA. {Connections: Nucleotides in mRNA after the stop codon were addressed in Chapter 2.} The paper by Hylenius and colleagues discovered that preeclampsia fetuses often contain a 14 bp insertion in exon 8.

Which of the five traits in Table 15.5 showed a significant difference between the low and high adult fly mortality treatments? Which of these differences followed the predictions of disposable soma theory? Summarize the disposable soma theory using the fly data.

In Table 15.5, all five fly traits showed significant differences between the two treatments. The male and female flies of the high mortality group developed faster, weighed less, and produced about 50% more offspring. All of these physical traits responded as you would have predicted from the disposable soma theory as a result of the strong selection pressure applied to the adults.

Why is cro produced instead of cI when no transcription factors are bound to either PRM or PR? What aspect of the cro protein prevents it from exhibiting the emergent behavior of cooperativity found in cI? What is the consequence when a second operator is occupied by cro dimers?

In the absence of cI, cro is produced because RNA polymerase has a higher affinity for the DNA of PR than PRM. PR is active unless RNA polymerase is blocked by one of two conditions. cI binds operator 2 very quickly after binding to operator 1 due to the cooperativity of two cI dimers interacting. However, cro lacks cooperativity because cro dimers are composed of a single domain that binds DNA. An infected cell either contain cI dimers blocking access to PR or the cell contains high concentrations of cro dimers that bind to operator 3.

What tissues produce leptin and the leptin receptor proteins? Support your answers using the data in this section.

In the brain tissue of Figure 15.23C, you might have expected to see the hypothalamus labeled brightly with radioactive leptin, but instead you see the choroid plexus contains the highest concentration of leptin receptor protein. The choroid plexus is composed of capillary blood vessels within the spaces, called ventricles, of the brain. The cerebrospinal fluid is produced by the choroid plexus and provides a physical barrier to exclude many blood-borne molecules from the brain and spinal cord. The presence of leptin in cerebrospinal fluid indicates some cells of the brain and spinal cord probably express leptin receptors on their plasma membranes, too. The current understanding of why the choroid plexus has a very high concentration of leptin receptors is that this capillary bed transports leptin from the blood into the cerebrospinal fluid. Leptin is produced in fat cells, circulates through the blood, and is transported to the cerebrospinal fluid, which supplies the leptin ligand to cells of the hypothalamus. It would be reasonable to predict at least some hypothalamus neurons produce leptin receptors on their plasma membranes, but you are not able to see the hypothalamus in Figure 15.23C.

Using the data in Table 15.4 below, which traits are significantly different between the two populations besides longevity? Which of these traits indicate the island population was subjected to harsher conditions than mainland opossums, even though island opossums lived longer?

Island opossums have much smaller litters and carry many more ticks than their mainland counterparts. Despite the higher load of ticks, the island opossums have glucose levels comparable to mainland individuals. In short, the disposable soma hypothesis predictions held true, and Austad's data supported the explanation for the evolution of senescence, which is an emergent property at the organismal level. With experimental support from Austad and many others, the disposable soma hypothesis is now referred to as a theory. {Connections: The scientific use of the term theory, reserved for hypotheses that are extensively validated, is discussed in Ethical, Legal, Social Implications 4.1.} For Homo sapiens, our life history in Figure 15.29A was initially determined by survival in the wild, but now humans have altered the selection pressures and we tend to live longer and reproduce later. People allocate many of their resources to live long lives, but Kirkwood comically summarized his opinion when he wrote, "The best recipe for a long life is to choose your parents well."

Compare and contrast lytic and lysogenic lifestyles for λ phage. What environmental signals can cause the virus to switch from one to the other?

Lambda has two choices of lifestyles that it can execute inside E. coli. One choice is to live a lysogenic lifestyle, which means the virus inserts its genome in the bacterial chromosome, and no new viruses are produced. Lysogenic λ phage is a latent stage, meaning the infected cell does not produce any progeny viruses. Lysogenic phage are like Trojan horses in that the lethal phage lay dormant but are ready to spring into action when the time is right. Until then, the viral genome is replicated as part of the bacterial chromosome every time the cell reproduces. In other words, every 30 minutes, the number of lysogenic viral genomes doubles, which leads to exponential growth of latent viruses. However, when the time is right, λ phage can switch from lysogenic to a lytic lifestyle, which results in the production of over one hundred new viruses that rupture, or lyse, the infected cell (Figure 13.7C). When the host cell dies and releases the new viruses, one hundred new bacteria can be infected, and with each new infection, the virus has a choice of lytic or lysogenic lifestyles. When groups of neighboring bacteria lyse within a lawn of bacteria (a plate completely covered with bacteria), the cells leave a clear circle of dead cells called a plaque. When a virus chooses the lysogenic lifestyle, the viral genome is inserted into the host chromosome, and the combined genomes are replicated every time the infected cell divides (Figure 13.8A). When a virus chooses a lytic path, the virus takes over the bacterial transcription and translation processes to produce about 100 new viruses that emerge from the host when it lyses. In the 1950s, André Lwoff discovered that ultraviolet (UV) light could switch λ phage from lysogenic to lytic (Figure 13.8B). Lwoff exposed lysogenic bacteria to UV light and then sampled their ability to remain lysogenic and form bacterial colonies or their ability to produce new viruses and plaques on a fresh lawn of uninfected E. coli. When plating E. coli to produce a lawn, Lwoff spread cells that if left uninfected would have formed a lawn after 16 hours of growth at 37º C. During this growth period, λ phage can infect the bacteria and choose between lytic and lysogenic lifestyles.

Explain what a lawn of cells is and how plaques are formed within lawns.

Lawn of cells: Explain what a lawn of cells is and how plaques are formed within lawns. Plaques: a clear circle of lysed host cells on a lawn of bacteria. Plaques are clear zones formed in a lawn of cells due to lysis by phage.

To quantify the cooperativity of cI binding, you need to estimate the slope of the corresponding Hill plot. Bio-Math Exploration 13.1 told you that for any point (x,y) on a repression curve, the value on the horizontal axis of the Hill plot is given by s = log10(x), and the value on the vertical axis of the Hill plot is given by t = log10(y/(1-y)). Compute the values of s and t for the two points where the λ PR repression curve intersects the green area in Figure 13.15C. Estimate the slope of the Hill plot between these two points by dividing the change in t by the change in s (rise over run).

Looking at the math in Integrating Question #20 may seem intimidating at first, but a calculator makes it easy to solve. First, you have to decide what the x and y values are for the two points in Figure 13.15C: (4.5*10-9, 0.2 and 4.5*10-8, 0.95). Now plug in each of the paired numbers into the formulas in question 20, and you should get: (-8.35, -0.6) and (-7.35, 1.28). From these paired values, you can calculate the rise (1.28 - (-0.6) = 1.88) and run (-7.35 - (-8.35)) = 1. When you divide 1.88 by 1, you get an estimated Hill plot slope of 1.88. The theoretical maximum Hill plot slope would be 2 for any two-component system (binding of two cI dimers), and 1.88 is very close to 2. Therefore, the cooperativity of cI is nearly as high as theoretically possible, which helps explain why operator 2 is bound with the second cI dimer almost immediately after operator 1 is bound with the first cI dimer.

