missed exams
under the 5 kingdom classification members of the kingdom monera are generally seperated from the members of other kingdoms by having heterotrophic nutrition vs autotrphic nutrition unicellullar organization vs multicellular organization microscopic size vs macroscopic size prokaryotic cells vs eukaryotic cells parasite hos releationship vs predator rpey relationship
prokaryotic cells vs eukaryotic cells Image result for 5 kingdoms Traditionally, textbooks from the United States used a system of six kingdoms (Animalia, Plantae, Fungi, Protista, Archaea/Archaeabacteria, and Bacteria/Eubacteria) while textbooks in Great Britain, India, Australia, Latin America and other countries used five kingdoms (Animalia, Plantae, Fungi, Protista and Monera). monera is for prokarotic including archae bacteria and eubacteria
Which of the following is true of eukaryotic protein translation and its components? A. A codon is four bases long B. The ribosome is made up of four subunits C. Codons bond to noncodons D. Aminoacyl-tRNA synthetase requires ATP for its function E. The stop codon codes for the amino acid serine
. Aminoacyl-tRNA synthetase requires ATP for its function
false? the pancreas can function as an endocrine and exocrine gland NE can act as a hormone and a neurotransmitter the adrenal gland produces hormones such as epi, alsosterone, and cortisol in insects molting and metamorphosis are regulated by the hormone ecdysone prostaglandins are modified proteins involved in inflammation, pain and fever
prostaglandins are modified proteins involved in inflammation, pain and fever prostaglandins are modified FA which help induce fever pain sensation and inflammation. aspirin may inhibit prostaglandin activity.
A student observes a population of 100 eagles in Hardy-Weinberg equilibrium. The eagle's allele for a pointed beak is dominant, and the allele for a rounded beak is homozygous recessive. If 9 eagles were found to have a rounded beak in the population, what is the allele frequency of the pointed beak? A. 0.09 B. 0.30 C. 0.70 D. 0.81 E. 0.91
.7 The equation for the Hardy-Weinberg equilibrium is: p2 + 2pq + q2 = 1. We also know that p + q = 1, where p (pointed beak) and q (rounded beak) are the dominant allele and recessive allele frequencies, respectively. p2 = homozygous dominant (pointed beaks) 2pq = hybrids (pointed beaks) q2 = homozygous recessive (rounded beaks) 9 eagles display rounded beaks, which means they are homozygous recessive for the allele. q2 gives us the number for the homozygous recessive individuals, so if 9% of the population is recessive, then we know that: 0.09 = q2; q = 0.30. Thus, according to p + q = 1, the frequency of the dominant allele must be p = 1 - 0.30 = 0.70.
A specific X-linked recessive allele is lethal. Males that inherit the recessive allele and females that are homozygous recessive will abort before birth. What is the probability that a heterozygous mother will give birth to a child with only one copy of the recessive allele? A. 0 B. 1/4 C. 1/3 D. 1/2 E. 1
1/3 In this example, displaying the X-linked recessive phenotype will result in death. Therefore, the father must display the dominant X-linked allele, otherwise he would not be alive. Using XA to represent the dominant allele and Xa to represent the recessive allele, we can construct the following Punnett square. XAXa × XAY XA Y XA XAXA XAY Xa XAXa XaY(death) From this square, we see four possible genotypes, but the child with the genotype XaY will not be born! Thus, there are only 3 possible phenotypes, XAXA, XAY, and XAXa. Only one of these has one copy of the allele Xa, thus there is a 1/3 probability that a child will be born with only one copy of the recessive allele.
A scientist wishes to create bacterial colonies on an agar plate for observation of growth. Unfortunately, directly adding bacteria from the original culture results in too many bacterial colonies to count. In order to create a solution of proper bacterial concentration for observation, the scientist performs a three-step 1:100 serial dilution of the original bacterial culture. What is the dilution factor of the final solution? A. 10-9 B. 10-3 C. 3.0 D. 103 E. 10
106 Since the scientist used a 1:100 dilution, this means with every dilution the concentration was multiplied by 10-2. In this problem, the scientist performed a three-step 1:100 dilution, meaning they diluted the original concentration first to 10-2, then to 10-4, and finally to 10-6. The question asks for the dilution factor, which is the final volume divided by the original volume (or the inverse of the concentration), so we flip 10-6 to 106 to get answer choice [E] as our final answer.
dino the dinosaur most likely lived 230 million years ago 770 million years ago 50 million years ago 1 billion years ago 4.5 billion years ago
230 million years ago dinosaurs first appeared during the triassic period 230 milllion years ago and became extinct 6.5 million years ago. dinosaurs did not have hair because only mammals grow hair and nurse their young.
Pyruvate has 2 carbon atoms. 3 carbon atoms. 4 carbon atoms. 5 carbon atoms. 6 carbon atoms.
3 carbon atoms. Pyruvate has three carbon atoms. During glycolysis, a six-carbon molecule of glucose is split into two, three-carbon, molecules of pyruvate.
An experiment is designed to track the movement of DNA within a cell. Which radioactive isotope would be the most useful to mark a nucleic acid? A. 32P B. 35S C. 15N D. 13C E. 15O
32P The structure of DNA contains a phosphate backbone, making the radioactive isotope 32P ideal for tracking DNA as it is not found in substantial amounts in other biomolecules, so the answer choice is correct. Recall that in the Hershey-Chase experiment, DNA was proven to be the genetic material of bacteriophages by incorporating 32P into phage DNA and observing that the radioactive phosphorus (as part of DNA) was transferred inside infected bacterial cells (while 35S, incorporated in the protein coat of the bacteriophage, remained in the liquid surrounding cells).
Cytosine (C) accounts for approximately 16% of an organism's DNA. What percentage of its bases would consist of adenine (A)? A. 16 B. 32 C. 34 D. 68 E. 72
34 the amount of adenine must equal the amount of thymine, and the amount of cytosine must equal the amount of guanine. Hence: A%=T% ; C%=G% Therefore, 16% of the DNA consists of C, and 16% must also consist of G. So 32% of the DNA consists of C and G. The other 68% must consist of A and T. We divide this by 2 and find that 34% of the DNA consists of A and 34% also consists of T. Checking our work, we see that C + G + A + T = 16% + 16% + 34% + 34% = 100%.
which statement is false sarcoma occurs only in connective tissue carcinoma occurs in epithelial tissue 35S would be a suitable radioactive tracer to study nucleic acid metabolism satellite DNA is found contrntrated at centromeres and at the end of chromosomes all are true
35S would be a suitable radioactive tracer to study nucleic acid metabolism nucleic acids such as RNA and DNA do not contain S, thus to study nnucleic acid metabilism a researcher would use isotop 32-p. the region of a chromosome to which the microtubules of the spindle attach, via the kinetochore, during cell division.
Assume a population of cows on a farm is in Hardy-Weinberg equilibrium. 400 cows have the genotype CC 400 cows have the genotype Cc 400 cows have the genotype cc. A parasite spreads through this population. Half the cows with the homozygous dominant genotype die, and all cows with the homozygous recessive genotype die. What will be the genotypic frequency of the heterozygotes in the remaining cows' offspring? A. 1/5 B. 2/9 C. 1/3 D. 2/3 E. 4/9
4/9 before parasire CC 400 Cc 400 cc 400 after parasite CC 200 Cc 400 cc 0 p is the dominant allele C q is the recessive allele c p=(2*200)+400=800 q=400 total # alleles= 800+400 total #alleles= 1200 p allele freq: 800/1200=2/3 q allele fred: 400/1200=1/3 genotypic freq of heterozygote =2pq 2x1/3x2/3 =4/9
If nondisjunction occurs during the first meiotic division of an oocyte (egg), which of the answer choices below could be a possibility for the number of chromosomes in the zygote? A. 22, 24 B. 22, 23, 24 C. 45, 47 D. 45, 46, 47 E. 90, 92, 94
45,47 nondisjunction during meiosis 1 will lead to 2 gametes with an extra chromosomes and 2 gametes missing a chromosome. 2 daughter cells will be 23+1=24 2 daughter cells will be 23-1=22 zygote will be 2x23+1=47 (24+23=47) or zygote will be 2x23-1=45 (22+23=45) nondisjunction in meiosis 2 will yeild one gamete with an extra chromosome, and one gamete missing a chromosome and 2 normal gametes one of the daughter cells will be n+1 one of the daughter cells will be n-1 two daughter cells will be normal n zygote will be 2x23+1=47 (24+23=47) or zygote will be 2x23-1=45 (22+23=45) or zygote will be normal 46.
If a human cell has 46 chromosomes at the beginning of meiosis, then at the end of meiosis I each cell would have: A. 23 chromatids B. 46 chromatids C. 46 chromosomes D. 92 chromatids E. 92 chromosomes.
46 chromatids Meiosis will take a diploid cell and produce two haploid cells. This means that our chromosome number will decrease to 23 chromosomes for each daughter cell. However, each of these chromosomes have two chromatids, so they will also have 23 x 2 = 46 chromatids. Memorizing the number of chromosomes/chromatids in meiosis and mitosis is definitely worth your time for the DAT.
A woman goes for genetic counseling, prior to conceiving a child. She is concerned that she may be a carrier for a lethal autosomal recessive condition that kills homozygous recessives by 1 year of age; her fraternal brother died of this condition. What is the probability that this woman is a carrier for this lethal condition? A. 25% B. 50% C. 66% D. 75% E. 100%
66% let A = normal a = lethal allele parents are Aa Aa because made an aa brother Aa x Aa= 1 AA : 2Aa : 1aa aa is lethal therefor not an option because if she was aa she would have died. remove aa from genotypic options for this woman 1AA : 2Aa therefore 2/3 (66%) chance this woman is a carrier
Gregor Mendel self-fertilizes 1200 pea plants with the genotype Yy (Y = dominant yellow allele and y = recessive green allele). Which option below most closely reflects the number of yellow seeded pea plants that he could expect to see produced in the F1 generation? (Assume one offspring per pea plant) 0. 306. 601. 894. 1187.
894 Yellow seeded phenotype is likely to be represented in a 3 to 1 ratio over the green seeded phenotype in the F1 generation. After 1200 fertilizations, roughly 900 yellow seeded pea plants and 300 green seeded pea plants are likely to be produced (a 3 to 1 ratio). Choice D is the closest approximation to the anticipated quantity and is therefore the correct answer choice. F1 Generation Y y Y YY Yy y yY yy Yellow Seeded Phenotype = YY, Yy and yY Green Seeded Phenotype = yy Ratio 3:1
In regards to human skeletal muscle, neural stimulation of an increasing number of motor units is referred to as tonus. recruitment. tetanus. an EPSP. contraction.
recruitment. In regards to human skeletal muscle, neural stimulation of an increasing number of motor units is referred to as recruitment. Recruitment is one way that animals can control the appropriate level of force needed to complete a task.
All of the following are methods plants use to reduce water loss EXCEPT for one. Which one is the EXCEPTION? A. trichromes and hairs on the leaves B. needle-like leaf structure C. thick waxy cuticles D. reflection of green light E. guard cells in stomata
reflection of green light Plants have evolved many different mechanisms to decrease the amount of water they lose to transpiration. Among them include growing little "hairs" on their leaves to retain more moisture and a thin needle-like leaf structure. Christmas trees and other conifers have needle leaves as an adaptation to growing in a dry environment. Desert plants have thick waxy cuticles as a way to minimize the amount of moisture the plant transpires, and guard cells in the stomata close off the holes in the leaves when activated. The reflection of green light is irrelevant to reducing water loss.
Which of the scenarios best represents the actions of a bacterial facultative anaerobe in the presence of oxygen? A. undergoes substrate level phosphorylation to produce all of its ATP B. releases carbon during Krebs cycle as CO2 molecules C. generates ATP across its mitochondrial membrane D. regenerates NADH through fermentation E. will die in the presence of oxygen
releases carbon during Krebs cycle as CO2 molecules ** NO MITO IN PROK
hypoventilation
respiratory acidosis increase in CO2 increase in H+
hyperventilation
respiratory alkalosis decrease in CO2 decrease in H+
In the carbon cycle, carbon dioxide is fixed during photosynthesis in plants. by phytoplankton in the ocean. during combustion of fossil fuels. during aerobic respiration. A and B.
A and B. In the carbon cycle, carbon dioxide is primarily fixed by phytoplankton in the ocean and during photosynthesis in plants. Choice C and Choice D are incorrect because combustion of fossil fuels and aerobic respiration are sources of carbon dioxide rather than sinks. It is critical to have a basic understanding of how carbon flows through the various reservoirs of the carbon cycle. Note: As a helpful secondary exercise, put your pencil to the paper and draw out the water cycle, carbon cycle, and nitrogen cycle.
Mendel's law of segregation refers to which of the following? A. A pair of alleles will separate to individual gametes. B. Chromosomes will independently align on the metaphase plate. C. Only one allele will be expressed in a phenotype. D. An individual's genotype has no influence on their phenotype. E. Alleles never appear on the same chromosome.
A pair of alleles will separate to individual gametes. The law of segregation refers to the separation of alleles to individual gametes. Essentially, each member of a chromosome pair will migrate to an opposite pole. Choice [B] refers to the law of independent assortment, where each chromosome pair does not influence the position of another chromosome pair on the metaphase plate. Choice [C] is false because multiple alleles can be expressed together, such as in co-dominance. Choice [D] is false, because a genotype usually determines a person's phenotype. However, the environment and genotype usually work together to display the phenotype. Choice [E] is false because linked genes can appear on the same chromosome.
Red-green color blindness (RGCB) is an X-linked, recessive condition. Rudy and Sarah have two sons, one of which is RGCB and one that is not RGCB. If Sarah's father was RGCB, what was Sarah's mother's genotype? (* denotes RGCB) XX. X*X. X*X*. X*Y. A single genotype cannot be determined exclusively.
A single genotype cannot be determined exclusively. A single genotype cannot be determined exclusively from the information given. The problem states that Rudy and Sarah have two boys, one boy that is RGCB and one that is not RGCB. In this problem, the genotype of Rudy's parents is irrelevant because each son only receives a single Y chromosome, which does not confer anything in an X-linked condition. Since RGCB is an X-linked condition, Sarah must have the carrier genotype X*X (See Generation 2 table below) in order to have one son that is RGCB and one son that is not RGCB. The problem also states that Sarah's father was RGCB, therefore his genotype was X*Y (See Generation 1 tables below). The two possible Generation 1 tables below show that Sarah's mother could have had either have the X*X or XX genotype and still produced a daughter (Sarah) with a genotype of X*X. With more than one possibility for Sarah's mother's genotype, Choice E is the correct answer.
