Water has many unique properties that favor life, including high specific heat, high surface tension, and high intermolecular forces due to hydrogen bonding.
Water is an excellent solvent and makes (C) inaccurate and the correct answer.
Each difference of 1 on the log scale is indicative of a tenfold difference in the hydrogen concentration on the pH scale.
According to the question, the pH values of the cells and juices are 7.4 and 1.5, respectively.
The hydrogen (H + ) ion concentrations have to vary by 10 orders of ten, or 1,000,000-fold.
The lower the pH, the more acidic it is.
The carboxylic acid group COOH is labeled (B) and the NH 2 is labeled (D).
There are variations in the fourth position called the R-group.
The absence of nucleic acids does not support the hypothesis that life may have arisen from the primordial soup of Earth.
No nucleic acid molecule was detected.
The hypothesis, so eliminate (A), (C), and (D), is supported by the presence of carbon molecule, amino acids, and sugars.
The choice is correct.
The sulfur element is found in the cysteine and methionine.
The Miller-Urey experiment did not include sulfur-based compounds, so it was impossible to form these two amino acids.
It's not a common component of life-forms, and is largely chemically inert, but it's a mineral form of glass.
Since oxygen is already present in several compounds included in the experiment, the addition of this compound does not provide any additional elements or chemical substrates, which would permit generation of additional amino acids or synthesis of nucleic acids.
The addition of sulfur compounds and phosphorus is necessary to create some acids and nucleic acids.
Water is added to a polymer to break the linkages.
Free water will not result because water is being broken down in the reaction.
The opposite reaction would result in free water.
Adenine is a result of the hydrolysis of DNA.
Choice, cholesterol, is a steroid and does not undergo hydrolysis.
Dipeptides could be created from the hydrolysis of polypeptides.
Phospholipids have a long tail and a head group.
The water at the boundary of the membrane is associated with the tails of the fat acids.
Most of the nucleieotides are found in DNA andRNA.
Water is not found in the membranes.
They have a positive and a negative charge.
There are no free amino acids in the cell.
The correct answer is (C).
By definition, all of the hydrogen will have a single protons.
All hydrogen atoms have one protons.
There are three particles in the nucleus of tritium.
The choice is correct.
Two protons means that the element is not hydrogen.
The atomic number, one, will never change without changing the identity of the atom.
(D) is incorrect because radioactive atoms don't give off electrons.
Active transport is the movement of substances against their concentration through the use of energy.
The critical structure of the sodium-potassium pump is its ability to move sodium and potassium against each other.
Simple diffusion can occur without the need for channels or pores.
Water does not require energy because it uses aquaporins to travel across the membrane.
An example of facilitation is the movement of sodium ion by a voltage-gated ion channel.
The form of active transport that this does not represent is the diffusion of the sodium ion from high to low concentrations.
Light microscopes can be used to see cells.
In the 17th century, some of the earliest studies using primitive microscopes recorded the shape and organization ofbacteria.
The choices are incorrect because the structures of the virus and cell organelle are too small to be observed using light and electron microscopes.
The cell wall is absent in animal cells.
This structure is targeted by several leading classes of antibiotics and would be an effective target of therapy.
All of the structures that are present in animal cells are called cytoplasm, plasma membrane, and ribosomes.
There is a cell wall and no mitochondria in the organisms.
The presence of a cell wall suggests that the organisms are not a group of organisms.
The absence of mitochondria is indicative of prokaryotic structure.
The most likely conclusion is that this is a bacterium.
The only correct choice is (B) because many of the mentioned organelles work closely together.
The Golgi bodies are filled with the proteins and then they are fused with the plasma membrane.
The nuclear envelope and nucleolus don't interact with vacuoles, so choice is incorrect.
The choice is incorrect because ribosomes don't often interact with mitochondria and lysosomes don't interact with mitochondria when they degrade them.
The nucleolus doesn't interact with the lysosomes or centrioles, so choice is incorrect.
Estrogen is said to be stable.
This means it can slip through the cell wall.