Explain what MHC I is and its role in the immune system.

MHC I molecules are integral membrane proteins as illustrated by the thin line passing from outside the cell into the cell's cytoplasm. The vast majority of MHC I protrudes into the extracellular world. MHC I proteins look like moose heads and between their antlers is an empty space. However, the thousands of MHC I molecules found on the surface of each cell never have empty spaces; the space is always occupied by a protein fragment, or peptide, that was translated inside the cell displaying the MHC I molecule. MHC I plus peptide is a cell's way of defining "self." In other words, MHC I molecules display fragments of every protein made inside that cell like a proud grandparent showing photos of grandchildren. All of the cells of an organism define "self" by displaying peptide fragments from every protein produced inside each cell of an individual. With the discovery of MHC I molecules, the rejection of allografts and acceptance of autografts made sense. MHC I presents self peptides on the surface of every grafted skin cell (Figure 15.11). Autograft cells display MHC I molecules on their surfaces with peptide fragments to inform the recipient's T cells that the skin graft is self, so the cells are never rejected. Allografts display MHC I plus peptides from proteins made inside genetically non-self cells and thus are recognized as foreign by the recipient's T cells. However, the data in Figure 15.11A uncovered an important characteristic of the MHC I molecules. Through careful breeding of mice, immunologists produced different strains of mice that were genetically identical at every locus except MHC I. Conversely, immunologists bred mice that were genetically identical only at the MHC I locus and varied at all other loci. Therefore, immunologists could distinguish the impact of MHC I differences versus the impact of the peptide differences on allograft rejection (Figure 15.11B).

Search the nucleotide database at NCBI using the accession number NM_002127.5. What is the chromosomal position of MHC IG? How many exons are in this gene?

MHC IG is within the MHC locus at 6p.21.3 (6p = short arm of chromosome six; band 21.3), and the gene is composed of eight exons. Only the first six exons encode amino acids and the last two exons are non-coding.

Which antibody in Figure 15.15 caused the most trophoblast cells to be lysed by NK cells? Why weren't 100% of the cells lysed when incubated with the MHC IG antibodies?

MHC IG protects trophoblast cells from lysis by NK cells in Figure 15.15, as indicated by the increased cell lysis when incubated with the anti-MHC IG antibody. The two irrelevant antibodies in Figure 15.15 did not prevent NK cells from ignoring non-self cytotrophoblasts. However, the anti-MHC IG antibody masked the protective MHC IG proteins and allowed the NK cells to attack the non-self cells. Only about 20% of the cytotrophoblasts were lysed when incubated with anti-MHC IG because in vitro experiments are never 100% efficient.

Based on the data in Figure 15.20A, speculate on whether you think human and mouse leptin proteins would function in the opposite species. Which tissue produces leptin as measured by mRNA? Is a leptin gene found in evolutionarily diverse animals?

Mammals share very similar leptin gene and protein sequences (Figure 15.20A) and all animals appear to have orthologs of leptin genes as measured by DNA Southern blotting (Figure 15.20C). In fact, mice can use human leptin protein when injected experimentally, and mouse leptin protein functions in human cells grown in tissue culture. Perhaps the biggest insight gained by the cloning and sequencing of the leptin gene was that its mRNA is detected only in fat cells and not any other tissue type, including the brain. The positive control of actin (produced in all cells) in the RT-PCR experiment indicates the mRNA was of high quality for all tissues and thus a negative result in the top gel indicates leptin mRNA was not detectable in these other tissues.

Describe differences between first-set and second-set rejections in Figure 15.6B. Although the data lack error bars, do you think there is a significant difference between a second-set rejection at the same site versus a second-set rejection at a different site (low dose)?

Medawar performed hundreds of skin graft experiments to quantify the rate of rejection and found that large skin patches were rejected faster than small patches as long as both were first-set rejections. Second-set rejections always occurred faster, and the time course of the rejection was not significantly influenced by whether the graft was transplanted to the same site or a different site.

Why would natural selection lead to the conservation of two α and two β subunits in vertebrate hemoglobin? What aspect of its emergent property of cooperativity requires four subunits rather than just one? What impact has hemoglobin's structure-function relationship had on conservation of hemoglobin amino acid sequences?

Natural selection has favored two globin genes instead of only one, presumably because it takes two different subunits working as dimers within the four subunit hemoglobin tetramer to produce allosteric interactions. You have learned much of what is known about hemoglobin function, and now you have hit the limits of the field's understanding. Biologists are still trying to determine the exact steps required for cooperativity and, specifically, why two different subunits work but α alone or β alone is insufficient for the emergent property of wild-type hemoglobin. It is clear from the highly conserved amino acid sequences that evolution has restricted the variation of the hemoglobin amino acid sequences. If you look closely in Figure 13.6A, you can see two termini near each other when oxygen is bound, but these termini are farther away when no oxygen is bound. The entire length of hemoglobin proteins has barely mutated over the 340 million years since the divergence of the most recent common ancestor of birds and mammals. Natural selection has purged changes that produced hemoglobins with reduced cooperativity. You can see the consequences of the reduced variation in hemoglobin in Figure 13.5 and in every vertebrate that breathes to live.

Search the Internet for the term "nude mouse" to see how the mouse strain got its name. Given the immunity phenotype of nude mice, do you think they live a healthy life in a normal environment? Search OMIM using the identifying code 242700. What is the consequence of this human mutation, and how does it compare with the immune system phenotype of the nude mouse?

Nude mice and humans with the mutation described by OMIM code 242700 lack T cells due to a nonfunctional thymus. T cells are called " T " because they mature and become "educated" in the thymus to recognize self. Without a thymus, animals are doomed to die from viral infections because T cells are primarily responsible for protecting us from viruses.

Apply the data in Figures 15.25 through 15.27 to explain why obesity often leads to other health-related complications.

Obesity can lead to a decrease in male fertility.

Convert the demographic data in Figure 15.17 into statements that could be used to support the need for physical activity (organized sports or regular exercise) for all students in high school.