From a fragment of DNA, which of the following molecules contains the least amount of nucleotides? A. The primary RNA transcript from the original DNA B. A single strand of the original DNA C. A single strand of the original DNA after a nonsense mutation D. A single strand of cDNA from the original DNA E. A single strand of the original DNA after a missense mutation
A single strand of cDNA from the original DNA After RNA is processed it must undergo several modifications. These include the addition of a 5' cap and a poly-A tail. The important modification is the removal of introns, or non-coding regions of the gene. Reverse transcriptase can translate this processed mRNA back to cDNA, or complementary DNA, which stands for a gene that lacks introns. This is useful in genetic engineering because prokaryotes cannot splice introns themselves. So we remove the introns from the gene before we insert it into the prokaryote.
The nucleus of an egg cell from a black-faced sheep is removed. Mammary cells from a white-faced sheep are placed in the denucleated egg cell. The manipulated egg is than placed into a surrogate black-faced female sheep. Of the options below, what would likely occur? A white-faced sheep will be born. A black-faced sheep will be born. A gray-faced sheep will be born. A brown-faced sheep will be born. A blue-faced sheep will be born.
A white-faced sheep will be born. It is reasonable to assume that a white-faced sheep will be born. This question directly details the experiment that resulted in the cloned sheep we now know as Dolly. If you are not familiar with this experiment it would be well worth your time to revisit it. Besides demonstrating that a large mammal could be cloned, the experiment helped researchers solidify the notion that all of the genetic information necessary to form a complete individual is contained in the nucleus.
During transcription, the nucleotide sequence of the nascent RNA is most similar to that of the sense strand. antisense strand. template strand. non-coding strand. None of the above.
sense strand. During transcription, the nucleotide sequence of the nascent RNA is most similar to that of the sense strand. Several synonyms are used to describe the complementary nucleotide sequences of double stranded DNA undergoing transcription. After the DNA double helix is opened, the single stranded DNA from which the RNA is directly transcribed is referred to as the template, antisense, or non-coding strand. The single stranded DNA strand that does not serve as a template for RNA transcription is known as the non-template, sense, or coding strand. The nucleotide sequence of the nascent RNA is the same as the sense strand except that U pairs with A instead of T.
What is the function of lacteals in the digestive system? A. Absorb fats in the small intestine. B. Neutralize acidic chyme from the stomach. C. Mechanically separate food in the stomach. D. Absorb water in the large intestine. E. Emulsify fats in the small intestine
A. Absorb fats in the small intestine Explanation - critical information to remember Lacteals are lymphatic capillaries that absorb dietary fats in the villi of the small intestine. Villi are small finger-like projections that increase the surface area of absorption in the small intestine. Bile (which is not an enzyme because it does not change anything) is released from the gallbladder to emulsify fats. This means it breaks up large fat globules into smaller fat globules for easier absorption by lacteals. A. Absorb fats in the small intestine The lacteals (found in the villi of the small intestine) are lymph vessels surrounded by capillary networks that function in the absorption of nutrients and fats in the digestive system, so the answer choice is correct. Fats are packaged into chylomicrons which move into the lymph vessels for transport to the rest of the body where they are stored as adipose tissue. B. Neutralize acidic chyme from the stomach The lacteals of the digestive system do not function in the neutralization of acidic chyme from the stomach; therefore the answer choice is incorrect. Neutralization of chyme is a function of alkaline bicarbonate secreted from the pancreas, and the hormone secretin is responsible for triggering its release. C. Mechanically separate food in the stomach Mechanical separation of food in the stomach is a function of the smooth muscle movements. The lacteals are located in the villi of the small intestine and function in the absorption of fat and nutrients; therefore the answer choice is incorrect. D. Absorb water in the large intestine The lacteals are located in the small intestine (not the large intestine) and function in the absorption of fats, not water; therefore the answer choice is incorrect. The major functions of the large intestine (or colon) are the absorption of water and electrolytes. It consists of four parts: ascending, transverse, descending, and sigmoid. Feces are stored in the end of the large intestine in the rectum, and excreted through the anus. E. Emulsify fats in the small intestine The lacteals of the digestive system do not function in the emulsification of fats in the small intestine; therefore the answer choice is incorrect. Emulsification of fats is caused by the secretion of bile, produced in the liver and released from the gallbladder. Fats (which are hydrophobic) clump up in the presence of aqueous intestinal fluid, but the emulsification due to bile breaks the fat up into small particles which exposes them to a greater surface area for lipase to break them down.
A scientist wants to detach a peripheral membrane protein from the exterior of a cell membrane. Which is the best method to do so? A. Change the salt concentrations. B. Perform a light centrifuge. C. Add a detergent. D. Change the temperature of the solution. E. Run a small current through the solution.
A. Change the salt concentrations. Peripheral membrane proteins are held in place by electrostatic interactions and hydrogen bonding. They are generally hydrophilic. Changing the salt concentration (a "salt wash") or the pH would disrupt both of these types of bonds and release the peripheral membrane protein from the cell membrane. A light centrifuge will precipitate cells and nuclei at most, only heavy centrifuges extract small soluble proteins. Adding a detergent is a method used to extract integral proteins because they are more hydrophobic in nature. The hydrophobic detergent will destroy the membrane and expose the hydrophobic integral protein. Changing the temperature will denature the protein. Running a current through the cell isn't a common or viable method. A. Change the salt concentrations. Peripheral membrane proteins are held in place by hydrogen bonding and electrostatic interactions. To detach such a protein from a cell membrane's exterior, one would need to disrupt these forces. There are two methods that will accomplish this task: changing the salt concentration, or changing the pH (altering the H+ concentration). The answer choice is correct. B. Perform a light centrifuge. Performing a light centrifuge would most likely disrupt the cell membrane and cause it to break apart, releasing its contents and fragmenting the membrane itself, with no specificity for detaching membrane proteins; therefore the answer choice is incorrect. Centrifuging is often used to break apart cells to separate their organelles on the basis of size and shape, as seen in the technique of differential centrifugation. C. Add a detergent. Adding a detergent would destroy the membrane, not detach peripheral membrane proteins; therefore the answer choice is incorrect. Detergents are useful when one needs to expose integral proteins embedded in the membrane (as integral membrane proteins are hydrophobic). D. Change the temperature of the solution. Changing the temperature of a solution could have several consequences, but efficiently detaching peripheral membrane proteins would not be one of them. A raise in temperature could potentially denature the peripheral membrane protein, or even disrupt the cell membrane itself. Sufficiently lowering the temperature could similarly result in disruption of the membrane. Neither of these effects is ideal for detaching peripheral membrane proteins; therefore the answer choice is incorrect. E. Run a small current through the solution. Running a small current through the solution - a technique known as electroporation - is used to create temporary holes in the plasma membrane. This does not detach peripheral membrane proteins from the external surface; therefore the answer choice is incorrect. Electroporation is often used on bacterial cells to introduce foreign DNA.
In birds, which of the following extra-embryonic membranes allows gas exchange? A. Chorion B. Yolk C. Allantois D. Somite E. Amnion
A. Chorion The chorion acts as an early pair of lungs for birds and reptiles. This allows them to excrete CO2 and absorb O2. In mammals, the chorion fuses with a part of the endometrium to form the placenta. The allantois in birds and reptiles stores wastes like uric acid. In mammals, the allantois develops into an umbilical cord. A. Chorion In birds, the chorion functions as a membrane for gas exchange, so the answer choice is correct. The chorion is the outermost extra-embryonic layer, located between the egg shell and albumin. It later fuses with the allantois. In contrast, the chorion of mammals implants into the endometrium and later fuses with maternal tissue to form the placenta (across which gases, nutrients, and wastes are exchanged). B. Yolk In birds, the yolk functions as a source of nutrients for the embryo; therefore the answer choice is incorrect. It is enclosed in the yolk sac. In contrast, the yolk sac of mammals is empty and encloses no yolk. Instead, nutrients are supplied by the placenta and umbilical cord to the developing embryo. C. Allantois In birds, the allantois initially keeps nitrogenous waste away from the embryo in the form of uric acid; therefore the answer choice is incorrect. It later fuses with the chorion, providing a membrane for gas exchange with the blood vessels below - since this is not the original function of the allantois prior to fusing, it is still not the best answer choice. In mammals, the allantois transports waste products to the placenta. D. Somite The somite is a structure that arises from the mesoderm during embryonic development, and is not an extra-embryonic membrane; therefore the answer choice is incorrect. In vertebrates, the somites go on to form parts of the spinal vertebrae, rib cage, and muscles. E. Amnion In birds, the amnion functions to cushion the developing embryo; therefore the answer choice is incorrect. The amnion is a fluid-filled cavity (of amniotic fluid), which can act as a shock-absorber for the developing embryo. Unlike the other extraembryonic membranes, its function is identical in mammals. Note that amphibians (unlike the other terrestrial vertebrates) are not amniotes, and therefore do not possess any of the extraembryonic membranes - including the amnion.
All of the following statements about the effects of activation of the sympathetic nervous system are true EXCEPT one. Which one is the EXCEPTION? A. Constriction of the pupil B. Increased heart rate C. Decreased intestinal peristalsis D. Dilation of bronchi and bronchioles E. Increased stimulation of the adrenal medulla
A. Constriction of the pupil the sympathetic nervous system results in the dilation of the pupil which allows more light to enter the eye therefore providing the brain with more info about the external world
What is the name of the structure that allows bacteria to stick to a surface? A. Flagella B. Clathrin C. Nucleoid region D. Teichoic acids E. Pili
A. Flagella A. Flagella The flagella of a bacteria does not have any adhesive capabilities; therefore the answer choice is incorrect. The flagella functions as a means of cell motility. In bacteria (which are prokaryotes), the flagella is constructed from the protein flagellin rather than microtubules (which are found in eukaryotic flagella). B. Clathrin Clathrin is not a structure that functions in sticking to surfaces; therefore the answer choice is incorrect. Clathrin is a protein that coats vesicles and plays a critical role in their formation, as well as their incorporation into the cell membrane during exocytosis and endocytosis. C. Nucleoid region The nucleoid region of a cell is the area within prokaryotic cells that contains its genetic material; therefore the answer choice is incorrect. The nucleoid region is irregularly shaped and typically contains the genome of a prokaryote in the form of a double-stranded, circular DNA molecule. It is important to note that the nucleoid is a region and not a membrane-bound organelle (in contrast to the eukaryotic nucleus). D. Teichoic acids Teichoic acids, found on the cell walls of gram-positive bacteria, provide cell wall rigidity and are used as recognition and binding sites by bacterial viruses, but do not function in sticking to surfaces; therefore the answer choice is incorrect. E. Pili The pilus (plural pili), an appendage found on the surface of some bacterial cells, is used for cell adhesion, so the answer choice is correct. The pilus also functions in bacterial conjugation, during which it acts as a bridge to connect the "donor" bacteria (which contains the F plasmid that allows for pilus formation) to a recipient bacteria. This allows for an exchange for genetic material and can introduce genetic variation.
A sample of verotoxin, which destroys blood vessels, is exposed to an earthworm and flatworm. It is found that the earthworm died while the flatworm remains unaffected. Which of the following explains why the flatworm survived? A. Flatworms lack a circulatory system. B. The earthworm has an open circulatory system. C. The earthworm could not excrete the toxin fast enough. D. Verotoxin cannot penetrate the flatworm. E. Flatworms have a closed circulatory system.
A. Flatworms lack a circulatory system. This is an application question regarding the circulatory system of different organisms. Flatworms have no circulatory system. All cells can diffuse with the environment. Oxygen can be absorbed and carbon dioxide can be excreted without the need of a medium, like blood. Thus, they lack blood vessels, and are immune to verotoxin. On the other hand, earthworms are annelids, which have closed circulatory systems and do have blood vessels, which explains why they die when exposed to verotoxin. Open circulatory systems pump blood into sinuses and bathe all tissues with nutrients and oxygen. Open circulatory systems occur in arthopods (insects) and some mollusks. Further Details - useful if you had trouble with the question
Which of the following is able to split a molecule of water during plant photosynthesis? A. P680 B. ATP synthase C. RuBisCO D. Magnesium E. NADP+ reductase
A. P680 able to splut water to liberate e- for use in the liht dependent reactions. this is photosystem II. this is a powerful oxifixer, which can oxidize a water molcule releasing electrons from water. this is called photolysis. a special type of chlorophyll A. the other one is photosystem I which is made of p700 mg is the ion that is complexed within the chlorophyll pigment and is importabt in photon capture. it is a coffactor to the chlorophyl molecule. is a cofactor is organic it is called a coenzyme but MG and fe are inorganic. chlorophpyll is a type of poryphin which have 4 joined pyrolle rings (hemoglobin as well matched with Fe). poryphorins are lipid derivatives.
what baceria is circular shaped and grows in linear form? A. Streptococcus B. Staphylococcus C. Streptobacillus D. Spirilla E. Diplococcus
A. Streptococcus Note: Important pieces of knowledge on appearance of bacteria from naming: Diplo-: pairs Strep-: chain Staph-: grape-like clusters -coccus: spherical -bacillus: rod-like Spirilla/spirochetes: spirals A. Streptococcus There are two important pieces of information about appearance given to us in the name of the bacteria: Strep indicates that these bacteria grow in a chain along a single axis, while coccus indicates a spherical shape (to each individual bacteria in the chain). These characteristics match the circled image above, so the answer choice is correct. B. Staphylococcus Bacteria from the genus Staphylococcus do not match the appearance of the circled image above; therefore the answer choice is incorrect. Even if you have never seen an image of Staphylococcus bacteria, the name of the genus gives the information necessary to deduce its appearance: Staph indicates these bacteria grow in clusters (similar to grapes), but the circled image shows bacteria growing in a single axis chain (indicating this is in fact Streptococcus). The second part of the name, coccus, indicates a spherical shape to each individual bacteria and is consistent with the image. C. Streptobacillus Bacteria from the genus Streptobacillus do not match the appearance of the circled image above; therefore the answer choice is incorrect. Deduce the appearance of this bacteria from the name: the second part of the name, bacillus, indicates a rod-like shape to the bacteria (you may have thought that the bacteria in the image are not perfectly sphere-like and mistakenly selected this answer instead, but the bacillus shaped bacteria are far more rod-like than spherical: see the image below for an example of streptobacillus). The first part of the name, Strep, indicates that these bacteria grow in a chain along a single axis which is consistent with the image. D. Spirilla Bacteria from the genus Spirilla are spiral shaped, while the circled image demonstrates a spherical shape (coccus); therefore the answer choice is incorrect. In addition, the image indicates bacteria growing in chains, whereas spirilla do not form chains. E. Diplococcus Bacteria from the genus Diplococcus do not match the appearance of the circled image above; therefore the answer choice is incorrect. Deduce the appearance of this bacterium from the name: the first part of the name, diplo, indicates that these bacteria appear in pairs (rather than the longer Strep chains seen in the image above). The second part of the name, coccus, indicates a spherical shape to each individual bacteria and is consistent with the image.