The effectiveness of the binding would be reduced by a noncompetitive inhibitor.
The choice is incorrect due to the fact that estrogen is binding to the cells in the body.
The choice is incorrect because testosterone and estrogen have different effects.
It's incorrect because wiping out the ovaries would eliminate estrogen levels.
Since the ER moves things out of the cell, you would need a cell that makes a lot of hormones.
The bacterium, (C), does not have something in its body.
Blood cells are not the best answer because they don't have large amounts of hormones.
The best answer is if you choose neurons, but the best answer is if you choose the pancreas, which makes a lot of hormones.
Water accumulates in cells in a hypotonic environment because contractile vacuoles expel it.
The strategy is wrong because it would increase water intake.
Choice does not help with water gain.
It would find more dangerously hypotonic environments if it had choice.
The Na + K + pump is consuming ATP.
The pump pumps both ion.
One has to be active and passive in order to be a cotransporter.
The choice is not true.
It is not produced.
The choice is not true.
The choice is incorrect since the ATP is hydrolyzed.
The matrix of the mitochondria is where the Krebs cycle occurs.
The inner and intermembrane space are used in the process of oxidation.
The activity of the active site is being altered by the binding of Y at a site outside the active site.
It must be an allosteric inhibitor if it is binding outside the active site.
It is considered a noncompetitive inhibitor because it is binding outside the active site.
The energy threshold that must be met to proceed from reactant to product is called the activation energy.
The diagram must show a decrease in the activation energy.
The energy of the reactants and products is not altered by the enzyme.
The pathway provided shows that if you consume one glucose and two ATP, you will produce four ATP.
There would be a net gain of four ATP.
pyruvic acid can be converted into either alcohol or acid.
During this process, NADH is recycled.
It is an exergonic process that results in the production of energy.
bacteria live in hot environments There is a DNA polymerase.
Part three of the cycle, at 72degC, is the stage in which DNA is shortened by a DNA polymerase.
The answer is the growth ofbacteria in the vents between 70-75degC.
The data that focuses on temperature is not reflected by the conditions described in (A) and (D).
The temperature of hot springs does not reflect the activity of Taq polymerase.
The free energy does not change between catalyzed and uncatalyzed reactions.
It is a correct statement, but it does not answer the question.
The choice is incorrect because of the reactions.
The treated mitochondria will produce more ATP because of the low pH.
Oxygen provides a terminal electron acceptor.
Choices are clearly wrong as they defy physics.
The choice is not correct because the catabolic reactions are not always 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- The answer is Choice (D) because it explains the increase in entropy.
heat is lost continuously even though organisms build and develop as ordered systems In addition, organisms exhale gases and produce waste products.
The chromatids are separated by shortening of the fibers.
The cell would be arrested in metaphase if the shortening of these fibers were blocked.
The cells are arrested in metaphase by aligning the chromosomes in the center of the cell and their attachment to the spindle fibers.
The chromosomes seem to be attached to the fibers, so there doesn't seem to be a separation.
The S phase of the cell cycle is when the genetic material is duplicated.
The experiment only has one phase labeled as an increase.
The phase lasts 30 minutes based on the time scale.
The phase that occurs right before the amount of genetic material should decrease is called Anaphase.
The correct answer is Phase D, it's the phase before the genetic material drops.
Choices don't occur, but if they did, they wouldn't affect the gametes.
Choice is not likely to happen because the kinetochore is only at the centromere.
The answer is D.
Sperm or ovum is the first type of cell for both A and B.
Half of the DNA of other types of cells can be found in these two gametes.
Choices can be eliminated.
In G 0 arrest, muscles and neurons are differentiated.
The taste buds of the liver and taste buds of the taste buds of the taste buds of the taste buds of the taste buds of the taste buds of the taste buds of the taste buds of the taste buds of the taste buds of the taste buds of the taste buds of the taste buds of The taste buds are most likely to be in the G 2 phase because they divide at a higher rate.
The answer is (B).
Major changes to the genome can be made by nondisjunction.
Deletion of an enhancer could affect the genes.