Obesity is a growing problem worldwide, especially in the United States. Over the past 35 years, the percentage of teens categorized as obese has tripled (Figure 15.17). As illustrated in the 2010 White House report, gender and ethnicity correlate with the frequency of obesity in teens between the ages of 12 and 19. What has changed in the last 35 years to cause a substantial shift in the body weight of Americans? Isn't body weight constrained by organismal homeostasis? {Connections: The Big Idea of Homeostasis was the focus of Chapters 10 to 12.} For many years, physiologists demonstrated the importance of exercise in regulating weight. From Figure 15.17C, you can see that about half of all high school boys exercise less than 1 hour a week, whereas 75% of high school girls exercise less than 1 hour a week. Is exercise the only factor that regulates body weight? Obesity is defined by the Centers for Disease Control and Prevention (CDC) as a body mass index (BMI) of 30 or greater. You can calculate your BMI using the CDC's online tool. The physiological trait of body weight may not seem like an emergent property, but you will be surprised by the interconnectedness of fat regulation. Therefore, it is important that you consider the experimental results that explain how mammals regulate their body fat.

What effect has evolution had on the amino acid sequences of α and β hemoglobin across species? What can you conclude about the need for α and β subunits instead of relying on just one of the two subunits?

One hallmark of evolution is that a mechanism which works well is often conserved in a wide range of species (Figure 13.5). You can validate this hallmark by examining protein sequences for α and β hemoglobin from diverse species. Because each hemoglobin tetramer is composed of two α subunits and two β subunits, you might wonder why the ratio is 2:2. Why didn't evolution select for a molecule composed of four identical subunits, or three α and one β, and so on? When you see amino acid sequences conserved in many species that diverged from each other hundreds of millions of years ago, these particular amino acids are unlikely to change without destroying the molecule's function. All of these animals use two α subunits and two β subunits, but the subunits do have some variable amino acids. However, functionally critical amino acids are conserved across all species, indicating their positions are essential to hemoglobin's function in general, especially hemoglobin's cooperativity of binding oxygen.

Watch how students at your school interact. Describe three emergent properties of human behavior that you could not predict if you only saw people living in isolation. You can choose different settings such as athletic events, parties, class or lab.

People saying hi to each other in the hall when they see each other. People asking others if they need help carrying items. People holding doors open for others during an assembly.

Use a medical dictionary website to define the terms preeclampsia and eclampsia. What medical conditions do these two terms describe?

Preeclampsia: a pregnancy-induced medical condition that results in protein in the urine due to high blood pressure. Eclampsia: a life-threatening medical condition brought on by pregnancy-associated high blood pressure and no other cause.

The disposable soma theory predicted individual plants that bolted quickly would allocate fewer resources into leaves and senesce sooner. Did the Arabidopsis data in Figure 15.31 support or refute the predictions? Was the converse prediction true (slow to bolt, slow to senesce, and more leaves produced)?

Similarly, the plants on the right side of Figure 15.31A put more energy into reproduction by bolting sooner, and as a consequence, produced fewer leaves and senesced sooner than their late blooming counterparts. Flexibility of response due to randomness is one of the themes of emergent properties. Before you consider mechanistic causes, you should confront a widely held misconception that bacteria do not age.

What can you conclude about the nature of the chemical signal in the two strains of V. fischeri? Do the two strains of V. fischeri secrete the same quorum signal? Support your answer with data.

Strain 1 cells glowed soon after they were added to the centrifuged media from bioluminescent strain 1 cells. The immediate glowing demonstrated that the quorum sensing molecule was present in the media and had been secreted by the previous population of strain 1 cells. Strain 2 cells responded similarly when given strain 2 media. However, strain 1 cells did not respond to strain 2 media and vice versa. The quorum sensing molecules allow cells to communicate between individuals of the same strain but not a different strain. Because filtering did not block bioluminescence, the quorum sensing molecule must be smaller than 2 µm. Boiling the media had differential effects; the secreted molecule of strain 1 was destroyed by heating, whereas the secreted molecule of strain 2 was not. Therefore, the two strains produced different molecules for quorum sensing as shown by their different sensitivity to boiling.

Go to the interactive Jsmol view of T cell receptor and MHC I, and determine where this T cell receptor touches the MHC I molecule and the displayed peptide.

T cell receptors interact with the MHC I molecule more than the peptide, but the exact number of peptide interactions varies with different T cell receptors and different peptides. Both MHC I and the peptide are in direct physical contact with the T cell receptor.

What significant differences do you see in Table 15.2? What impact does exercise have in rat body composition? What are the implications from Table 15.2 for school children shown in Figure 15.17C?

Table 15.2 revealed that exercise has a profound impact on the body composition of ob rats. ob rats that exercised had significantly less fat than ob rats which did not exercise. Similarly, the amount of protein was significantly reduced in ob rats compared to wt rats, but ob rats that exercised had significantly more protein, presumably muscle protein, than ob rats lacking exercise. Although the ob rats were heavier than their wt counterparts, the total body weight differences were not significant. If you compare only their dry matter, ob rats were heavier, which implies water was more abundant in wt rats than ob rats. Based on these findings, you could predict that American teens who exercise less have less muscle protein and more fat than their peers of the same weight who exercise more. These data and many others clarify the link between lack of exercise and obesity.

Use Figure 14.1B to speculate how squid benefit from hosting V. fischeri. Hypothesize how the bacteria benefit from their host animals. Does their own bioluminescence help the bacteria directly?

The bacteria do not generate light for their own benefit; they do so for the squid. Without bioluminescence, the nocturnal squid would be easier to see by predators below the squid. In addition, a glowing squid does not cast a moon shadow, which means it can sneak up on its prey better. The squid benefits from its bacterial symbiont and in exchange provides the bacteria with shelter and nutrients. Therefore, V. fischeri benefits indirectly from its ability to produce light inside squid.

What is the cellular cause of ob sterility in males in Figure 15.27? What effect does injected leptin have on the fertility of ob males?

The connection between ob and male sterility is pretty clear in Figure 15.27. ob males that received leptin injections regained their fertility because they produced sperm for as long as the injections continued. Pair-fed mice did not produce sperm, indicating the amount of food was not the important factor. Unfortunately, few studies have been published trying to connect leptin and female fertility

Interpret the weight differences for each of the rats weighed individually at the end of the experiment in Figure 15.18C. What role does the "single rat" play in this experimental design?

The data in Figure 15.18 were among the first to establish firmly the critical role the hypothalamus plays in body fat regulation. Parabiotic control rats gained less weight than the single rat, but you cannot determine if this difference is significant because the figure does not display any variance. The experimental parabiotic rats collectively gained more weight than the single rat, and the rat from the conjoined pair that had its hypothalamus surgically lesioned gained the most weight of all rats. However, the parabiotic rat with a functional hypothalamus gained the least weight. Perhaps the obese condition of the conjoined individual with a hypothalamus lesion was able to communicate biochemically its physiological status to the conjoined control rat, but you would need more data to confirm this hypothesis. Preliminary data are good for generating hypotheses, not reaching firm conclusions.

What trait was significantly affected in Figure 15.26A? What are the implications of these data with overall human health? What happened to leptin mRNA when adult male rat adipose cells were exposed to high levels of insulin? What sort of feedback loop may be established between insulin and leptin?