Which of the following has a similar function to the cristae of mitochondria? A. Thylakoid membrane in chloroplast B. Cisternae in Golgi complex C. Nuclear envelope D. Intermembrane space in chloroplast E. Endoplasmic reticulum
A. Thylakoid membrane in chloroplast The crista of mitochondria houses the electron transport chain (ETC) to create a proton force for ATP synthesis. The thylakoid membrane is the equivalent to this in photosynthesis; this is where the electron transport chain occurs to produce ATP and NADPH in photosynthetic organisms in the light-dependent reactions. The splitting of water in the plant provides two electrons to be excited by photons of light in Photosystem II, which will then travel through the ETC to produce some ATP. The pair of electrons can be excited again in Photosystem I, where they can go through the ETC again to produce ATP (cyclic photophosphorylation), or react with NADP+ and H+ from splitting water earlier to form NADPH. NADPH and ATP will be used in the light-independent reactions to reduce CO2 into glucose.
Which of the following is NOT a polymer? A. Uracil B. Trypsin C. A plasmid D. Glycogen E. Chitin
A. Uracil uracil is one of the four nitrogenous bases found in RNA (not a polymer trypsin is an enzyme in the small intestine and enzymes are proteins. proteins are polymers of AA plasmids are extrachromosomal DNA in bacteria. DNA is a polymer of nucleotides. glycogen and chitin are polymers of glucose. chitin is B 14 (we cant break it down same with cellulose. glycogen is alpha 14)
in mammalian developmental bio, the inner cell mass eventually develops into A. an embryo B. a protein C. a lipid D. a carbohydrate e. Amniotic fluid
A. embryo The inner cell mass does indeed develop into an embryo during mammalian development.
The atrioventricular node A. prevents the ventricles and the atria from contracting at the same time B. prevents the backflow of blood from the ventricles into the atria C. is known as the pacemaker as it sets the heart rate D. stimulates the left and right atria to contract E. is located at the base of the heart, at the bottom of the ventricles
A. prevents the ventricles and the atria from contracting at the same time -if the AV node was not prevents the atria and ventricles would contract at the same time.
The graph below shows the tension in a muscle fiber over a period of time. What is occurring in the plateau portion of the graph? (plateau at high tension) A. tetanus B. tonus C. Muscle Relaxation D. Recruitment E. None of the above
A. tetanus Repeated stimulation by nervous impulses from the brain can cause prolonged muscle fiber contraction. This extended state of contraction is known as tetanus and is indicated by the plateau in the graph. Choice B is incorrect because tonus refers to partial muscle contraction. Choice C is entirely incorrect because it is clear from the graph that the muscle fiber has reached a relatively high state of tension. Choice D is incorrect because recruitment refers to spatial summation rather than temporal (time) summation. Recruitment occurs when a greater quantity of muscle fibers are activated by the brain rather than an increase in frequency of action potentials that stimulate muscle fiber contraction.
Large tendons can be strongly incorporated with the _______________ of compact bone. A.periosteum. B.red bone marrow. C.yellow bone marrow. D.epidermis. E.synovial fluid.
A.periosteum. The periosteum is the most superficial layer of human bone. Tendons that are associated with powerful movements tend to be highly integrated with the periosteum of compact bone. The periosteum houses osteoclasts and osteoblasts and is nutrient rich due to being highly vascularized. Bone marrow fills the interior of long bones and is typically characterized as either being yellow (C) or red (B). The epidermis (D) is the most superficial layer of human skin. Synovial fluid (E) serves to cushion and lubricate the enclosed cavities of synovial joints.
Which of the following options best describes the wobble hypothesis? Abnormal base paring can occur between a codon and anticodon. A codon is composed of three amino acids. Only certain amino acid sequences can act as start codons. DNA -> RNA -> Proteins. Ribosomes move in an unstable fashion during translation.
Abnormal base paring can occur between a codon and anticodon. The wobble hypothesis asserts that Watson-Crick complementary base pairing rules can sometimes be broken between the anticodon of tRNA and the codon of mRNA. This untraditional pairing usually occurs between the 5' nucleotide of an anticodon and the 3' nucleotide of the codon. The wobble hypothesis was first proposed by Francis Crick and partially explains the degeneracy of the genetic code.
Which of the following is a function of the human integumentary system? Blood storage. Proprioreception. Body temperature regulation. Physical protection. All of the above.
All of the above. The integumentary system is an organ system consisting of the skin, hair, nails, and exocrine glands. All of the options listed are functions of the human integumentary system. The versatility of the integumentary system is evident in the variety of functions it performs. The integumentary system plays a role as a blood reservoir in the human body by storing blood until active muscles or organs require increased circulation. More obviously, the integumentary system is important for physical protection from harmful pathogens and physical trauma. Additionally, the integumentary system provides thermal regulation by responding to both colder and warmer environments. Lastly, the human integumentary system plays a major role in proprioreception. Proprioreception refers to the sensation and awareness of the body's orientation in space.
Which of the following is able to move through the nuclear pore complex of eukaryotic cells? some proteins. RNA polymerases. mRNA's. dNTP's (deoxy nucleoside triphosphates) All of the above.
All of the above. The nuclear envelope in eukaryotic cells is studded with an abundance of nuclear pores which tightly control traffic into and out of the nucleus. A variety of proteins are allowed to pass through the nuclear pore complex such as RNA polymerases and histones. dNTP's are allowed to enter the nucleus to be incorporated during nucleic acid synthesis. More obviously, mRNA is allowed to pass through the nuclear pore complex to fulfill its role in protein synthesis in the cytoplasm.
Which of the following is a type of learning? Associative. Imprinting. Habituation. Insight. All of the above.
All of the above. Associative learning, imprinting, habituation, and insight learning all qualify as types of learning.
Which statement below is true regarding proteolytic digestive enzymes present inside a normally functioning pancreas prior to their release into the gastrointestinal tract? The active sites of the proteolytic enzymes are distorted. The proteolytic enzymes have not been activated. The proteolytic enzymes are in their zymogen forms. The proteolytic enzymes are in their proenzyme forms. All of the above.
All of the above. Prior to expulsion into the gastrointestinal tract, a proteolyic enzyme is in its inactive (B), zymogen (C) (or proenzyme (D)) form. In a normally functioning pancreas, the active site of the proteolytic enzyme is distorted (A) and cannot catalyze the digestion of proteins until expelled into the small intestine. Premature activation of proteolytic digestive enzymes can result in dangerous conditions such as acute pancreatitis. Inactive, precursor proteolytic enzymes (zymogens) can usually be identified because "ogen" is added to the end of the enzyme's name. For example, trypsinogen (zymogen form) in the pancreas becomes activated into trypsin (active form) upon entering the duodenum.
What functionality is served by the cytoskeleton in eukaryotic cells? Cell division. Cell crawling. Movement of organelles. Movement of cytoplasm. All of the above.
All of the above. The highly dynamic cytoskeleton in eukaryotic cells aids in cell division, cell crawling, and the movement of cytoplasm and organelles. A dynamic cytoskeleton is a defining characteristic of most eukaryotic cells. The cytoskeleton is primarily made of three components; actin filaments, intermediate filaments, and microtubules.
Which of the following factors can impact the permeability of a phospholipid bilayer? Temperature. Degree of hydrocarbon tail saturation. Cholesterol content. Hydrocarbon tail length. All of the above.
All of the above. The relative permeability of a phospholipid bilayer is influenced by temperature, cholesterol content, the degree of hydrocarbon tail saturation, and hydrocarbon tail length. Permeability generally increases with increasing temperature. However, permeability tends to decrease with increased hydrocarbon tail saturation and length. The affect of cholesterol content in a phospholipid bilayer depends on the temperature and can either increase or decrease the permeability.
Which is the most correct definition of competitive inhibition? A. Another enzyme completes the same reaction and uses the available reactant. B. An inhibitor binds directly to the active site and prevents the reactant from binding. C. A metal cofactor prevents the binding of the reactant by binding at the active site. D. An inhibitor binds to the enzyme at a binding site, and prevents the enzyme from catalyzing the reaction. E. ATP releases its energy and drives the reactant to form the product.
An inhibitor binds directly to the active site and prevents the reactant from binding. Metal cofactors and ATP are often used to help boost enzyme activity, and are not competitive inhibitors. Non-competitive inhibition (D) occurs when an inhibitor is able to prevent the enzyme from binding with the reactant by binding to the enzyme at a site away from the active site, and change the enzyme's conformation so it cannot bind to the reactant. Competitive inhibition occurs when the inhibitor competes directly with the reactant at the active site, and this substrate takes the place of the reactant and prevents the reaction from occurring.
Genetic diversity is increased by each of the following EXCEPT one. Which one is the EXCEPTION? A. Prophase 1 B. Metaphase 1 C. Anaphase 2 D. Fertilization E. Gene flow between populations
Anaphase 2
What is the name of the smooth muscle cells that are primarily responsible for causing goose bumps in a human hair follicle? Hair papilla. Subcutaneous adipose tissue. Sebaceous gland. Hair bulb. Arrector pili.
Arrector pili. The arrector pili are smooth muscle cells that surround hair follicles. These cells cause the hair to become erect and move into a position more perpendicular to the surface of the skin. The contraction of arrector pili causes the human skin to dimple (goosebumps).
Where would you be least likely to find an aquaporin in a mammalian nephron? Distal convoluted tubule. Proximal convoluted tubule. Descending loop of Henle. Ascending loop of Henle. Collecting duct.
Ascending loop of Henle. The ascending loop of Henle is the only segment of a mammalian nephron that is impermeable to water. Aquaporins are transmembrane proteins that allow water to traverse through lipid bilayers. Considering the function of aquaporins is to allow water to pass through a membrane, it would be highly unlikely to find any in the ascending loop of Henle. Note: This question is an excellent example of the verbal application based biology questions routinely found on the official. Instead of just asking the examinee "What is the function of an aquaporin?" or "What region of a mammalian nephron is impermeable to water?" the modern official is more likely to incorporate both concepts into a single question. Remember to ask yourself: How does the smaller structure (aquaporin) serve the function of the larger structure (ascending loop of Henle)?
which is false a drug that inhibits microtubule polymerization may help to fight cancerous cells because cells given this drug would be unable to separate chromosomes during cell division a person is born with a genetic defect that produces abnormal microtubules. sperm cells, cells of the larynx, and trachea would be greatly affected signals between the cytoskeleton and the ECM involves collagen the SER is involved in the synthesis of phospholipids and steroids.
signals between the cytoskeleton and the ECM involves collagen integrins are cell adhesion molecules. they couple the extracellular matrix from the outside to the inside of the cell (cytoskeleton). integrins are heterodimers of alpha and beta subunits. these cell surface receptors that interact with the ECM are responsible for the mediation of intracellular signals. wound repair is a critical even for our survival, the process of wound coverage is called re-repithelializaiton. which can occur about 24 hours after trauma. Integrins help mediate the movement of epidermal cells into the area that was injured. this is but one of many examples in which integrins play a major role in cell signaling.
DNA microarrays are most useful in measuring the size of an organism's genome. nucleotide sequence of a particular gene. speed at which DNA is replicated in a cell. speed at which transcription occurs in a cell. simultaneous expression of numerous genes.
simultaneous expression of numerous genes. DNA microarrays are most useful in measuring the simultaneous expression of numerous genes. The advent of microarrays has provided researchers with a tool for analyzing gene expression on a massive scale. DNA fixed to a microarray matrix can be analyzed via various labeled probes to indicate the relative gene expression at the time and place from which the probe originated.
which hormone will inhibit many functions of the GI tract estrogen somatostatin thyroxin glucagon gastrin
somatostatin somatostatin supresses the reelase of GI hormones such as gastrin secretin and CCK. this supression will decrease the rate of empyting along with reducing blood flow within the intestinesq
which is incorrectly matched spermatogonium : diploid primary spermatocyte : diploid spermatid:diploid oogonium: diploid secondary oocyte:haploid
spermatid:diploid spermatids are haploid cells which can differentiate into mature sperm cells.
Secretin A. stimulates secretion of gastric juices into the stomach B. stimulates secretion of salivary amylase C. stimulates secretion of bicarbonates into the small intestine D. stimulates secretion of pepsin into the kidneys E. stimulates secretion of salts from the bloodstream into the colon
stimulates secretion of bicarbonates into the small intestine
Which of the following best defines a notochord? A. A component that eventually develops into the spinal cord. B. A length of cartilage extending along the body, which will become the spine. C. That which will become the pharynx, gills, or other feeding system in the animal. D. A muscular tail extending behind the anus. E. A groove located in the underside of the pharynx that produces mucus in filter-feeding species.
B. A length of cartilage extending along the body, which will become the spine. The notochord is that feature that will eventually develop into the spine. The spinal cord begins as the dorsal neural tube. Pharyngeal slits are present in all chordates at some point during their development. For fish they will become gills, in other animals it will evolve into the pharynx, and in urochordates they are part of the filter-feeding system. The muscular tail that extends behind the anus is known at the post-anal tail. The endostyle is what produces mucus for filter feeders. Because it also stores iodine, it is believed to be the pre-cursor for the thyroid gland. Chordates share four main features during their development that are important to know: Notochord Dorsal hollow nerve cord Pharyngeal gill slits Muscular post-anal tail
Dandruff sheds from the epidermal layer known as the
stratum corneum
Which of the following are best characterized by brackish water? A. A river emptying into an inland lake. B. A mangrove leading to an estuary. C. A stream in a tropical forest. D. A freshwater marsh and lake. E. Seawater washing onto a beach.
B. A mangrove leading to an estuary. Brackish water means a mix between salt and fresh water, which would be found in an estuary. An estuary is a specific area where freshwater meets seawater. A mangrove swamp often grows near an estuary and is characterized by a mix of salt and freshwater. Some marshes can be brackish, but D specifically stated that this marsh was freshwater, and thus not characterized by brackish water. A. A river emptying into an inland lake. A river (a source of fresh water) emptying into an inland lake (also a source of fresh water) would not be characterized by brackish water, which is a mixture of both fresh water and seawater; therefore the answer choice is incorrect. B. A mangrove leading to an estuary. A mangrove (a type of tree that can grow in salty water) leading to an estuary (a body of water that connects to the open sea, but has rivers or streams flowing into it) would be expected to have a mixture of both fresh water and seawater (brackish water). Estuaries are characterized by brackish water, which is higher in salinity (saltiness) than fresh water, but lower in salinity than seawater. Since an estuary is a mixture of both types, the answer choice is correct. You may not have been familiar with the term "brackish water" (or know what an estuary is), but you should still try to answer questions like these via process of elimination by recognizing that each of the other answer choices describes only one type of water, but not both. C. A stream in a tropical forest. A stream in a tropical forest is purely fresh water so it would not be characterized by brackish water, which is a mixture of both fresh water and seawater; therefore the answer choice is incorrect. D. A freshwater marsh and lake. A freshwater marsh and lake contain only one type of water: freshwater. This combination would not be characterized by brackish water, which is a mixture of both fresh water and seawater; therefore the answer choice is incorrect. Some marshes can have brackish water, but the answer choice specifically describes a freshwater marsh, so it should be eliminated as an answer choice. E. Seawater washing onto a beach. Seawater washing onto a beach is characterized by only one type of water (seawater) so it would not be characterized by brackish water, which is a mixture of both fresh water and seawater; therefore the answer choice is incorrect.