As the same information exists in the genome, it's not likely that translocation will produce any effect.
The father and mother have the same blood type.
It is not possible for a child to be O blood type if neither parent has a recessive allele.
This is an example of incomplete dominance when the phenotype associated with two traits is mixed.
In this case, neither red nor blue is dominant, and the resulting offspring exhibited a mixture of the traits.
The following combinations should occur 25 percent of the time if you cross the pea plant that is Heterozygous for both traits.
There is a chance of it being tall and also a chance of it being yellow.
If you add them together, you can get 1/3.
The woman must have one normal and one sick copy of the X chromosomes.
There is a 50 percent chance that the boy will receive a disease-causing X chromosomes from his mother.
If he gets the disease, he must have a second X chromosome.
They must have a normal copy of the X chromosomes.
There is a chance that the woman may be affected with the disease, but it is very unlikely because there are two copies of the X chromosomes involved.
The conditions by which a boy would receive the disease would be very similar to the conditions by which a girl would receive the disease.
Turner syndrome has only one copy of the X chromosomes.
They both have a 50 percent chance of getting the disease.
Choice doesn't explain why the parents are normal.
The choice is a better explanation.
Carriers of genetic diseases often don't have a recognizable phenotype because one normal allele provides enough functioning protein to avoid the ill effects of the disease allele.
The choice is possible, but the answer is better.
It is not correct because CLN3 is an autosomal gene.
The testcross that Choice describes will let the breeder know if the male is male or female.
If any white-spotted pups result from that cross, the male would have contributed a spotted allele.
It's incorrect that Choice describes a male that is gay.
Stop breeding speckledle is a way to remove spots from the line, but it won't help you identify males with the spot allele.
It will not help you determine which black Labrador males have a spot allele.
This question does not apply to choice.
The genes are referred to as Choices and D because they are not following the Law of Dominance.
It is normal for certain genes to not follow the law of independent assortment.
Choices can be eliminated if they are unlikely to be true.
Choice does not answer the question.
The most probable explanation is choice.
When the DNA polymerase needs 5' to 3' to create short DNA segments, okazaki fragments are generated.
The yeast cell has 3' to 5' activity, so there would be no lagging strand and likely no Okazaki fragments.
The stop codon must have been introduced by mutagenesis since the gene is much shorter than expected.
This is a nonsense example.
There is an order for DNA replication.
If a codon is an anticodon, the AUG segment is on the codon.
The polypeptide undergoes a change.
Choices, intron excision, and poly(A) addition are examples of post-transcriptional modifications.
During translation, the formation of C occurs.
There would be 7 codons and a maximum of 7 amino acids if 21 nucleotides compose a sequence and 3 nucleotides compose each codon.
The choice is incorrect due to the fact that thebacteria make the proteins that reside on the plasma membrane.
Choice is incorrect becausebacteria use transcription factors.
The best answer is A.
It is likely that choice is incorrect.
It would be bad for the cell to have choice.
The choice is incorrect as it only occurs during or preceding the detonation of a nuclear device.
Viruses do not have their own ribosomes, flagella, or independent metabolism, so they are incorrect.
There is only one possible answer.
The transformation occurs whenbacteria take up their surroundings.
Thebacteria did not come back to life.
The transforming agent was notProtein.
Genes can't turn into other genes simply by being in a different cell context.
The strands have been opened if there are DNA/RNA fragments.
The D is also present because the primers have been added.
There are small chunks and long chunks.
If there was no polymerase, there wouldn't be any new ones.
The short and long leading strand fragments could not be attached to the missing DNA ligase.
Both mammals and cephalopods have similar eye structures.
This is an example of convergent evolution.
The data shows a transition from white to black.
This is an example of selection.
Between 1802 and 1902, the number of white-bodied pepper moths decreased and the number of black-bodied pepper moths increased.
The longtail moths were included in the experiment as a control to compare the effects that are not associated with color.
If the color of ash or soot produced by the Industrial Revolution were white or light gray, this would likely reverse the trend observed.
The mole rats live in the same location, which means there is no geographic barrier.