The emergent properties of weight regulation and the many functions of leptin are apparent in Figure 15.26. Elevated leptin in wt rats led to increased blood pressure, which is a contributing factor in heart attacks and strokes. Added insulin produced elevated leptin mRNA in isolated rat cells, indicating glucose regulation influences leptin production and fat accumulation. It makes sense that glucose and fat would have overlapping cell signaling mechanisms, but their overlap makes it very difficult for investigators to distinguish cause and effect for obesity and diabetes. Regardless of the mechanisms, obesity and leptin are linked to high blood pressure and diabetes. Because obesity results in elevated insulin levels and more leptin, it appears the two protein ligands may spiral upwards in a connected positive feedback loop—more insulin produces more leptin and more leptin stimulates more insulin.

What role does cooperativity play in the λ switch to a lysogenic lifestyle?

The emergent property of cooperativity assures that cI always binds to operator 2 before operator 3. With one cI dimer sitting on operator 1, the second cI dimer can interact with the DNA at operator 2 and the cI dimer already bound to operator 1. This cooperativity of protein and DNA interaction assures that operator 2 fills almost immediately after operator 1 is filled. Once the concentration of cI dimers increases substantially, operator 3 is occupied by a third cI dimer, but this dimer does not interact with the previous two dimers already bound to operators 1 and 2. The emergent property of cooperativity of cI binding to operator 2 contributes to the positive feedback loop and neither of these emergent properties would have been evident if cI were studied in isolation. After biding to operator 3, cro dimers can bind to either operator 1 or 2 with equal probability, because cro does not use cooperativity for the second binding of a cro dimer.

What emergent properties are involved in the switch to lysis from lysogeny?

The emergent property of cooperativity of cI binding to operator 2 contributes to the positive feedback loop and neither of these emergent properties would have been evident if cI were studied in isolation. Hemoglobin and λ phage both exhibit emergent properties of switching between two states. Hemoglobin is a toggle switch that stays in the cooperative O2-binding shape as long as O2 is abundant and the pH is within a particular range. However, outside of these conditions, hemoglobin rapidly reverts to its no-oxygen bound state. The self-sustaining bistable toggle switch is possible because cro and cI both produce positive feedback loops, which are emergent properties. However, cro also produces a negative feedback loop that you can detect by studying Figure 13.13C.

Describe the data that demonstrated that pregnant mammals have a functional immune system but do not reject their non-self embryos.

Did rabbit a retain her immune function while pregnant? Did rabbit a's immune system prevent her from carrying fetuses that were 100% non-self? Support your answer with data from Figure 15.7.

The experiment in Figure 15.7 directly tested whether pregnant females had weakened immune systems. If pregnant females had reduced immune function, you would not expect them to reject skin allografts, but they did. About 75% of the foster mothers were able to carry the implanted b c embryos to term and deliver healthy offspring even though these transplanted offspring are 100% non-self. The immune systems of both rabbits a and d in Figure 15.8 rejected the allografts from the b c offspring as well as the new rabbit e graft. Rabbit d rejected both allografts at about the same rate (6.5 and 7 days), but rabbit a responded differently. Rabbit a rejected the allograft from foster b c offspring significantly faster (p < 0.01) than it rejected the allograft from rabbit e, even though both patches of skin were transplanted at the same time. {Connections: p-values were considered in BME 0.1.} The b c allografts were rejected faster because the skin patches were second-set transplants, whereas the allograft from rabbit e was a first-set rejection. Rabbit a rejected allograft from her b c foster offspring faster because rabbit a had previously been exposed to skin from rabbit b, as shown in Figure 15.7. Rabbit a rejected allograft e slowly because rabbit a had not been exposed to tissue from rabbit e before. Immune responses are specific for the source of non-self tissue, and immunity does not uniformly affect all allografts equally. {Connections: Specificity of the immune response was first seen in Section 5.3.}

Summarize the results in each parabiotic experiment in Figure 15.19. Does one mutant mouse produce what looks like a dominant effect even though mutant alleles of ob and db are recessive?

The experiments outlined in Figure 15.19 produced some unexpected results. A parabiotic db mouse caused its conjoined partners (ob and wt) to stop eating and eventually die from starvation. Therefore, it appears that something produced by the db mice was capable of overriding appetite control in both types of mice. When ob was conjoined to wt mice, the wt mice were unchanged, but the ob mice reduced their weight and other phenotypes associated with the ob phenotype. It appears the wt mouse produced something that was dominant to the ob phenotype. The dominant aspects of db and wt mice indicate the dominant factor was present in the blood, because they co-circulated blood in parabiotic experiments. Because ob mice were homozygous recessive, they must lack the leptin protein. Loss of functional leptin in ob mice explains why wt mice produced leptin and complemented the ob deficiency and restored normal eating and fat deposition in conjoined ob mice. What was unclear at that time was why db, a recessive mutation, overwhelmed both ob and wt mice to the point of starvation.

cI and cro proteins initiate two different positive feedback loops. Compare and contrast these two feedback loops. Are the three operators and two promoters in Figure 13.14 symmetrical, or is there a physical bias in one direction?

The final elegance of the λ bistable toggle switch that you will study is the subtle but significant asymmetry in the positioning of operator 2. The shift is barely detectable in Figure 13.14 but more apparent if you count the bases in Figure 13.15A. If you look closely, you will see that operator 2 is skewed by one base toward PR and away from PRM. The skew is easiest to see if you count the number of bases covered by operator 2 and part of either -35 box of the two adjacent promoters. The difference of one base means that when cI dimers are bound to operator 2, RNA polymerase is allosterically stimulated to begin transcription of cI. When cro dimers are bound to the same operator 2, the dimers obstruct the PR's -35 box and RNA polymerase cannot bind to PR. Now compare the DNA sequences of the -35 and -10 boxes of the two λ promoters (Figure 13.15B). Neither promoter perfectly matches the E. coli consensus sequence for RNA polymerase binding, but PR varies by only one base in each binding site; and these bases are located on the extreme edges of the boxes. In contrast, PRM has two mutations in both RNA polymerase binding sites, and these mutations are in the middle of the -10 and -35 boxes. Once you compare the two promoter sequences to the consensus binding sites, you can understand why PR does not require a transcription factor to be active but PRM does. To summarize, the two promoters vary by only six bases and the operators are skewed by only one base, but the two promoters have different properties due to their slightly different structures. The variation in sequence and symmetry produces a repression curve for PR that is reminiscent of the hemoglobin saturation curve, because both curves illustrate the cooperativity inherent in both processes (Figure 13.15C). You can see the consequences of cooperative binding of cI to operator 2 compared to a hypothetical repression curve in the absence of cooperativity. The green area highlights a tenfold change in cI concentration along the logarithmic scale on the x-axis, from 5 × 10-8 M to 5 × 10-7 M. Over this range of cI, repression due to cooperativity changes from about 15% to over 90%, whereas the hypothetical non-cooperativity repression would change only from about 80% to 99%. As with hemoglobin, cooperativity by cI produces a very sudden transition within a narrow range of concentration. Cooperativity converts a subtle input change to nearly a digital output for a quick transition between two states, the signal to flip the bistable toggle switch from lytic to lysogenic.