A species of desert plant secretes a chemical into the surrounding soil that kills seeds from any other species of plant that attempts to germinate in that area. What is this an example of? A. Commensalism B. Allelopathy C. Exploitation competition D. Apparent competition E. Intraspecific competition
B. Allelopathy The correct answer is B. Allelopathy is type of interference competition, where the establishment of other individuals who would compete for a mutual resource is prevented, in this case using a toxic chemical. Intraspecific competition (E) occurs when competition occurs within the same species; in this case the plant is competing with all the other species of plants in the area. A. Commensalism In commensalism, a form of symbiosis, one of the two organisms benefits while the other remains unaffected. The question describes a situation where the desert plant benefits but other species of plant are harmed (as their seeds are killed); therefore the answer choice is incorrect. Examples of commensalism include barnacles and whales (the barnacle gets wider feeding opportunities as a result of being attached to the whale, while the whale is unaffected). B. Allelopathy Allelopathy is the production of biochemicals by an organism that influences the growth, survival, and reproduction of other organisms. The desert plant in the question secretes a chemical that prevents the survival of seeds from other plant species, so the answer choice is correct. Allelopathy is a form of interference competition, which occurs directly between individuals via aggression. In interference competition, other individuals are directly prevented from physically establishing themselves on a shared habitat. C. Exploitation competition Exploitation competition is a type of competition that occurs indirectly through depletion of a common resource. For example, lions and cheetahs face exploitation competition in Africa as both hunt for a common resource: the gazelle. If cheetahs were more successful and ate all the gazelles, lions would suffer from depletion of the food resource. The question describes secretion of a chemical into the soil that is harmful to seeds from other species of plants, but not the depletion of a common resource; therefore the answer choice is incorrect. D. Apparent competition Apparent competition is a type of competition that occurs between two species preyed upon by the same predator. For example, say a species of spider and a species of beetle are both hunted by owls and the amount of spiders suddenly increased. This would lead to survival of more owls (due to the increased food resource of spiders), which would in turn hunt more of the beetles, ultimately decreasing their overall number. The question does not describe a predator hunting two species; therefore the answer choice is incorrect. E. Intraspecific competition Intraspecific competition is a type of competition that occurs between members of the same species. The question describes a situation where the secretions of the desert plant are harmful to other plant species; therefore the answer choice is incorrect.
The deregulation of apoptosis may lead to the development of various diseases. Which of these may develop from insufficient apoptosis? A. Alzheimer's disease B. Cancerous growth C. Parkinson's disease D. Turner Syndrome E. Huntington's disease
B. Cancerous growth Alzheimer's, Parkinson's, and Huntington's disease are neurodegenerative diseases, which are characterized by excessive apoptosis. Turner Syndrome is a genetic disease that does not have a direct connection to apoptosis. However, a lack of apoptosis would lead to cell proliferation, as the body would not be able to successfully eradicate damaged cells. This cell proliferation often manifests as cancer. Fun fact: During early development (as fetus), humans have webbing between their fingers and toes. It is the process of apoptosis that eliminates this webbing - when this process fails to occur, the webbing can remain into adulthood. A. Alzheimer's disease Alzheimer's disease is a chronic neurodegenerative (loss of structure, function, and potentially death of neurons) disease with a prominent symptom of memory loss. One would not expect a disease that results from cell atrophy and death to be a result of deregulated apoptosis, as apoptosis itself leads to programmed cell death; therefore the answer choice is incorrect. B. Cancerous growth Cancer, and its resulting growths, is the result of uncontrolled cell proliferation. Apoptosis is programmed cell death, which functions both in normal cell development but is also a necessary part of destroying damaged and mutated cells that multiply uncontrollably. If apoptosis were to be deregulated, uncontrolled cell growths, including cancer, would be an expected result. The answer choice is correct. C. Parkinson's disease Parkinson's disease is a neurodegenerative (loss of structure, function, and potentially death of neurons) disease with symptoms of motor control loss. One would not expect a disease that results from cell atrophy and death to be a result of deregulated apoptosis, which itself leads to programmed cell death; therefore the answer choice is incorrect. D. Turner Syndrome Turner syndrome is an aneuploidy (specifically a monosomy) that is the result of a missing X chromosome due to nondisjunction in gamete formation during meiosis. Failure of the sex chromosomes to segregate correctly during meiosis is in no way related to the deregulation of apoptosis (which leads to programmed cell death); therefore the answer choice is incorrect. Rather than having the expected sex genotype of XX, a female with Turner Syndrome would be XO, indicating the absence of an X chromosome. E. Huntington's disease Huntington's disease is a neurodegenerative (loss of structure, function, and potentially death of neurons) disease with symptoms of declining mental (and eventually physical) ability over time. One would not expect a disease that results from cell atrophy and death to be a result of deregulated apoptosis, which itself leads to programmed cell death; therefore the answer choice is incorrect. During apoptosis, cells do not rupture or cause an inflammatory response. They continue to shrink, eventually resulting in nuclear collapse and the formation of apoptotic bodies that are eventually lysed. Apoptosis is an essential part of normal development.
Which two structures form the placenta? A. Chorion and allantois B. Chorion and endometrium C. Yolk sac and allantois D. Oviduct and yolk sac E. Ovary and amnion
B. Chorion and endometrium The chorion is an extraembryonic membrane formed during development. The chorion allows for gas exchange in birds and reptiles, while in mammals it attaches to the endometrium and develops into the placenta. The placenta is a structure that allows for the exchange of gases, nutrients, and wastes from the embryo and the mother. A. Chorion and allantois While the chorion eventually forms part of the placenta, the allantois does not; therefore the answer choice is incorrect. In mammals, the allantois eventually forms the umbilical cord between the embryo and placenta, which functions in the transport of gases, nutrients, and waste. B. Chorion and endometrium In mammals, after the chorion implants into the maternal endometrium (specifically tissue known as the decidua basalis), the structures fuse and begin to form the placenta, so the answer choice is correct. The placenta typically completes formation by the end of the first trimester, and takes over production of human chorionic gonadotropin (HCG) so that secretion of progesterone and estrogen remain constant. It also allows for gas, nutrient, and waste exchange between the mother and fetus. C. Yolk sac and allantois Neither the yolk sac nor the allantois form the placenta; therefore the answer choice is incorrect. In mammals, the yolk sac plays a role in formation of developing RBC's, and eventually disappears. The allantois eventually forms the umbilical cord between the embryo and placenta, which functions in the transport of gases, nutrients, and waste. D. Oviduct and yolk sac Neither the oviduct nor the yolk sac form the placenta; therefore the answer choice is incorrect. The oviduct (Fallopian/uterine tube) is a part of the female reproductive system that allows eggs to move from the ovary to the uterus (there are two oviducts, one for each ovary). In mammals, the yolk sac plays a role in formation of developing RBC's, and eventually disappears. E. Ovary and amnion Neither the ovary nor the amnion form the placenta; therefore the answer choice is incorrect. The ovary is a part of the female reproductive system responsible for the production of eggs. Each female has two ovaries. The amnion encloses the amniotic cavity, a fluid-filled cavity that cushions the developing embryo. These structures do not contribute in any way to formation of the placenta.
Which biome is characterized as the largest biome on Earth with long harsh, snowy winters, short rainy and humid summers and with a landscape dominated by conifers? A. Deciduous forest B. Savanna C. Tundra D. Taiga E. Chaparral
taiga Deciduous forests do have winters, but they are not characterized as harsh or very snowy and this biome is known for having 4 distinct seasons each year. The savanna biome has small grasses, small shrubs, scattered trees, and have both dry and rainy seasons. The tundra biomes are best defined as regions that have short growing season, cold temperatures all year round and have a layer of permafrost over the soil. The taiga biome has long, cold winters and very short rainy seasons. Taiga does have vegetation with most of the tree covering being provided by conifers like spruce, fir and pine trees. Tropical deserts are characterized as having extremely high temperatures with sporadic rainfall. The only biome that accurately fits the description in the question is D.
Regarding the lymphatic system, all of the following are true EXCEPT one. Which one is the EXCEPTION? A. Contains valves which ensures unidirectional lymph flow B. The main white blood cell types in lymph are neutrophils, and macrophages C. Lymph is propelled by contraction of smooth muscle as well as skeletal muscle D. Connects into the blood circulatory system E. Controls interstitial-fluid levels
B. The main white blood cell types in lymph are neutrophils, and macrophages lymphatic system is a low pressure system therefore valves are required to make sure lymph only travels in the forward direction. the main WBC types in the lymphatic system are the lymphocytes. lymphocytes are T cells and B cella the lymphatic system relies on both smooth muscles in the vessels and contraction of the vessels by adjacent skeletal muscle in order to be propelled the lymphatic system captures interstitial fluids and returns it to the blood system lymphatic system is a low pressure system which is similar to veins to ensure unidirectional flow.
Which of the following is correct regarding the trp operon? A. Humans utilize the trp operon to manufacture tryptophan B. The trp operon is an example of a repressible operon C. The trp operon codes for proteins which digest tryptophan D. The trp operon is the site where trp repressor binds E. The trp operon does not have an upstream promoter sequence
B. The trp operon is an example of a repressible operon the typ repressor protein in constitively expressed. however the repressor is inactive in the ABSENCE of tryptophan. bacteria use tryptophan for their function but if they cant get it from the environment they will make it. tryptophan is managed by 5 structural genes in the operon which code for trptophan synthetase.
Which of the following is true of natural killer immune cells? Natural killer cells A. help the immune system retain immunological memory B. are recruited to the sites of action after T lymphocytes C. induce apoptosis in cancerous cells D. are precursors to monocytes E. contain a high amount of histamine granules
B. are recruited to the sites of action after T lymphocytes natural kiler cells are part of the innate immune system ,whereas cytotoxic t cells are part of the adaptive immune response. the innate immune system is quicker acting as it does not require activation. the adaptive immune system requires activation and this means it is slower to begin acting. natural killer cells defensive actions include creating holes in the cellular membrane of its target which causes target cell lysis, and inducing apoptosis in rapidly dividing cell. share similarities with cytotoxic T cells (CD8) which are part of the adaptive immunity. NK and CD8 attack similarly: -use perforin = poke hoels in pathogenic cell membranes -granzymes = protease which stimulates a target cell to undergo apoptosis. helper T cells are called CD4. CD4 cells can not recognise antigens that are freely floating they require activation by the MHC of an antigen presenting cells. APC's are cells that can take an antigen from the pathogen to be destroyed, process the antigen and attatch the antigen to a MHC on the APC's membrane... MHC 1 is on all cells MHC2 is only on antigen presenting cells and have adaptvie immunity activation process innate immunity is the body's first line of defense and is a non secific response. innate imunity begins with physical and physiological barriers that prevent infection from entering the body like the skin mucous membranes, cilia, stomach acidity, lyzozyme in tearssaliva and mucus. then continues with an inflammatory response. which is caused by neutrophils monocytes/ macrophages/ eisinophils and basophils/mast cells. the final main aspect of innate immmunity is the complement system, which is a series of blood proteins which turn on eachother in a scascading series of activations through the release of cytokines (intercellular signaling molecules). the complement system can lyse bacterial membranes trigger inflammation and activate the adaptive immune response. the membrane attack coplex (MAC) is formed by the complement system/ the MAC can punture the cell membrane of infecting cell.s monocytes turn into macrophages. immune cell called a monocyte nut whne it enters tissues it differetiates and becomes macrophage acronym for remembering the relative number of WBC's : never let moneys eat bananas N=neutrophils (most numerous leukocyte usually first to respond) L= lymphocytes (b and t) M= monocyes/macrophages E-eosinophils B= basophils which form mast cells
The hormone atrial natriuretic peptide A. is secreted by the brain B. generally acts as a vasodilator C. is activated by the enzyme renin D. raises blood pressure E. limits blood flow to the brain
B. generally acts as a vasodilator -a hormone released by the atria of the heart in response to increased BP which functions to reduce the blood pressure. -ANP performs the opposite function to the renin angiotensin aldosterone system which raised blood pressure. -when the blood pressure is high the atria of the heart are distended and expanded by the blood the atrial stretching is a sign to the atria that th eBP is too high and this causes cardiac muscle cells in the atria to release ANO. -vasodilation is widening of the lumen of blood vessels (mainly act son the arteries and arterioles because these vessels are the ones that have high amount of smooth muscle). -ANP acts to vasoconstrictor the efferent arteriole and vasodilate the afferent arteriole to increase the glomerular filtrate rate and therefore decreasing the amount BP.
Which of the following changes would decrease the activity of an enzyme? A.Increasing temperature moderately. B.Decreasing substrate concentration. C.Increasing enzyme concentration. D.Maintaining the surrounding environment at the enzymes optimal pH. E.None of the above.
B.Decreasing substrate concentration. Decreasing the substrate concentration generally reduces the activity of an enzyme. Likewise, increasing the concentration of an enzyme generally results in an increase in overall activity (C). A moderate increase in temperature generally increases the activity of an enzyme (A). However, at very high temperatures proteins become denatured and therefore the activity of enzymes generally decreases.
An ecological footprint is a phrase used to describe the relative impact of introducing a non-native species into an ecosystem. an impression left by an animal into the soil. the average annual carbon emissions from a single nation. the dollar equivalent needed to restore an ecosystem to its original condition following a natural disaster. the amount of raw land necessary to sustain an individuals lifestyle habits.
the amount of raw land necessary to sustain an individuals lifestyle habits. An ecological footprint is a phrase used to describe the amount of raw land necessary to sustain an individuals lifestyle habits. The calculation typically considers factors such as eating, traveling, and housing habits. For example, a person who tends to eat processed meat, travels frequently, and lives alone in a large home will more likely have a larger ecological footprint than a person that is a vegetarian, travels rarely, and lives in a small home with other people.
A bird inhales radioactive oxygen molecules while breathing. Which exhalation would the same radioactively labeled oxygen molecules first reappear? A.The first exhalation. B.The second exhalation. C.The third exhalation. D.The fourth exhalation. E.The fifth exhalation.
B.The second exhalation Avian respiration is drastically different than human respiration. Due to the unique anatomy of birds, respiration is both continuous and unidirectional. Air sacs allow birds to exchange gas during both inhalation and exhalation. Because air is stored in air sacs prior to entering the lungs for exhalation, the radioactively labeled oxygen molecules described in this question would not appear until the second exhalation. In contrast, humans employ tidal breathing, a respiratory process that inhibits gas exchange during exhalation. In humans the labeled oxygen would appear in the very first exhalation due to tidal breathing. Tidal breathing is drastically more inefficient then the continuous type of breathing described for birds.