They do not attempt to breed.
They have formed two different species in the same area.
This is a small population so genetic drift is likely.
This situation does not describe convergent evolution and the second part of the answer choice does not accurately describe the change of the population.
The number of people with red hair will likely decrease as the redhead allele is mixed with the more dominant hair colors.
The answer is D because it is the only choice that does not predict evolution on the island.
Choices prevent a Hardy-Weinberg equilibrium and predict evolution.
Humans do not mate randomly and the population is small.
The population will evolve or drift.
The trait is being selected based on female preferences.
Bigger tails indicate reproductive fitness.
Peahens choose males who are healthy and strong enough to grow the big tails and hopefully produce the best offspring.
Choices B, C, and D do not describe the female preferences.
The behavior displayed by the Chimpanzees is indicative of insight because the Chimpanzees have figured out how to solve the problem without external influence or learning.
Viruses display reproductive strategies similar to r-strategists because they aim to reproduce as fast as possible and create as many offspring as possible in order to increase their odds of transmission to other hosts.
They would be considered secondary consumers because they consume the primary consumers.
brook trout can tolerate pH values as low as 4.9 and do not appear to diminish, so we can eliminate if the creek's pH is greater than 4.9.
The stream's pH must have dropped because the crayfish can't tolerate lower levels.
Only (C) is in this range.
Acid rain would explain why the creek's pH would drop over time.
To return to the creek environment, snails must have a pH of at least 6.1.
Light and gravity responses of plants are referred to as choices.
Choice is a made-up word.
The way in which plants respond to touch is referred to as Thigmotropism.
The answer is correct.
If (D) is correct, cutting out beef and lamb from your diet can reduce your carbon footprint even if you don't eat vegan.
Choices (A, B, and C) are all statements about reducing the carbon footprint of food.
A community that has been destroyed and rebuilt is a result of secondary succession.
If there were no life before the new growth, choice would be the answer.
Only a mature forest has choice.
The species taking over the field is native to the rainforests on the fringes of the field.
There is nothing in the question that suggests an invasion.
B is correct.
If the population is too small, there won't be any room for pathogenicbacteria to grow in the intestines.
The choice is not accurate because thebacteria do not need a niche prepared for them.
Choice does not answer the question.
The choice is incorrect and you will get A as your answer.
To answer this question, you must first calculate how much weight each subject lost and then divide by the number of subjects.
The subject gained four pounds.
The total weight lost is 60 pounds, divided by 5 subjects is 12 pounds.
The average weight loss is 12 pounds.
There is a plant in this picture.
The answer must start with "Yes" because the median values are smaller than this.
In order to determine the median height for all six plants, their heights must first be organized in ascending order.
The middle two numbers are 67 and 68; when averaged, this produces a median of 67.5 inches.
If all the data points are whole numbers, the median value must end in either.0 or.
The value is for mean, not median.
The most frequently recurring number is 75, so this is the mode.
The smallest number is 32 and the largest is 81.
The range is the difference between the two.
The product rule is being tested.
The sum rule is being tested, but you also need to use the product rule.
In chi-square tests, a calculated kh 2 value is compared to a critical value from a chi-square table.
You accept the null hypothesis if kh 2 CV.
You reject H o if kh 2 > CV.
The observed and expected data match and the experimental plants have a 3:1 ratio of yellow to green plants.
The only possible answer choices are (B) and (C); the other choices mix up how kh 2 values and critical values are compared.
Three-fourths of the plants studied are yellow and one-fourth are green.
We compare expected and observed data and come up with a value.
The degrees of freedom in this test are determined by the two possibilities being compared.
You don't need to look up because the critical value is listed in all answer choices.
You reject H 0 since kh 2 > CV.
The observed data does not match the expected data.
The first thing you need to do is sort out all the genes.
The F 1 generation had long phenotypes, so they must be dominant to the short phenotypes.
All of the F 1 plants were LlSs.
The null hypothesis is that the two genes are segregating.
The expected ratio of F 2 phenotypes will be 9 LS :3 Ls :3 lS.