Describe the mutant MHC IG allele depicted in Figure 15.16. What effect does this mutation have on the MHC IG protein?

The human MHC IG gene is composed of eight exons with the last two exons being part of the 3' untranslated region of the mRNA. {Connections: Nucleotides in mRNA after the stop codon were addressed in Chapter 2.} The paper by Hylenius and colleagues discovered that preeclampsia fetuses often contain a 14 bp insertion in exon 8. The 14 bp insertion into exon 8 of MHC IG mRNA does not affect the primary amino acid structure of the MHC IG protein. Because the protein is not altered, there must be another mechanism for reduced MHC IG function. MHC IG mRNA that contains the 14 bp insertion was degraded faster than mRNA lacking the insertion. Therefore, fetuses that produce the mutant mRNA produce less MHC IG protein and are more vulnerable to maternal immune system attack.

Use the graphs in Figure 15.29 to discredit Weismann's hypothesis that death and aging evolved to make room for the young.

The hypothesis that aging and death made room for the young was discredited because of data similar to the "constant" graph in Figure 15.29. Most organisms die prior to senescence, so aging to make room for the next generation was inconsistent with the data.

Did rabbits a and d in Figure 15.8 reject skin grafts from rabbit e at about the same rate? Were the skin grafts from b c offspring rejected at the same rate by both rabbits a and d? Explain any significant differences you see in the rejection rates by rabbits a and d.

The immune systems of both rabbits a and d in Figure 15.8 rejected the allografts from the b c offspring as well as the new rabbit e graft. Rabbit d rejected both allografts at about the same rate (6.5 and 7 days), but rabbit a responded differently. Rabbit a rejected the allograft from foster b c offspring significantly faster (p < 0.01) than it rejected the allograft from rabbit e, even though both patches of skin were transplanted at the same time. {Connections: p-values were considered in BME 0.1.} The b c allografts were rejected faster because the skin patches were second-set transplants, whereas the allograft from rabbit e was a first-set rejection. Rabbit a rejected allograft from her b c foster offspring faster because rabbit a had previously been exposed to skin from rabbit b, as shown in Figure 15.7. Rabbit a rejected allograft e slowly because rabbit a had not been exposed to tissue from rabbit e before. Immune responses are specific for the source of non-self tissue, and immunity does not uniformly affect all allografts equally. {

Summarize the evolutionary conservation of leptin orthologs across many animal species.

The investigators predicted that because leptin regulates a fundamental aspect of physiology, the genomes of many animal species would encode orthologs of leptin. The team of biologists tested genomic DNA from many species using the mouse leptin gene as a probe (Figure 15.20C). Mammals share very similar leptin gene and protein sequences (Figure 15.20A) and all animals appear to have orthologs of leptin genes as measured by DNA Southern blotting (Figure 15.20C). In fact, mice can use human leptin protein when injected experimentally, and mouse leptin protein functions in human cells grown in tissue culture. Perhaps the biggest insight gained by the cloning and sequencing of the leptin gene was that its mRNA is detected only in fat cells and not any other tissue type, including the brain. The positive control of actin (produced in all cells) in the RT-PCR experiment indicates the mRNA was of high quality for all tissues and thus a negative result in the top gel indicates leptin mRNA was not detectable in these other tissues.

Based on the lipostat model in Figure 15.24, explain why ob and db mice are obese. Extend your explanation to make sense of the parabiotic mice data in Figure 15.19. Does the lipostat concept explain the emergent properties of diabetes and infertility found in ob and db individuals?

The lipostat predicted that leptin initiates a negative feedback loop, which signals the organism to become lean. More fat produces more leptin, which results in less fat and less leptin. ob mice fail to produce functional leptin, so they cannot benefit from the leptin negative feedback and therefore continue to accumulate more fat. In parabiotic mice, ob mice benefited from the wt leptin protein that circulated in the shared blood. However, db mice lack the leptin receptor, so their cells cannot bind leptin protein and thus continue to accumulate more fat, resulting in more leptin production (a positive feedback loop). When conjoined to other mice, the db mouse overwhelms its parabiotic partner with excess leptin protein, which causes the partner to become fatally thin. Unfortunately, the lipostat model does not explain the emergent properties in ob and db mice that develop diabetes and are less fertile than their wt counterparts. All you can conclude at this time is that fat and energy regulation is a ancient trait, shared by all animals, and one that is connected to sugar homeostasis and reproduction.

What is the purpose of the negative control media in Table 14.1? What can you conclude from the negative control results?

The negative control experiment confirms that the cells will not immediately glow when placed into fresh media. "Delayed" indicates cells did not glow until their own cell density increased to a quorum sensing concentration. The negative control media never contained any cells, so it lacked quorum signaling capacity.

Propose a mechanism to explain the induction of bioluminescence in V. fischeri. What can you conclude about the regulation of bioluminescence in V. fischeri, given the two types of bioluminescence data in Figure 14.3?

The next research question was obvious: How do bacteria sense their population density? Ithaca College chemist Anatol Eberhard hypothesized that the population used a chemical signal for quorum sensing. Eberhard predicted V. fischeri cells secreted a chemical that simultaneously induced the luciferase gene in all of members of the population.

How old was the oldest human, and how many species are listed as living longer than humans? How many trees are in the top 20? Where do most of these top 20 species live?

The oldest human was 122 years and 164 days. Perhaps you were also surprised to realize that 14 of the oldest species live in the water and that 14 species are known to live longer than humans. The oldest silver-haired gorilla was 55.4 years old, lived in the Dallas, Texas zoo, died of cancer, and was diagnosed with cataracts in her eyes. Humans live a long time compared to other primates, but we cannot compete with sponges for longevity.

What statistical aspect is lacking in Figure 15.17 that would help you know if the upward trend is significant or not? Analyze any obesity differences between girls and boys of all ethnic groups.

The percentage of obese American teens is increasing, but you cannot tell from Figure 15.17 whether the yearly differences are significant or not because the data lack error bars. The data lack any indication of variance, so you cannot evaluate whether changes are significantly different. Variance was probably omitted from the data to keep it simple for the typical American who does not understand the use of error bars. {Connections: The importance of studying males and females in research are in Ethical, Legal, Social Implications 28.1.} Non-Hispanic black girls have the highest incidence of obesity, followed by Hispanic boys. Non-Hispanic boys and girls have the lowest levels of obesity nationally, but regionally this trend may not hold true everywhere. For example, the state with the highest percentage of obesity in 2019 was Mississippi with 40.8%, whereas Colorado had the lowest percentage with 23.8%. You can mouse over the interactive US map to see historical upward trends for each state.