The most heavily vascularized layer of human skin is the A.hypodermis. B.dermis. C.epidermis. D.All of the above are equally vascularized. E.None of the above are vascularized.
B.dermis. The most heavily vascularized layer of human skin is the dermis. The dermis provides nourishment to the deepest cells of the epidermis. However, the more superficial cells of the epidermis receive little nourishment and are resultantly dead. The hypodermis is not nearly as vascularized as the dermis and is composed mainly of adipose tissue to cushion and insulate the body.
The phylogenetic tree below represents the evolutionary relationships of various types of land plants. Which clade is not a monophyletic group? Land plants. Vascular plants. Seed plants. Bryophytes. None of the above. All choices are monophyletic groups.
Bryophytes are not a monophyletic group. A group is considered monophyletic if the most recent common ancestor and all of its descendents are included in the group. The most recent common ancestor to all bryophytes is located at the lowest junction shown on the phylogenetic tree. The descendents of the most recent common ancestor to bryophytes include angiosperms, gymnosperms, pterophytes, lycophytes, mosses, and liverworts. Since angiosperms, gymnosperms, pterohytes, and lycophytes are not a part of the group, bryophytes are not considered monophyletic. In addition to bryophytes, another non-monophyletic group are the seedless plants.
The network of fat and connective tissue that serves the digestive system through structural support, fat storage, and by providing the framework through which blood supply, nerves, and the lymph vessels can reach various locations are known as digestive glands. endocrine glands. exocrine organs. the pleural cavity. the mesenteries.
the mesenteries. Mesenteries are a network of fat and connective tissue that serves the digestive system through structural support, fat storage, and by providing a framework through which blood supply, nerves, and the lymphatics can reach various locations.
Which of the following is a structural component of every amino acid and is not found in most monosaccharides? A. Hydrogen B. Carbon C. Oxygen D. Nitrogen E. Phosphorus
there is nitrogen in all amino acids but is not found in most unmodified monosacharides
In a flower, the alleles for plant height are tall (T) and short (t), and the alleles for petal color are red (R) and white (r). A cross between two parents, both short and red, produces a progeny of 3/4 short and red and 1/4 short and white. What are the genotypes of the parents? A. ttRR B. ttRr C. ttrr D. TtRr E. TtRR
ttRr If all of the progeny is short, then we know that both parents must be homozygous recessive for plant height, or tt, so we can eliminate choices [D] and [E]. Some of the progeny produced red flowers, so the parents cannot be homozygous recessive (rr) for this trait, so we can eliminate Choice [C]. Lastly, some of the progeny was white, and some of it was red, so therefore we cannot have a homozygous dominant parent (RR), so Choice [A] can be eliminated. This leaves us with Choice [B] as the final answer.
which is not a viral disease hepatitus small pox chicken pox rabies tuberculosis
tuberculosis tuberculosis is a disease caused by bacteria that spreads through microscopic droplets released into the air.
Which type of cells in human alveoli produce surfactant? dendritic cells. b cells. type I epithelial cells. t cells. type II epithelial cells.
type II epithelial cells. There are two types of epithelial cells in human alveoli. Type I epithelial cells provide structural support while type II cells produce surfactant.
In classical conditioning, a neutral stimulus becomes the conditioned stimulus when paired with the fixed action pattern. conditioned response. unconditioned response. conditioned stimulus. unconditioned stimulus.
unconditioned stimulus. In classical conditioning, a neutral stimulus becomes the conditioned stimulus when paired with the unconditioned stimulus.
What letter of the alphabet does the shape of the hyaline cartilage rings of the human trachea resemble most? A. B. C. D. O.
C. When a cross section of the human trachea is viewed, the cartilage rings appear C-shaped. The rings are not entirely connected in a circular shape because of the close proximity of the esophagus. The flexible muscle and connective tissue between the trachea and esophagus is allowed to expand toward the windpipe whenever a sizable bolus is moving through the digestive system. Without the unique shape of the hyaline cartilage rings the esophagus would have more difficulty safely passing food on to the stomach.
A color blind man without hemophilia (both X-linked traits) marries a woman who is a carrier for both traits. What is the probability they will have a son with both color blindness and hemophilia? Hemophilia and color blindness are unlinked genes. A. 1/1 B. 1/16 C. 1/8 D. 1/4 E. 3/16
C. 1/8 The man's genotype is XcY and the woman's genotype is XcX / XhX. The traits of hemophilia and color blindness are unlinked, meaning they are on either different X chromosomes or far enough apart on the same X chromosome so that they assort independently. In this case, since the mother can only donate one X chromosome (since the father will donate his Y chromosome for a male child), the genes must be far enough apart on the same X chromosome so that they assort independently, and are thus unlinked. We want three things to happen, the man must donate the Y chromosome (1/2 probability), the woman must donate Xh (1/2 probability), and she must also donate Xc (1/2 probability). We just multiply these chances together to find (1/2)(1/2)(1/2) = 1/8.
In total, how many autosomes does the nucleus of a human cell have? A. 23 B. 36 C. 44 D. 46 E. 53
C. 44 you have 44 autosomes and 2 sex chromosomes. each gamete contains 22 autosomes and 1 sex chromosome.
Which of the following eukaryotic organisms would contain coenocyte cells? A. A tree B. A flower C. A fungus D. A human E. A frog
C. A fungus a coenocyte is seen in fungi and protists. it is a large multinucleate cell which occurs when a fungal or protists cell divides its nucleus but does not undergo cellular division (cytokinesis). no septae
What is the best description of a genetic haplotype? A. Two genes that have a recombination frequency of 51% or greater B. When many different forms of the same gene exist in the population C. A group of genes that are linked and are often inherited together D. When one copy of a functional gene has insufficient protein production to result in the wild type condition E. A single gene that is X linked
C. A group of genes that are linked and are often inherited together -haplotype are close together so they have small recombination frequency -haploinsufficient means that it required two copies of the functional gene to produce enough protein to show the trait. one copy of the gene is not enough.
In humans, an embryo develops within a woman, ultimately resulting in the live birth of a newborn. This birthing process is known as vivipary. ovovivipary. ovipary. parthenogenesis. binary fission.
vivipary. Vivipary is a birthing mechanism through which an embryo develops within a mother and ultimately results in the birth of a live newborn. Ovovivipary (B) is a reproductive mechanism by which an egg develops and hatches within the body of a mother. This process is common to many reptilian and fish species. Ovipary (C) is a birthing strategy correlated to hatching from an egg outside of the mother. Ovipary is common to many bird species. Parthenogenesis (D) corresponds to the development of an embryo without fertilization by a male. Binary fission (E) is most obviously incorrect because the process is related to cell division in prokaryotes
Which of the following best describes an ecological community? A. A specific species, that live in one location B. A specific species, that live in one location and the abiotic factors that species interacts with C. All organisms that live in one location D. All organisms that live in one location and the abiotic factors the organisms interact with E. The abiotic factors in one location
C. All organisms that live in one location - a population is a specific species that lives in one location. a species is a group of organisms able to interbreed and have viable fertile offspring. - a population is a specific species that lives in one location. only at the ecosystem level do we consider abiotic factors an ecosystem represents all the organisms in one location (community) and the abiotic factors (temp sun and water ) that interact with them.
The embryo of an animal developing in an egg outside of the mother's body is more likely to A. not require oxygen B. have more abundant yolk C. be classified as a placental animal D. not undergo embryonic cleavage E. not possess a chorion
yolk if the egg is developing external to the mothers body, the mother can not directly provide nutrients once the egg has been laid. therefore the egg needs larger yolk
Which of the following is true for the gametes of an individual with the genotype AaBb? Assume independent assortment but no crossing over. A. ¼ AABB + ½ AaBb + ¼ aabb B. ½ Aa + ½ Bb C. ½ AB + ½ ab D. ¼ AB + ¼ Ab + ¼ aB + ¼ ab E. ¼ AB + ½ Ab + ¼ ab
¼ AB + ¼ Ab + ¼ aB + ¼ This question is asking us what types of gametes an individual with the genotype AaBb will create. To produce gametes, we know the individual must undergo meiosis. Meiosis normally ends in four unique haploid cells. So we just list out all the possible combinations, which includes AB, Ab, aB, and ab.
Which of the following would result in a "shift to the right" in the hemoglobin dissociation curve? A. Decreased carbon dioxide. B. Decreased body temperature. C. Decrease in pH D. Increase in pH E. Fetal hemoglobin
C. Decrease in pH CADET face right CO2, acid, 2,3diphosphateglycerate, exercise temperature increase in CO2 causes it to enter plasma more, and it is converted into HCO#- and H+ by carbonic anhydrase which causes a decrease in Ph. central chemosreceptors in the medulla regulate and the H+ concentration in the cerebrospinal fluid indirectly. the peripherial chemoreceptors are in the carotid and aortic arch. that monitor CO2 O2 and H+ concentration. 2,3DPG is produced form intermediates of the glycolysis therefore it shows that we are actively aming ATP for use. at low ozygen levels, an enzyme catalyzes the syntheses of 2,3DPG therefore the concentration increases and the addinity of Hb for ozygen decreases. thsi is helpful for unloading oygen during anemia and at high altutudes. at high oxygen levels, oxyhhemoglobin (hemoglobin attatched to exygen) inhibits enzyme that synthesises 2,3DPG. fetal hemoglobin has a higher affinity for O2 than regular hemoglobin. this helps grab the oxygen from the maternal blood. myoglobin does not respond to cooperatively. it is a single subunit (hemoglobin in 4) and saturates quickly and releases in very low oxygen "emergency muscle" situations
If a gene is haploinsufficient, how would a null allele in a heterozygote be expected to behave assuming the heterozygote organism has the genotype wild type / null allele? A. Incomplete dominance B. Codominance C. Dominant D. Recessive E. Showing pleiotropy
C. Dominant haploinsufficiency means that one copy of a wild type gene is not sufficient to achieve a normal phenotype. the normal trait requires two functional copies of the gene. therefore in the presence of a null allele the organism will have a non wild type phenotype which means that the null allele is dominant. the null allele is a version of a gene that completely lacks normal gene function. the wild type is haploinsufficient. basically the null acts like the dominant so NW = null NN = null WW=trait
Which of the following is correct regarding the lac operon? A. The amount of lactose present directly affects the levels of cAMP B. Expression of lac repressor protein is increased or decreased depending on glucose levels C. Elevated allolactose inhibits the repressor protein D. E. coli uses lactose as it's preferred energy source E. The lac operon is an example of a repressible operon
C. Elevated allolactose inhibits the repressor protein allolactose is an isomer of lactose so it can bind the repressor protein and remove it. the lac operon is an example of a INDUCIBLE operon. it is repressed and needs to be induced to be turned on . Camp is inversely related to glucose levels. if glucose is high CaAMP levels are low. CAMP binds to the cartabolite activator protein (CAP) and will help attract RNA polymerases. lactose absent, glucose absent camp is high -->off lactise absent, gluccose present, camp is low --> off lactose present, glucose present, camp low --> very low basicallly off lactose present, glucose absent, camp high --> high transcription TRP operon is a repressible operon
Parathyroid hormone has what direct effect on the nephron? A. It increases reabsorption of potassium ions B. It increases secretion of sodium ions C. It increases reabsorption of calcium ions D. It increases secretion of chloride ions E. It increases reabsorption of water
C. It increases reabsorption of calcium ions it increases blood calcium. by stimulating osteoclasts as well as calcium reabsorption in the kidneys. calcitonin is released by the thyroid and inhibits osteoclasts which stimualting osteoblasts. this causes a decrease in blood calcium/
What type of cell junction is most common between digestive tract cells? A. Desmosomes B. Gap junctions C. Tight junctions D. Plasmodesmata E. Anchor junctions
C. Tight junctions Tight junctions prevent the passage of materials between digestive tract cells in animals. This way, the body must absorb the food through intracellular space. Desmosomes are a type of anchoring junctions that provide mechanical stability and are often seen in animal skin cells. They consist of the protein keratin to bind adjacent cells together. Gap junctions are narrow protein channels between animal cells that allow for the exchange of ions and small molecules, but not cytoplasm. They are often used in cell-to-cell communication by the exchange of materials or electrical impulse. Plasmodesmata are narrow protein channels between plant cells.
If a person has circulating anti-A and anti-B antibodies, what is their blood type? A. Type A B. Type B C. Type O D. Type AB E. Type ABO
C. Type O universal acceptor : type AB+ universal donor: type O- a person with type O will produce antibodies to non self antigens if a person has circulating anti-A and anti_-B antibodies then they can not have these blood cell surface antigens. otherwise they would mount an immune response to their own blood.
During the mammalian fertilization process, what component of the oocyte is responsible for the binding of the sperm? A. Vitelline membrane B. Centriole C. Zona pellucida D. Progesterone E. The uterine wall
C. Zona pellucida The correct answer is [C]. Vitelline is a similar component in the egg cells of sea urchins, and is responsible for much of the same function. Progesterone plays a role in leading the sperm to the egg, but not in the binding. The zona pellucida is the outer layer of the egg that is responsible for the initial binding of the sperm to the egg. The ZP3 protein is responsible for the species-specific binding of the sperm to the egg, but is more important for helping galactosyltransferase (GaIT) activate the acrosome reaction. Without it, the sperm will not be able to bind to the egg. Once the egg has undergone the acrosome reaction and fused with the sperm, it can then bind to the uterine wall. A. Vitelline membrane Choosing the vitelline membrane is at first a tempting answer, as it is an oocyte component that functions in the binding of sperm. However, the vitelline membrane (or vitelline layer) is the term used to describe this structure in non-mammals (for example, a common model used to describe the process of egg fertilization is the sea urchin - the sea urchin egg would be said to have a surrounding vitelline membrane); therefore the answer choice is incorrect. B. Centriole Centrioles are not a component of oocytes that function in the binding of sperm; therefore the answer choice is incorrect. A pair of centrioles make up a centrosome, found in the nucleus of animal cells which serve as microtubule organizing centers during cell division. C. Zona pellucida The zona pellucida (specifically the ZP3 protein) is the component of the oocyte responsible for the binding of sperm in mammalian fertilization, so the answer choice is correct. The zona pellucida is a glycoprotein membrane that surrounds the oocyte. It binds sperm, and is required to initiate the acrosome reaction (the sperm releases the contents of its acrosome as it approaches the egg which contributes to charge-based fast block of polyspermy). Other zona glycoproteins include ZP2, which helps in oocyte-sperm recognition and in the prevention of polyspermy, and ZP1, which cross-links ZP3 and ZP2. Without ZP1, the zona pellucida cannot form. D. Progesterone Progesterone is a gonadal steroid hormone, not a component of the oocyte; therefore the answer choice is incorrect. It does not function in any way to bind sperm. Progesterone is produced by the ovaries (later in pregnancy, the placenta also produces progesterone) and functions in the menstrual cycle and the development and maintenance of the endometrial wall and fetus. Birth control pills frequently use high dosages of progesterone (or progesterone and estrogen together) to cause negative feedback that suppresses LH and FSH levels, which in turn prevents ovulation from occurring. E. The uterine wall The uterine wall is not a component of the oocyte, nor does it function in the binding of sperm; therefore the answer choice is incorrect. The oocyte is already fertilized by the time it makes contact with the uterine wall. In mammals, the blastocyst embeds into the uterine wall approximately 6 days into pregnancy.
The sensation of touch is transmitted to the human brain via A. efferent nerve fibers B. motor nerve fibers C. afferent nerve fibers D. autonomic nerve fibers E. ventral root nerve fibers
C. afferent nerve fibers memorize SAME SA = sensory afferent ME= motor efferent
The proteome is best described as A. the protein component of the ribosome B. a nuclease which can degrade mRNA C. all the proteins expressed by one type of cell under one set of conditions D. cellular machinery for recycling proteins E. the extracellular export of proteins modified and packaged through the Golgi apparatus
C. all the proteins expressed by one type of cell under one set of conditions the proteome is all of the proteins expresssed by an organism, tissue, or a cell under specific organismal or cellular conditions. proteomics is the study of the proteins within the proteome. dont get confused with proteasome. which is a protein molecule which can destroy and degrade damaged or surplus proteins within the cell.
A grandmother and her granddaughter are the only two members in a family to have a genetically inherited disease. Neither parent of the granddaughter has the disease. The disease is probably inherited as a(n): A. Y-linked dominant disease. B. autosomal dominant disease. C. autosomal recessive disease. D. X-linked dominant disease. E. X-linked recessive disease.
C. autosomal recessive disease. The question does not specify whether the grandparents gave birth to the mother or father of the granddaughter. Let's test each scenario. First, we examine the possibility of the disease being X-linked. If the grandparents gave birth to the father, then the father would be diseased because he would inherit a Y chromosome from the grandfather and a diseased X chromosome from his mother, therefore inheriting the disease. However, the question says that both parents do not have the disease. So the disease must be autosomal recessive. If we assume the grandparents gave birth to the mother, then the mother would inherit one normal X chromosome from the grandfather, and one diseased X chromosome from the grandmother, making the mother a carrier. However, for the daughter to get the disease, the father must also supply a diseased X chromosome. Since both parents do not have the disease, the disease cannot be X-linked. Therefore, we conclude the disease is autosomal recessive.
To separate a homogeneous mixture of salt water, which of the following pieces of laboratory equipment is necessary? A. buret B. pipet C. condenser D. filter E. separatory funnel
C. condenser Homogeneous mixtures are often separated using distillation. In distillation, the different boiling points of the substances are used to separate the mixture. The mixture is boiled, until one of the compounds evaporates. The evaporated compound is then immediately cooled through the condenser and collected on the other end in a flask. The only piece of laboratory equipment given in the answer options that coincides with this process is [C]. A pipet is used for transferring substances of small volumes, a buret is used during titration experiments to determine concentrations, and a filter would be used to separate a heterogeneous mixture. A separatory funnel is useful for separating heterogeneous mixtures, like an organic solvent and aqueous solvent.
Which area of the eye has the highest density of photoreceptors? A. sclera B. optic nerve C. fovea D. lens E. cornea
C. fovea FOVEA IS PART OF THE RETINA YOU JACKASS A. sclera The sclera is a protective connective tissue layer (containing collagen and elastin) that surrounds the eye. It does not contain any photoreceptors; therefore the answer choice is incorrect. The sclera is the white portion of the eye. B. optic nerve The optic nerve is the bundle of axons of cells that are indirectly synapsed to photoreceptors. The optic nerve itself exits the back of the eye at the blind spot, where there are no photoreceptors; therefore the answer choice is incorrect. When light enters the eye, it hits the photoreceptors (rods and cones). These transmit signals to amacrine and bipolar cells, which in turn synapse to the ganglion cells of the optic nerve fibers. C. fovea The fovea is a section in the back of the eye that is densely packed with the highest concentration of photoreceptors, so the answer choice is correct. The fovea is responsible for high acuity vision, allowing for sharp details to be discerned visually. In fact, approximately half the information carried by the optic nerves comes from the relatively small area that is the fovea. D. lens The lens is controlled by ciliary muscles and functions to focus images. It is transparent and does not contain any photoreceptors; therefore the answer choice is incorrect. Contractions of the ciliary muscles alter the curvature of the lens, which in turn changes the distance of the focal point. When opacities form in the lens of the eye, cataracts can result as light is blocked from entering. E. cornea The cornea (the transparent front portion of the eye) functions primarily to focus light. It does not contain any photoreceptors; therefore the answer choice is incorrect. Unlike the lens, its curvature cannot be altered. Since the cornea is entirely transparent, it contains no blood vessels. Instead, it must receive nutrients via diffusion from the aqueous humor behind it.
Which of the following will have a significant effect in stimulating development of cancer? A. A base-pair mismatch during DNA replication which has been repaired B. A cell that has one mutated allele of a tumor suppressor gene C. p53 failing to activate D. A cell that has two copies of proto-oncogenes E. A decrease in reactive oxygen species
C. p53 failing to activate tumor supressor genes follow a "TWO HIT HYPOTHESIS". meaning that mutations are required in BOTH of their genes in order to be cancer causing. tumor supressor genes required two mutated alleles in order to have loss of function (cancer causing) this is because tumor supressor genes are haplosufficient. one tumor suporessor gene has suffficient function to produce enogh protein to have a tumor supressor effect. tumor supressor genes make protein roducts that protect the cell from cancer. a loss of that protective function causes cancer p53 is the tumor supressor guardian of the cell. its failure to activate will encourage ecancer. another important gene is p21. it is a cyclin dependent kinase (CDK) inhibitor. CDK stimulates cell division. if p21 inhibits CDK then p21 prevents cell division increasing the activity of p21 will increase the amount of inhibition of CDK and p21 will decrease the amount of celll division. proto-oncogenes are the norma, noncancerous state of genes which regulate growth and division. oncongenes are the altered allele version of the protooncogene where the regulation of growth function has been dysregulated. the oncogene can cause cancer through a gain of funtion mutation. it only takes ONE COPY of a gain of function mutation in an oncogene to cuause cancer.
A queen honeybee's unfertilized egg developing into a viable organism is an example of A. sexual reproduction B. budding C. parthenogenesis D. regeneration E. fragmentation Click here to view the genius solution
C. parthenogenesis parthogenesis si an unfertilized egg developing into a viable living organism.
Which of the following options would find it most difficult to diffuse across the phospholipid bilayer of a cells plasma membrane? A.O2. B.CO2. C.K+. D.H20. E.Glycerol.
C.K+. The hydrophobic interior of the phospholipid bilayer of a cells plasma membrane tends to allow smaller, non-polar, uncharged molecules and atoms across by diffusion. However, larger, polar, or charged molecules or atoms typically traverse the bilayer through mechanisms other than diffusion, such as active transport. K+ ions do not readily diffuse across the plasma membrane of a cell due to their charge. However, O2, C02, H20, and glycerol can diffuse more easily across the plasma membrane.
Which of the following pairs is incorrectly matched with the location from which it is secreted? A.Growth hormone--anterior pituitary gland. B.Norepinephrine--adrenal medulla. C.Serotonin--pineal gland. D.Gonadotropin releasing hormone--hypothalamus. E.Aldosterone--adrenal cortex.
C.Serotonin--pineal gland. In humans, serotonin is primarily secreted from the enterochromaffin cells of the gastrointestinal tract, not the pineal gland. Serotonin has a variety of functions including affecting hunger, mood, and body temperature. It may have been easy to confuse Choice C with the hormone melatonin, which is secreted from the pineal gland. Melatonin is primarily known for its involvement in the sleep-wake cycle in humans.
What word is best described by the quotation below? "A shared trait derived from a common ancestor." A.Clade. B.Phylogeny. C.Synapomorphy. D.Taxon. E.Analogous trait.
C.Synapomorphy. Synapomorphies are shared traits derived from a common ancestor. By studying synapomorphies of organisms taxonomists are able to reconstruct evolutionary relationships. A clade (A) is a term used to describe an evolutionary ancestor and all of its descendents, and only its descendents (monophyletic group). Phylogeny (B) is a term that refers to the study of evolutionary relationships and therefore does not fit the definition provided in the question. A taxon (D) is a term that refers to a single organism or group of organisms. Analogous traits (E) are similar characteristics as a result of convergent evolution, and therefore are not derived from a common ancestor.
If two species are competing for the same resources, which of the following is NOT a likely outcome?
Convergent evolution
How many molecules of ATP are needed during the Calvin Benson Cycle to produce a disaccharide such as sucrose? A. 9 ATP B. 18 AT C. 27 ATP D. 36 ATP E. 45 ATP
D. 36 ATP for disacharide 36 molecules of ATP are needed to produce a single disaccharide, such as sucrose, during the Calvin Benson Cycle. To produce a single 3-carbon G3P molecule a net total of 9 ATP, 3 CO2, and 6 NADPH must be consumed during the Calvin Benson Cycle. However, since the disaccharide sucrose contains 12 carbon atoms it should not be surprising that four molecules of G3P are needed for its synthesis. Therefore, 4 G3P/sucrose X 9 ATP/G3P = 36 ATP/sucrose.
What similarity would allow an identical twin to donate a kidney to his identical twin brother with no need to suppress the recipient brother's immune system? A. Because they have an identical blood type B. Because they have identical circulating antibodies C. Because they have an identical thymus and spleen D. Because they have identical major histocompatibility complex molecules E. Because they have the same set of parents Click here to view the genius solution
D. Because they have identical major histocompatibility complex molecules blood type just represents what surface antifgens are present on the blood cell. blood type does not determine transplant compatability. identical twins do not have identical circulating antibodies. circulating antibodies are a reflection of what pathogens the body has been exposed to . identical twins have exact same DNA and MHC. MHC are a group of cell surface proteins that bind to antigens and present the antigen to immune cells. only monosygotic identical twins have the exact same MHC>
Holoblastic cleavage would be expected in all of the following animals, EXCEPT one. Which one is the EXCEPTION? A. Humans B. Sea urchins C. Gorillas D. Birds E. All of these animals undergo holoblastic cleavage
D. Birds holoblastic cleavage is cleavage occurring through the entire embryo. it is a complete cleavage that evenly divides the entire embryo into distinct cells (blastomeres). it is typical of eggs that do not contain a lot of yolk (sea urchins and humans) or only moderate amounts of yolk (frog). *** frog embryo is classified as holoblastic cleavage even though it does contain animal and vegetal poles meroblastic cleavage is cleavage that does not occur through the entire embryo, the entire egg does not evenly divide. there are a bunch of dividing cells at the animal pole on top of a yolk that i not dividing (vegetal pole) . this is typical in large eggs that contain `alot of yolk like fish birds reptiles and monotremes.
If a host cell is infected by a virus, what might that host cell release to warn other nearby host cells of the viral infection? A. Epinephrine B. Immunoglobulin C. Major histocompatibility complex (MHC) molecules D. Interferon E. C5b and C6 from the complement system
D. Interferon is a cytokine that infected ost cells can release to warn other nearby cells that a pathogen has infected that cell. interferons help ready neighnoring cells for defense. it is a soluble protein that cells secrete to other cells. it is a form of cell to celll communication. as an aspect of innate immunity is the complement system. which is a series of blood proteins which turn on eachother in a cascading series of activations through the release of cytokines, the complemet system can lyse bacterial emmbranes ,trigger inflammation, activate the adaptive immune response and help clear immune complexes which are antibodies bound to an antigen the membrane attack complex (MAC) is formed by the complement system which is 5 complement proteins jointogether and can puncture the cell membrane of infecting cells.
Which of the following best describes the function of a gizzard? A. It acts as the site of oxygen exchange in Cnidaria B. It is a sex organ in Porifera C. It is the voice box in Arthropoda D. It is a site of digestion in Annelida E. It forms a rudimentary backbone in Echinodermata
D. It is a site of digestion in Annelida the gizzard is the site of mechanical digestion in a group of animals including the earth worm. the gizzard is a part of the digestive tract that has muscular walls which function of pulverize food. often the food is mixed with other organic matter like soil hat aids in breaking down (digesting) food.
All of the following statements about fixed action patterns are true EXCEPT one. Which one is the EXCEPTION? A. It is an instinctive behavior B. A releaser is a simple element within a stimulus that elicits this response C. The are typically preserved amongst all members of a particular species D. Neuronal networking means the response cannot ever be altered E. Once the stimulus elicits a response, the response cannot be stopped until the action is completed
D. Neuronal networking means the response cannot ever be altered a release is the simple recognizable feature of a complex stimulus that stimulates a FAP (can also be called sign stimule). while a FAP is an instinctual response that has neuronal circuity supporting it, animals are capable of learning and altering this response.
Which of the following is a component of a human's axial skeleton? A. Tibia B. Tarsals C. Pelvis D. Rib cage E. Humerus
D. Rib cage The human skeletal system is divided into two components: the axial skeleton and the appendicular skeleton. The tibia is the large shin bone in your lower leg, tarsals are the bones in your feet (and carpals are the bones in your hands), and the humerus is the arm bone from your shoulder to your elbow. The axial skeleton system is named because it lies on the "axis" of the human body.
All of the following statements about eukaryotic organelles are true EXCEPT one. Which one is the EXCEPTION? A. The rough endoplasmic reticulum is studded with ribosomes B. Mitochondrial genes are inherited from the mother C. Peroxisomes contain the enzyme catalase D. The centrosome adds carbohydrates to proteins E. The Golgi apparatus packages vesicles for export from the cell
D. The centrosome adds carbohydrates to proteins centrosomes are MTOCs made of two centrioles 90degrees to each other. centrioles are arranged in 9x3 microtubule orientation. is where the spindle fibers of metaphase and anaphase emerge from. peroxisomes contain the enzyme catalase which breaks down reactive oxygen species. without this function, relative oxygen species would accumulate and induce DNA mutations.