Use the data in Figure 13.2 to determine what percentage of oxygen is released by hemoglobin when it reaches your muscles at 4 kPa oxygen concentration. Assuming 100 mL of blood in the lungs carries 13 mL of oxygen, how many milliliters of oxygen would be delivered to muscles by hemoglobin? How much oxygen would be delivered to muscles by the high affinity or low affinity binding proteins?

The percentage of oxygen released by hemoglobin is approximately 40% (1.0 - 0.6 divided by 1.0 in Figure 13.2) when moving from 13 kPa to 4 kPa. The low affinity molecule releases about 55% of its oxygen when it moves from the high to low oxygen concentration, (0.45 - 0.2) ÷ 0.45 = 55.5%. However, the low affinity molecule binds less than half as much oxygen as hemoglobin. Instead of releasing about 5.2 mL of oxygen (13 mL in blood × 40% released = 5.2 mL O2 released) the way that hemoglobin does, the low affinity molecule would only release 3.2 mL of oxygen (13 mL O2 in blood × 45% bound × 55% released = 3.2 mL O2 released). The emergent property of being able to switch from high to low affinity is what gives a selective advantage to species with hemoglobin.

Here is a philosophical question to make you think about the nature of emergent properties. If emergent properties are evident only when two or more components interact, what is the source of the emergent property? Does each component possess an equal portion of the property itself, or are emergent properties separate from each component?

The source of the emergent property is the two or more things contributing two it. Each component possesses an equal portion of it as both are having the interaction between one another.

How can two proteins bind to the same promoters but have very different amino acid sequences? Use the data in this section to support your answer.

The two promoters span three numbered operator sites, which are the binding locations for the two critical transcription factor proteins. Each operator is 17 bp long; each promoter incorporates one full operator and one partial operator. The choice between lysogenic and lytic is made by determining which promoter is activated, PR or PRM. Figure 13.9 depicts two RNA polymerases bound to the λ genetic switch, but this would not happen in a living cell. Only one RNA polymerase binds to a single promoter. Once bound, the RNA polymerase could move only in one direction to transcribe one of the two critical genes adjacent to the PRM or PR promoters. The choice between lytic and lysogenic is determined by which promoter binds the RNA polymerase and whether cro or cI is transcribed. Transcription factors bind to DNA and stimulate the adjacent RNA polymerase to produce mRNA. If a transcription factor were bound to operator 1, then RNA polymerase could not bind to PR, and it would have to bind to PRM and transcribe cI. Conversely, transcription factors bound to operator 3 would block access to PRM and result in the RNA polymerase binding to PR and initiating transcription of cro mRNA.

Because cro and cI bind to the same operators, do you think their amino acid sequences are highly conserved? Perform the experiment by aligning the two protein sequences using BLASTp. Use accession numbers ABD60142 for cro and NP_040628.1 for cI (also called λ repressor).

Therefore, if you wanted to search for potential common evolutionary origins, you could align the amino acids as you did for Integrating Question #22. Your BLASTp results showed the two proteins have very little amino acid sequence similarity (Figure 13.16). The lack of similarity indicates these proteins probably have different evolutionary histories rather than evolving from a common ancestral gene. Perhaps the lack of amino acid conservation should not be surprising given that the proteins have different affinities for the three operators. However, you could not have known until you aligned the sequences. Proposing hypotheses and testing them is how scientists work and contribute to the collective knowledge. Now you are doing science too.

Which graph in Figure 13.2 most closely resembles the sigmoidal curve of hemoglobin? Which curve most closely resembles water's capacity to dissolve oxygen? Explain why a molecule with high affinity for oxygen would lead to death if this molecule were used in blood instead of hemoglobin.

Unlike a solution of hemoglobin, water does not contain any molecules that increase its capacity to dissolve oxygen, so oxygen dissolves linearly in water. But hemoglobin can bind to oxygen with a high affinity once the oxygen reaches about 2 kPa. Hemoglobin has a hybrid binding affinity, whereas water is more closely related to the low affinity curve. The high affinity curve would be lethal if it were in your red blood cells. Your muscles and other cells would never receive any oxygen because this molecule would not release oxygen until the surrounding oxygen concentration was almost zero.

How many days does it take a recipient mouse to first-set reject an allograft when the MHC I molecules are different between the donor and the recipient? How many days does it take a recipient mouse to reject an allograft when only the MHC I molecules are identical between donor and recipient? Which molecule plays the bigger role in determining the rate of rejection—the peptide or MHC I? Support your answers with data in Figure 15.11.

Unlike autografts, which are never rejected, allografts containing different alleles of MHC I are first-set rejected in 4 days. However, the immune system takes 20 days to first-set reject allografts with identical MHC I alleles but displaying non-self peptides as defined by the recipient's T cells (Figure 15.11B). Therefore, T cell receptors use MHC I as the primary means for rejecting non-self tissue, which is how the name major histocompatibility was chosen for the protein. As with mice, human organs and tissue with different MHC I alleles are not compatible for transplantation and are rapidly rejected as non-self. T cells more slowly reject a graft if the peptides are non-self and the MHC I is self. Identifying non-self peptides within MHC I is how viruses are detected and destroyed by T cells when your cells become infected. These data highlight the fact that T cells must be "educated" to distinguish self peptides from non-self peptides. The education of T cells is beyond the scope of this book but is a key element of an immunology course.

Describe the physical movement in Figure 13.6 that causes the allosteric modulation of hemoglobin to switch its binding affinity for oxygen.

When O2 binds to the Fe2+ in one of the four hemes, a new hydrogen bond is formed between O2 and one of the histidine amino acids conserved in both α and β protein subunits (Figure 13.6B). Before the heme binds an O2, it is slightly domed (Figure 13.6C), but when O2 binds to the concave side of the curved heme, the heme flattens (Figure 13.6D). The heme Fe2+ moves about 0.45 angstrom (Å) when it binds O2. When the iron moves, the amino acid histidine on the opposite side of the oxygen moves about 0.6 Å. Histidine movement starts a rapid chain reaction of movement by the other amino acids in the subunit. To give you a relative distance comparison, the diameter of one oxygen atom is 1.3 Å, which means the Fe2+ in the heme moves a distance that is one-third as wide as one oxygen atom. When the heme in one hemoglobin subunit moves in response to O2 binding, the other three subunits also move and rotate in response, thus increasing access to the hemes and their affinity for O2. The movement of Fe2+ by the first heme generates an allosteric modulation among the other three subunits, which is the cause of hemoglobin's cooperativity, its Hill plot slope of nearly three, and the emergent property that allows you to breathe and live. {Connections: Allosteric modulation was examined in Section 7.2.} It is difficult to visualize the movement of four protein subunits in response to O2 binding one heme. Therefore, you can view the animated allosteric modulation of hemoglobin below Integrating Question 12 because this shape change is the heart of hemoglobin's emergent property of cooperativity. You can see in the animation to the left and in Figure 13.6 that a tiny movement by heme when it binds O2 causes that first protein subunit to move, thus allosterically modulating the other three subunits. Once modulated, the other three heme groups have a higher affinity for O2 so that all four hemes bind O2 molecules in rapid succession. This chain reaction is similar to setting up a long line of dominos and then tipping one over. Once the cooperativity begins, it ensures that a hemoglobin molecule carries four O2 molecules. As the pH changes in muscles, the shape of hemoglobin changes, and once the first heme group releases its O2, the other three follow suit in milliseconds.

How can λ plaques be big enough to see in Figure 13.7C if E. coli is too small to see? Why don't the plaques continue to grow in size and cover the whole plate and lyse the entire lawn of cells?

When a λ phage is lytic, it produces progeny and ruptures the host cell. You could not see the consequence of the first host cell lysing, but with each round of lysis, more cells become infected and lyse. As the viruses diffuse in a circle surrounding the original lysed cell, more and more bacteria rupture, producing a clear plaque in a lawn of bacteria that have not ruptured. If you can see a plaque, then you are looking at the consequences of hundreds of thousands of E. coli lysing. If a plaque had a radius of 0.5 mm, or 500 µm, then the plaque would occupy 785,398 µm2. As you learned in Section 8.2, a single E. coli is 2 µm by 0.5 µm and would occupy about 1 µm2 on the lawn of cells. Therefore, one virally infected cell with a lytic λ phage results in the lysis of approximately 785,398 bacteria. Once bacteria stop growing, they stop transcribing and translating their own genes, which means lytic λ reproduction essentially stops when the host bacterium stops growing. Because λ needs growing bacteria to produce phage efficiently, once the bacteria form a lawn and stop growing, the plaques grow very, very slowly.

How do positive and negative feedback loops influence the bistable toggle switch of λ phage?

When cI dimers bind to operators 1 and 2, a new emergent property becomes evident (Figure 13.11). Activation of PRM by cI leads to production of more cI proteins and more cI dimers. More cI dimers stimulates the production of more cI monomers. The λ genome is now locked into a positive feedback loop, which is a distinctive emergent property. A positive feedback loop exists when the product of a process results in more of the same product being produced. Once the virus has chosen lysogeny, cI ensures the lysogeny decision is maintained and reinforced through the positive feedback loop. The emergent property of cooperativity of cI binding to operator 2 contributes to the positive feedback loop and neither of these emergent properties would have been evident if cI were studied in isolation. However, cro also produces a negative feedback loop that you can detect by studying Figure 13.13C. As cro accumulates in an E. coli doomed to lysis, cro dimers eventually bind to operators 3 and 2 and turn off the production of more cro protein. Cro's only function is to switch λ from lysogenic to lytic, and cro does not form part of the new virus as seen in Figure 13.7.

Summarize the cellular circuit in Figure 14.4 that produces quorum sensing in V. fischeri. Cite evidence that prokaryotes can discriminate cell density for different microbial species.

When living in isolation, V. fischeri cells secrete a low level of the quorum sensing molecule called an autoinducer that is produced by the protein enzyme called LuxI (pronounced "lux eye"). As long as the cells live in isolation, the autoinducer diffuses into the surrounding environment and becomes very diluted. Simultaneously, V. fischeri cells produce a few molecules of an intracellular protein called LuxR (pronounced "lux are"). LuxI is the enzyme that synthesizes the autoinducer. LuxR first binds the autoinducer ligand and then binds to the promoter of an operon that initiates transcription of luxI and luxR genes. {Connections: Operons and promoters were considered in Section 2.2.} When LuxR protein binds both the autoinducer and the promoter of the lux operon, LuxR becomes a very effective transcription factor. As you can see from Figure 14.4A, the lux operon enhances its own production, which is an emergent property called a positive feedback loop. {Connections: Feedback loops have appeared in Section 13.3, Section 10.2, and Section 9.3.} In addition to producing more LuxI and LuxR proteins, the lux operon also encodes the luciferase enzyme, which is how V. fischeri produces light to help the bobtail squid. The positive feedback loop of autoinducer production induces glowing by V. fischeri, which is another emergent property of quorum sensing. When V. fischeri cells detect high cell density as defined by increased autoinducer concentration, the cells use their resources to make more LuxI and LuxR and emit a glowing light. V. fischeri is not the only species that uses quorum sensing, as you can see from the assortment of structurally similar autoinducer molecules secreted by three different species of bacteria (Figure 14.4B)

Use Figure 13.3 to determine how much more oxygen is released by hemoglobin at 40 mmHg when blood moves from pH 7.6 to 7.2. What is the functional consequence of hemoglobin changing its shape depending on the pH? Would myoglobin make a better or worse oxygen carrier for the blood? Support your answer mathematically using Figures 13.2 and 13.3.

When the concentration of O2 is 40 mmHg in Figure 13.3, hemoglobin releases about 20% of its O2 at pH 7.6 and about 40% at pH 7.2. Simply by moving to muscle cells with lower pH, hemoglobin will release more O2, which is ideal, because lower pH indicates an area of higher CO2 and thus low O2. At 40 mmHg O2 concentration where hemoglobin releases about 40% of its O2 at a pH of 7.2, myoglobin would only release about 10% of its O2. Therefore, myoglobin must have a different function from hemoglobin because it would be a bad alternative to hemoglobin in your red blood cells. Myoglobin is located in your muscle cells and serves as an O2 buffer or reservoir. You can see from Figure 13.3 that hemoglobin changes from 80% O2 saturation to 20% saturation when the amount of oxygen changes from 40 to 20 mmHg and the pH changes from 7.6 to 7.2. This large release of oxygen contrasts with myoglobin which still has about 75% of its O2 bound to the protein at 20 mmHg regardless of the pH. Myoglobin's O2 reservoir gives your muscles an emergency supply of O2 but only after hemoglobin is nearly depleted of O2.

Why would different bacterial species use different autoinducer molecules to communicate? Why are autoinducer shapes so similar to each other?

When you look at the shapes of autoinducers in Figure 14.4B, you can see the ring structures are similar but the portions on the left are distinct. You might imagine that if each species needed to speak in a unique chemical language, its autoinducer would share no structural similarities. The common ring portion of the molecules probably reflects a common evolutionary origin of quorum sensing. Furthermore, it may be that some bacteria benefit when their autoinducers are comprehended by other species. {Connections: Interception of communication by different species is in Chapter 18.} These different autoinducer structures may explain why strain 1 and strain 2 did not communicate to each other in the experiments summarized in Table 14.1.

Use the data in Figure 13.1 to explain why so little oxygen dissolves in serum compared to whole blood.