What is the fundamental niche of an organism? A. The organism's specific role in an ecosystem B. Environmental conditions where the organism lives C. The organism's place in an ecosystem's foodweb D. The environmental conditions where an organism can potentially live E. The habitat an organism occupies in the environment
D. The environmental conditions where an organism can potentially live Note that this question asks about the "fundamental niche" of an organism, not just the "niche". The fundamental niche of an organism is simply the range of environmental conditions an organism can live in, where there is no influence from predators. The realized niche is where the organism actually lives in the environment. A. The organism's specific role in an ecosystem An organism's specific role in an ecosystem describes its niche, which is a distinctly separate concept from a fundamental niche. The question asks specifically about a fundamental niche; therefore the answer choice is incorrect. It is important to pay close attention and read questions on the DAT carefully because it is a very easy error to see the word "niche" and automatically jump into the answer choices, which can lead to the selection of a wrong answer. B. Environmental conditions where the organism lives The environmental conditions where the organism actually lives is the realized niche; therefore the answer choice is incorrect. This should not be confused with the fundamental niche that describes the full range of environmental conditions where an organism could potentially live. The presence of competing species will usually restrict an organism to its realized niche so that it may coexist within the larger environment. C. The organism's place in an ecosystem's food web A food web is the expanded version of a food chain, which shows which organisms are consumed by one another, and their sources of energy. There are various levels (e.g. primary producer, primary consumer, secondary consumer, etc.) which depict an organism's place in the food web on the basis of where it acquires its energy. This concept is unrelated to the fundamental niche, which describes the full range of environmental conditions where an organism could potentially live; therefore the answer choice is incorrect. D. The environmental conditions where an organism can potentially live The environmental conditions where an organism can potentially live correctly describes the fundamental niche, so the answer choice is correct. An organism free of competing species or other forms of interference could live throughout the entire fundamental niche, but most organisms are forced to live within a smaller realized niche (where there is less competition for resources) due to presence of various outside pressures. E. The habitat an organism occupies in the environment The habitat an organism occupies in the environment describes the type of place an organism is living, including all other organisms present as well as the physical and chemical aspects of the environment. For example, the habitat of the bacteria of H.pylori is the digestive system of a human; the habitat of a pet lizard would be the tank it lives in. The question asks about the fundamental niche, however, which is not just where an organism is currently living but the entire range of environmental conditions it could potentially live in; therefore the answer choice is incorrect.
Which of the following is true regarding aquatic biomes? A. Aquatic biomes cover around 25% of the Earth's surface B. Freshwater biomes make up around half of all aquatic biomes C. Aquatic biomes consume most of the Earth's oxygen D. The size and diversity of organismal populations oceans are able to support is limited by abiotic factors E. The intertidal zone is where an estuary meets with an ocean
D. The size and diversity of organismal populations oceans are able to support is limited by abiotic factors -sunlight is an abiotic factor that influences ecosystems, as you go deeper into the ocean, there is less sunlight and this affects what can survive. oceans have a euphotic zone closest to the surface which has strong irradiance from the sun, plants are able to photosynthesize and survive. deeper down the sun is less able to filter through this is called the disphotic zone and is semi irradiated with sun. but there is not sufficient sunlight for most plants to survive here. beneath this layer and down to the oceans floor is the aphotic zone. essentially no sunlight can penetrate down to this layer. no plants can survive. and there are few fish species. -aquatic biomes cover around 75% of the earth and are the largest of the earths biomes -freshwater biomes make up around 3-4% of all aquatic biomes. -the aquatic biomes contributes most of the earth oxygen through the huge amount of algae that subsides within the vast and globe dominating aquatic biomes. -the intertidal zone is the zone where the ocean meets land. -an estuary is where freshwater from a coast (via river or stream) meets a salt water ocean.
If blood pressure is low, which of the following is true? A. Antidiuretic hormone release is inhibited B. Renin will partially cleave angiotensin II to activate it C. Angiotensin II production would further decrease blood pressure D. Vasoconstriction will increase total peripheral resistance E. The kidneys would lower plasma salts levels Click here to view the genius solution
D. Vasoconstriction will increase total peripheral resistance JGA cells in the glomerulus can detect changes in the blod and sodium levels. when blood presure or blood sodium is low these cells release renin. renin converts angioteninogen to angiotensin I which is not as strong as angiotensin II. angiotesin converting enzyme (ACE) converts angiotensin I to angiotensin II. angiotensin II has many effects on the body -stimulates secretion of aldosertone (mineralcorticoid_ from the adrenal cortex. -increases Na+ reabsorption from the proximal tubule and water will follow the salt. -a potent systemic vasoconstrictor causing cessels to constrict thereby increasing total peripherial resistance (TPR) the primary receptor for aldosterone is in the distal concoluted tubule and the collecting duct. this increases sodium reabsorption which water will follow. also does K+ secretion. MAP (mean arterial pressure) is calculated from CO (cardiac output) of the heart and the TPR (total peripherial resistance) CO = HR X SV MAP = (HRXSV) X TPR
Oligodendrocytes are best associated with: A. Beta cells of the pancreas B. G cells of the stomach C. Nerves in the peripheral nervous system D. White matter of the brain E. Spermatogonia of the testes
D. White matter of the brain Oligodendrocytes are responsible for laying down the myelin sheath surrounding neurons in the central nervous system, so choices [A], [B] and [E] can be eliminated. Remember, the central nervous system includes the brain and the spinal cord. Schwann cells perform the same function in the peripheral nervous system (PNS). Myelination insulates the neurons (like the plastic around an electrical wire) and allows the neurons to transmit electrical impulses more quickly and efficiently. Without myelination, your neurons will not be able to properly communicate together, resulting in a variety of diseases, such as multiple sclerosis.
The main extracellular buffer system in humans consists of: A. amphipathic proteins B. phosphate C. sulfate D. bicarbonate E. histamines
D. bicarbonate The main extracellular buffer is bicarbonate. This is how CO2 is transported throughout the body. The body uses pH sensors to determine how much we have to breathe. If we are exercising vigorously then our cells will produce large amounts of CO2, which will lower our extracellular pH (make it more acidic due to the CO2), which will make us breathe more to expel the excess CO2. Phosphate and other proteins are used as the main intracellular buffer inside cells.*****
Fluorescence recovery after photobleaching (FRAP) is a biological laboratory technique best used for A. separating a heterogenous sample into the individual, homogenous components B. counting the number of cells in a sample C. detecting the quantity of a specific protein in a sample D. detecting the movement of lipids and proteins within a cell E. simultaneously measuring a cell's DNA expression of thousands of different genes
D. detecting the movement of lipids and proteins within a cell basic lab techniques: centrifugation spins a heterogenous sample in a lab apparatus called a centrifuge, and separates a sample based on mass, density, and shape. chromatography apparatus involves three components (sample, mobile phase, and stationary phase. seperation is based on solubility and polarity western blotting is a technique for quantifying the amount of a target protein from a sample. first gel electrophoresis then transferred onto a membrane. then uses a primary antibody which is given a piece of the target protein as its antigen. a second antibody is then applies which binds to the first and will cause a glow. FRAP and FILM are visualization of live cells through flouresence. -FRAP = flourescence return after photobleaching. photobleaching is when a sample is bleached with light so that it can no longer flouresce. flouresce recovery after photobleaching is a measure of how and when a part of s studied samle recovers its fourescent properties. -FILM = floursecnece lifetime imaging microscipy. the amount of time it takes for an excited molecule to release all of its flourescence is known as fourescence lifetime. tells us concetrations of varoius ions molecules ande en fgases.
Sodium ions entering a neuron A. pass through the neuron's cell membrane via a peripheral membrane protein B. require ATP hydrolysis in order to be shuttled across the neuron's cell membrane C. liberate water molecules from the neuron's cell membrane D. have their transport influenced by an electrochemical gradient E. can diffuse directly through the neuron's cell membrane
D. have their transport influenced by an electrochemical gradient dont need active transporters because following the gradient in
Two plants that are both monohybrid are crossed. If the phenotypic ratio seen in the F1 offspring is in a 1:2:1 ratio, this is indicative of A. complete dominance B. true breeding parents C. genetic drift D. incomplete dominance E. homozygous parents
D. incomplete dominance
All of the following are methods plants use to reduce water loss EXCEPT for one. Which one is the EXCEPTION? A. trichromes and hairs on the leaves B. needle-like leaf structure C. thick waxy cuticles D. reflection of green light E. guard cells in stomata
D. reflection of green light Plants have evolved many different mechanisms to decrease the amount of water they lose to transpiration. Among them include growing little "hairs" on their leaves to retain more moisture and a thin needle-like leaf structure. Christmas trees and other conifers have needle leaves as an adaptation to growing in a dry environment. Desert plants have thick waxy cuticles as a way to minimize the amount of moisture the plant transpires, and guard cells in the stomata close off the holes in the leaves when activated. The reflection of green light is irrelevant to reducing water loss. Plants reflect green light because it is the wavelength of light their chlorophyll are not capable of absorbing - as a result it is reflected, and plants generally appear green in color.
What bone would you look at if you were interested in investigating an intramembrous ossification pattern of development? A.Ulna. B.Femur. C.Tibia. D.Clavicle. E. Humerus.
D.Clavicle Ossification, or bone formation, occurs by either an intramembrous or endochondral pattern of development. For the official, you should be familiar with both of these patterns of bone formation. Endochondral ossification typically occurs in the long bones of the skeleton and proceeds by the gradual replacement of cartilage with bone. Long bones such as the femurs, tibias, and humeri grow through an endochondral ossification pattern of development. Intramembrous ossification does not involve cartilage. Intramembrous ossification is typically associated with growth in flat bones such as the mandible or cranial vault.
Aequorin is a luminescent protein that fluoresces in the presence of Ca2+ ions. Where would you inject aequorin if you were interested in studying polyspermy in sea urchin eggs? A.On the surface of the flagella of the sperm. B.Inside the lysosomes in the egg. C.Inside the acrosome of the sperm. D.In the cytoplasm of the egg. E.Inside the cortical granules in the egg.
D.In the cytoplasm of the egg. The prevention of polyspermy in sea urchins begins when the acrosomal membrane of a sperm cell fuses with the plasma membrane of the egg. The contact initiates a signal transduction pathway through the secondary messengers DAG and IP3. Through the pathway, Ca2+ ions are released from the egg endoplasmic reticulum and move away from the point of sperm entry, through the cytoplasm, in what is typically described as a calcium wave. The high concentration of Ca2+ ions causes cortical granules to fuse with the egg plasma membrane, releasing their contents into the perivitelline space. This reaction allows water to flow out of the egg's interior, creating a fertilization envelope. The rapid formation of the fertilization envelope serves to prevent further fertilization from more than one sperm. Of the choices provided, injecting aequorin into the egg cytoplasm is the most logical location to research the calcium wave that prevents polyspermy in sea urchin eggs. Note: This question is an excellent example of a very difficult biology question that is quite representative of the verbal application based problems you will find on the official.
Which of the following statements is true regarding genetic mutations? A.Mutations only occur in coding regions of genetic material. B.Mutations only occur in non-coding regions of genetic material. C.Mutations cannot be induced by environmental stimuli. D.Somatic hypermutation is a key phenomenon of the protective response of the human immune system. E.All of the above.
D.Somatic hypermutation is a key phenomenon of the protective response of the human immune system. Most mutations are deleterious thus resulting in their elimination through evolution by natural selection. However, some mutations are beneficial, and essential, to the human immune system. For example, during a process known as somatic hypermutation, mutations are "intentionally" introduced into genes that are responsible for the eventual synthesis of antibodies. Without mutated genes our immune systems would not be capable of successfully battling the wide variety of, and rapidly changing, assortment of invading pathogens.
The table below outlines characteristics of several types of muscle cells. Which entry in the table is incorrect? Muscle Cell Type Striated # Of Nuclei Cardiac YES MULTIPLE Smooth NO 1 only Skeletal YES MULTIPLE A.The cardiac-striated box. B. The smooth-striated box. C.The skeletal-striated box. D.The cardiac-# of nuclei box. E. The smooth-# of nuclei box.
D.The cardiac-# of nuclei box The cardiac-# of nuclei box is incorrect in the table shown. The information contained in this question commonly appears on official exams. Cardiac muscle cells have one centrally located nucleus, are generally striated, contain intercalated discs, and function to serve in the rhythmic contraction of the heart. Skeletal muscle cells are generally striated, contain multiple, peripherally located, nuclei, and responsible for the movement of bones. Smooth muscle cells are generally non-striated, contain a single, centrally located, nucleus, and are located in places such as the muscular layer surrounding the digestive or reproductive tracts in animals. Unlike cardiac muscle cells, smooth and skeletal muscle cells do not have intercalated discs. Note: As a helpful secondary exercise add an additional column to the table that corresponds to the presence or absence of intercalated discs.
Which option most accurately completes the quotation below? "Pepsin cleaves ____________ in a _____________ manner." A.carbohydrates, highly specific B.carbohydrates, random C.proteins, random D.proteins, highly specific E.nucleotides, random
D.proteins, highly specific it is important to realize that many digestive enzymes such as pepsin, trypsin, and chymotrypsin cleave polypeptides in a highly specific manner. Each enzyme is conditional on the particular peptide bonds it will cleave in a polypeptide. The amino acid residues immediately surrounding a peptide bond, as well as other conditions, dictate the outcome of the cleaving.
Which of the following occurs during embryonic cleavage from zygote to morula? A. Decrease in morula size. B. Decrease in blastomere size. C. Differentiation of blastomeres to pluripotent cells. D. Increase in morula size. E. Invagination into zygote to form the neural tube.
Decrease in blastomere size. During cleavage of the zygote, each new blastomere becomes increasingly smaller. However, the size of the entire zygote/morula does not change. Each division gives less cytoplasm to the daughter cell in preparation for cell differentiation. A pluripotent stem cell is a cell that can differentiate into any of the three germ layers (ectoderm, mesoderm, endoderm), but cannot develop into an entire individual because it cannot produce extra-embryonic germ layers. Gastrulation occurs when a group of cells invaginate into a blastula.
Which of the following options represents the layer of skin where the majority of human sweat glands reside? Hypodermis. Dermis. Epidermis. Muscle. Bone.
Dermis. Human skin is composed of three primary layers. From deep to superficial the layers are: hypodermis, dermis, and epidermis. Human sweat glands primarily reside in the dermis layer of the skin. The dermal layer is abundant with a variety of structures, including blood vessels, nerves, and lymphatic vessels.
If a cell homogenate is spun in a centrifuge, the homogenate is fractionated into parts based on relative differences in A. mass B. density C. shape D. A and B E. A, B and C
E. A, B and C the sedimentation rate of a particle in a centrigufe tube will increase as the object increases in mass, density, and as the shape becomes more compact. therefore all three of these features show how a cell lysate is prepared in centrifugation. the process of centrifugation: if a scienctist wants to isolate a specific component of the cell within a sample of cells, use centriguation. first the cells must be splut open so that the components can be seperated. this process is called homogenization and is accomplushed with a lab blender or shear forces. the homogenate is the mixture of the split open cells and is stored in an inert buffer to preserve the cell components. a centrigure spins at high speeds. the HIGHER MASS, MORE DENSE, MORE COMPACT SHAPE, particles will sediment to the bottom and is called the pellet. whatever is not in the pellet is at the top and this is refered to the supernatant. cells and te nucleus come first, then dense organelles then ribosomes then cytosol.