Whole blood is about 45% red blood cells, 54% serum, and about 1% white blood cells and platelets. If you centrifuged the blood and removed all of the blood cells, you would be left with a straw-colored liquid called serum. One hundred milliliters of serum can dissolve only 0.3 mL of O2, but whole blood could dissolve 60 times more. Therefore, nearly all of the dissolved O2 is carried by red blood cells. Every milliliter of your whole blood contains about 5 million red blood cells, and every cell carries about 280 million hemoglobin molecules

Can male mice accept female skin (see Table 15.1)? Does prior exposure to male cells influence graft rejection in male female allografts? What might you expect a pregnant human mother to do to every male fetus given the data in Table 15.1?

Yes male mice can accept female skin. Prior exposure to male cells makes females reject the male skin faster. A pregnant human mother may reject a male fetus faster.

Compare and contrast the hemoglobin cooperativity switch with the genetic switch of λ phage.

You can identify some similarities and many differences between hemoglobin's emergent property of a toggle switch caused by cooperativity and the emergent property of the bistable toggle switch in λ phage. Hemoglobin is not bistable because it remains in the cooperative state only as long as O2 is abundant. The λ switch is flipped from one state to the other and then remains in that state until a new external signal causes the switch to flip to the other state. The self-sustaining bistable toggle switch is possible because cro and cI both produce positive feedback loops, which are emergent properties. However, cro also produces a negative feedback loop that you can detect by studying Figure 13.13C. As cro accumulates in an E. coli doomed to lysis, cro dimers eventually bind to operators 3 and 2 and turn off the production of more cro protein. Cro's only function is to switch λ from lysogenic to lytic, and cro does not form part of the new virus as seen in Figure 13.7.

What is the molecular shape change that increases binding affinity in the three heme groups, the cause of hemoglobin's cooperativity, and its emergent property?

You can see in the animation to the left and in Figure 13.6 that a tiny movement by heme when it binds O2 causes that first protein subunit to move, thus allosterically modulating the other three subunits. Once modulated, the other three heme groups have a higher affinity for O2 so that all four hemes bind O2 molecules in rapid succession. This chain reaction is similar to setting up a long line of dominos and then tipping one over. Once the cooperativity begins, it ensures that a hemoglobin molecule carries four O2 molecules. As the pH changes in muscles, the shape of hemoglobin changes, and once the first heme group releases its O2, the other three follow suit in milliseconds.

Cooperativity produces a profound change with a small change in ligand concentration. Explain how this happens and contrast the change produced with and without cooperativity. Use Figure 13.15C to support your answer.

You can see the consequences of cooperative binding of cI to operator 2 compared to a hypothetical repression curve in the absence of cooperativity. The green area highlights a tenfold change in cI concentration along the logarithmic scale on the x-axis, from 5 × 10-8 M to 5 × 10-7 M. Over this range of cI, repression due to cooperativity changes from about 15% to over 90%, whereas the hypothetical non-cooperativity repression would change only from about 80% to 99%. As with hemoglobin, cooperativity by cI produces a very sudden transition within a narrow range of concentration. Cooperativity converts a subtle input change to nearly a digital output for a quick transition between two states, the signal to flip the bistable toggle switch from lytic to lysogenic.

Explain why hemoglobin's hybrid affinity for oxygen is ideal for use as a carrier of oxygen in blood.

You rely on the ability of hemoglobin molecules to become completely saturated when blood is in your lungs and to release oxygen when the blood reaches the rest of your body, which is low in oxygen (Figure 13.2). Oxygen is the most concentrated, about 20 kPa, in the air sacs of your lungs. Blood in your lungs indirectly interacts with the air you breathe by way of your lung cells. Your flattened and thin lung cells have a very large surface area to volume ratio to improve their capacity to absorb oxygen from the air. {Connections: The consequences of different surface area to volume ratios were explored in Section 8.2. Bio-Math Exploration 8.1 revealed the geometric origin of these consequences.} Lung cells are about 65% efficient at bringing oxygen into your body, so the concentration of oxygen in your lung cells is only about 13 kPa (20 x 0.65 = 13) instead of the theoretical maximum of 20 kPa. Hemoglobin's sigmoidal oxygen solubility curve means that nearly every hemoglobin molecule will be saturated with oxygen at 13 kPa of oxygen (compare Figures 13.1 and 13.2). The concentration of oxygen in muscle tissue is lower, approximately 4 kPa. The 3.25-fold reduced concentration of oxygen from lungs to muscles produces an oxygen gradient and a chemical mechanism to trigger hemoglobin's cooperative behavior. As you can see in Figure 13.2, molecules with only high affinity (teal line) would bind oxygen efficiently in your lungs, but they would not release oxygen to your muscles. Low affinity molecules (purple line) would release oxygen to your muscles, but they could not bind very many oxygen molecules in your lungs. Only a cooperative, hybrid molecule (black line) with high affinity at 13 kPa oxygen and reduced affinity at 4 kPa can provide the required behavior to deliver oxygen where it is needed. The capacity of hemoglobin to switch from saturation to releasing oxygen is similar to the emergent property of a neuron's action potential, in that a small threshold depolarization produces a larger depolarization that initiates an action potential. Rapidly switching between two states of oxygen saturation is an emergent property. Hemoglobin has a hybrid binding affinity, whereas water is more closely related to the low affinity curve. The high affinity curve would be lethal if it were in your red blood cells. Your muscles and other cells would never receive any oxygen because this molecule would not release oxygen until the surrounding oxygen concentration was almost zero.

What is a Hill coefficient, and how is this related to an emergent property?

it is a measure of cooperativity in a binding process. a hill coefficient of 1 indicates independent binding, a value of greater than 1 shows positive cooperativity binding of one ligand facilitates binding of subsequent ligands at other sites on the multimeric receptor complex. The theoretical maximum Hill coefficient is the number of molecules involved (four for hemoglobin), but hemoglobin only has a Hill plot slope of about three. However, no biological molecule ever reaches the theoretical maximum Hill plot slope, because this would indicate all four oxygen molecules bind simultaneously rather than three binding in response to the first binding event. The Hill plot slope of almost three is close to the practical maximum, which indicates the high degree of cooperativity in hemoglobin binding.

Bio 113 Midterm #4 Chapters 13-15

Set pelajaran terkait

Ironies in "Possibility of Evil"

chapter 15

Life Insurance: Contract Law

Chapter 3 - The Economics of Money, Banking, and Financial Markets

Chp 30 PrepU Perioperative nursing

APUSH 24- 25

Ch 24 Calculations

Business Organization chapter 1

chapter 8

The Bronze Bow Sentences

Chapter 24 "The New Deal"

Chapter 5: Activity-Based Costing: A Tool to Aid Decision Making

Chapter 13

role of the Project manager

saunders respiratory

SOP #1-9

Durham, Chapter 10 High-Risk Labor and Birth

CS II Chapter 22 Quiz

1 Risk Treatment Basics

What's an organisms name