Which of the following is not a step used during the polymerase chain reaction (PCR) method of DNA amplification? A.Identifying the location of the gene of interest. B.Denaturation. C.Primer annealing. D.Extension. E. Enclosure.
E. Enclosure. "Enclosure" is not a step used during PCR. PCR allows scientists to replicate massive amounts of a particular gene of interest. By amplifying a gene, researchers can then further investigate the activities associated with the DNA sequence. After the location of the gene of interest is elucidated, scientists heat a solution containing double stranded DNA and other necessary components. The heat denatures the double stranded helix into two single strands of DNA. Next, the temperature of the solution is decreased, allowing the primers in the solution to bind to specific nucleotide sequences on both ends of the DNA sequence to be amplified. Lastly, DNA polymerase in the solution replicates both strands of the DNA duplex to produce an additional copy of double stranded DNA. Repeating the cycle can create abundant copies of DNA.
The mixture of organic molecules and water present on early Earth provided the environment necessary for the formation of the first living entity. These primitive oceans have also been called A. amniotic fluid B. the fluid mosaic model C. the air water barrier D. pangea E. The prebiotic soup
E. prebiotic soup The mixture of organic molecules and water present in primitive oceans is also called the prebiotic (or primordial) soup. Many scientists believe that the conditions of the prebiotic soup allowed the first living entity to form. Choice A is incorrect, amniotic fluid is the liquid that within the amniotic sac of a pregnant woman. Choice B is incorrect, the fluid-mosaic model pertains to the structure and function of cell membranes. Choice C is erroneous. Choice D is incorrect, Pangaea is the name of the supercontinent that existed about over 200 million years ago.
Atmospheric N2 gas is fixed into more usable forms by a.symbiotic bacteria B.lightning C.during the creation of fertilizers D.combustion of fossil fuels E.all of the above
E.all of the above Atmospheric N2 gas is fixed into more usable nitrogenous forms through a variety of mechanisms. Symbiotic bacteria, lightning, combustion of fossil fuels, and commercial fertilizer production are all ways in which nitrogen gas is fixed. The importance of the nitrogen cycle in ecology cannot be understated.
Genetic material is condensed within the confines of a nucleus partially through the electrostatic attraction between the DNA sugar phosphate backbone and the side chains of certain amino acids found within histones. Considering the information in the table below, it should not be surprising that histones contain an abundance of Amino Acid Side Chain Charge at Physiological pH Aspartate negative Glutamate negative Glycine neutral Arginine positive Lysine positive A.aspartate. B.glycine. C.aspartate and glutamate. D.aspartate and arginine. E.lysine and arginine.
E.lysine and arginine. The vast amount of genetic material found in eukaryotic cells is condensed within the confines of a nucleus partially through the electrostatic attraction between DNA and histones. The abundance of arginine and lysine residues found in histones allows the proteins to carry a positive charge at physiological pH. The negatively charged phosphate groups of DNA are attracted to the positively charged amino acid side chains found in the histones, ultimately allowing the efficient storage of genetic information.
At great risk to his own life, a baboon helps another totally unrelated baboon ward off a predator because he understands that his heroic act may benefit him by a returned favor in the future. This is an example of A.inclusive fitness. B.kin selection. C.altruism. D.mate choice copying. E.reciprocal altruism.
E.reciprocal altruism. Reciprocal altruism occurs when an individual aids another unrelated individual despite an elevated risk to one's own fitness. The individual who provides the "favor" must have the capacity for higher learning processes and be able to comprehend that a selfless act now for another individual may lead to a similar returned favor in the future.
Where do spermatozoa gain motility? Epididymis. Seminiferous tubules. Ejaculatory duct. Vas deferens. Seminal vesicle.
Epididymis. Seminiferous tubules (B) are the tightly coiled tubes in the testes where spermatogonia give rise to primary spermatocytes. Through meiosis and subsequent differentiation, primary spermatocytes eventually give rise to secondary spermatocytes, spermatids and ultimately immotile spermatozoa in the seminiferous tubules. In the pathway of tubules exiting the seminiferous tubules ciliated epithelial cells help move the immotile spermatozoa towards the epididymis. In addition to an increase in spermatozoa motility, testicular fluid is reabsorb by stereocilia in the epididymis, which serves to concentrate the ejaculate. After further maturation in the epididymis, the spermatozoa enter a long, slender, and very muscular duct called the vas (ductus) deferens (D). The vas deferens propels sperm forward during ejaculation through rhythmic smooth muscle contractions. The pair of ejaculatory ducts (C) are short ducts that results from the combination of the vas deferens and seminal vesicle (E) on both sides of the body. The seminal vesicle located along each ejaculatory duct is a gland that releases the majority of fluid contained in semen. The product excreted from the seminal vesicle contains many carbohydrates, including fructose, which provides energy for the nutritional demands of sperm propulsion. Seminal vesicles do not store sperm.
Which of the following organisms does not use gills to respire? Fish. Frog. Sharks. Gilled snails. Tadpoles.
Frog. Tadpoles live underwater and respire by the use of gills, but by the time the juvenile has developed into an adult frog successful respiration is accomplished by the use of lungs. Frogs undergo other drastic changes during their development including: losing their tail, growing limbs, and moving to more terrestrial habitats. Fish, sharks, and gilled snails also respire by use of gills.
layers of the epidermis
From deep to superficial, the layers of the human epidermis are: stratum basale, stratum spinosum, stratum granulosum, stratum lucidum, and stratum corneum. Dandruff sheds from the epidermal layer known as the stratum corneum. The stratum corneum is the most superficial layer of the epidermis. As new skin cells are produced and pushed upward from the deeper layers, cells from the stratum corneum are sloughed off.
what part of compact bone is tunnels that perforate the bone and run parralel to their longitudinal axis? Lamellae. Osteon. Caniliculi. Haversian canal. Volkmann canal.
Haversian canal. Haversian canals, as indicated by the arrows in the illustration, perforate compact bone, running parallel to their longitudinal axis. Haversion canals typically enclose blood vessels, nervous and connective tissue, serving to nourish and support surrounding cells. Haversian canals contain little, if any, lymphatic vessels.
Which of the following is a type of intramolecular force that is seen in the tertiary structure of a protein, but not in the secondary structure of a protein? A. Subunit bonding B. Hydrogen bonding C. Peptide bonding D. Covalent bonding E. Hydrophobic interactions
Hydrophobic interactions
What mechanisms protect the epithelial cells of the stomach cavity from the degradative affects of the enzyme pepsin? I. Mucus coats the interior of the stomach cavity. II. The stomach cavity continually regenerates cells. III. Pepsin is initially in a non-active form prior to entering the stomach lumen. I only. II only. III only. I and II. I, II and III.
I, II and III. All of the numbered statements contribute to the protection of the epithelial cells of the stomach lumen from the degradative affects of pepsin. Mucus coats the interior of the stomach cavity and limits both physical and chemical digestion of the stomach lining from occurring. The stomach lining also continuously regenerates new cells to compensate for the increased agitation associated with digestion. Lastly, the digestive enzyme pepsin is in its inactive form (pepsinogen) until it is safely within the confines of the lumen of the stomach.
A person is injected with a drug that is known to increase the concentration of parathyroid releasing hormone in the blood stream to higher than normal levels. What is the likely outcome due to administration of the drug? Increased osteoblast activity. Decreased osteoclast activity. Increased bone calcium uptake. Decreased blood calcium levels. Increased blood calcium levels.
Increased blood calcium levels. Parathyroid releasing hormone activates osteoclasts that break apart the mineralized matrix of bone, releasing calcium into the bloodstream. Thus, blood calcium levels rise due to parathyroid releasing hormone. Choice A, Choice B, Choice C, and Choice D are not logical choices and represent possible outcomes if additional calcitonin were injected into a person. Calcitonin decreases blood calcium levels by activating osteoblasts. Note: Calcitonin and parathyroid releasing hormone are commonly presented official hormones.
Which of the following is not part of the innate immune system? Human skin. Neutrophiles. Macrophages. Killer T cells. Lysozyme.
Killer T cells. Killer T cells are not part of the innate immune system; they are part of the adaptive immune system. The innate immune system provides an immediate, non-specific, response to invading pathogens through both mechanical and chemical mechanisms. Human skin (A) serves a protective function in the innate immune system by providing a physical barrier, separating the interior of the human body from external pathogens. Additionally, the innate immune system is composed of cells such as chemokines, neutrophiles (B), and macrophages (C) which work in unison to formulate a generalized recognition of foreign matter. Lysozyme, an enzyme found in substances such as tears or saliva, has the capacity to chemically degrade certain forms of bacteria. Structural differences between bacterial and eukaryotic cells permits lysozyme to harm pathogens without harming host cells. The adaptive immune response is longer lasting, takes longer to develop, and demonstrates an extreme level of specificity through the recognition and destruction of foreign matter.
A surge of which hormone causes ovulation in females? A. FSH B. GnRH C. Prolactin D. LH E. Estrogen
LH The female hormonal cycle begins in the hypothalamus, which monitors the amount of estrogen and progesterone in the blood. If there is a low amount of these hormones, the hypothalamus will secrete GnRH, which stimulates the anterior pituitary to secrete FSH (follicle stimulating hormone) and LH (luteinizing hormone). FSH will stimulate the development of the follicle and oocyte, and the follicle will secrete estrogen. Positive feedback from estrogen will cause a surge of LH, which will cause ovulation. The follicle now becomes a corpus luteum and develops from the help of LH, and will secrete both estrogen and progesterone. High levels of estrogen and progesterone will thicken the endometrium in the uterus for implantation of a fertilized egg. However, the high levels of estrogen will cause the hypothalamus to stop producing GnRH, which will stop the anterior pituitary from releasing FSH and LH. Now, if the corpus luteum is unfertilized, then it will deteriorate, as the body is no longer supporting it, and menstruation occurs. If the egg is fertilized, it will implant into the endometrium and produce HCG (human chorionic gonadotropin) to support the corpus luteum, which will continue to secrete estrogen and progesterone to maintain the endometrium, and pregnancy will continue.
Which of the following factors is involved in both specific and non-specific immunity? A. MHC II B. B cells C. Lysozyme D. Helper T cells E. Macrophages
Macrophages Macrophages can act as either non-specific or specific phagocytes. In non-specific defense, they engulf pathogens by phagocytosis. In specific defense, they engulf antibody-coated antigens by phagocytosis. Major histocompatibility complex II, or MHC II, and B cells are only used in specific immunity. Lysozyme is involved only in non-specific defense as an anti-bacterial enzyme. Helper T cells are lymphocytes that originate in the bone marrow like B cells, but mature in the thymus (hence why they're called T cells). They have antigen receptors on their surface and only used in specific immunity.
Which germ layer produces the notochord? A. Ectoderm B. Mesoderm C. Endoderm D. Epiderm E. Hypoderm
Mesoderm
A researcher wants to insert a eukaryotic gene, which currently contains introns, into a prokaryote by use of a plasmid to produce synthetic insulin. All of the following are steps that should be included in the procedure EXCEPT for one. Which one is the EXCEPTION? A. Use reverse transcriptase to create the cDNA of the eukaryotic gene B. Use a restriction enzyme that produces sticky ends C. Methylate the plasmid DNA to change its configuration D. Use ligase to seal the eukaryotic DNA into the plasmid E. Place the cell in a medium of CaCl2 and heat
Methylate the plasmid DNA to change its configuration All of these are steps in genetic engineering except for Choice [C]. Methylation is often used to differentiate between original and new strands in DNA replication. All of the other choices are necessary for the introduction of a new functional eukaryotic gene into a prokaryote. To insert the gene, we need to: Use the mRNA of the eukaryotic gene and reverse transcriptase to produce our cDNA (Choice A), which we will insert into the prokaryote. Use restriction enzymes (Choice B) to cut up the prokaryote plasmid and insert our cDNA into the plasmid. Use ligase to seal our cDNA into the prokaryote plasmid (Choice D). Use transformation (heat and CaCl2) to introduce the plasmid into the prokaryote (Choice E). Fun fact: Genentech was the company that started selling synthetic human insulin using this technique, and is considered to have founded the biotechnology industry.
During cell division, which of the following stages are most similar? A. Mitotic prophase and meiosis metaphase I B. Mitotic telophase and meiosis telophase I C. Mitotic anaphase and meiosis anaphase I D. Mitotic metaphase and meiosis metaphase II E. Mitotic metaphase and meiosis metaphase I
Mitotic metaphase and meiosis metaphase II During mitotic metaphase, all of the chromosomes (each with two chromatids) line up on the metaphase plate in the cell. During metaphase II in meiosis, half of the chromosomes (each with two chromatids) also line up on the metaphase plate. Metaphase II is more similar to mitotic metaphase because metaphase II only consists of chromosomes with 2 chromatids each, while metaphase I consists of homologous chromosomes (or tetrads, with 4 chromatids).
Which of the following is an example evolution by natural selection? A giraffe has long necked offspring due to years of reaching for leaves. A weight lifter has very muscular kids due to years of weight lifting. A man has skinny kids due to lack of physical activity during his life. A snake has offspring without venom due to lack of venom use during its lifetime. None of the above.
None of the above. All of the options in the question are examples of Lamarkian evolution, not evolution by natural selection. A common misconception of many introductory biology students is to disregard Lamarkian evolution as an obsolete theory that had little historical significance. It is important to note that prior to Lamark's ideas, virtually nobody had ever even considered that species were capable of change through time at all! Although his mechanism for explaining evolution was incorrect, Lamark's courageous and radical thinking most definitely paved the way for future scientists. All of the choices in the question are in line with a Lamarkian way of thinking. Remember, Jean-Baptiste Lamarck believed that the characteristics an individual acquired or neglected during their own life were capable of being inherited by the next generation. Note: The phrases "use-disuse", "theory of inheritance of acquired characteristics", and "Lamarkian evolution" are essentially synonymous. If any of these phrases appear on your official exam, think of Lamark!
Which of the following is not an example of sexual dimorphism? A male lion has a mane while a female lion does not. A male peacock has elaborate and colorful tail feathers when compared to a female peacock's tail feathers. A female hyena is considerably larger than a male hyena. A male baboon has larger canine teeth than does a female baboon. None of the above. All choices are examples of sexual dimorphism.
None of the above. All choices are examples of sexual dimorphism. All of the options are examples of sexual dimorphism. Sexual dimorphism literally means that there are traits that distinguish males from females. In many cases, since the fitness of males and females is primarily dictated by different constraints it is reasonable to assume that different environmental pressures influence the two sexes. For example, in many mammal species the males tend to be larger, more brightly colored, and equipped with defense features such as horns. Females, contrary to males, tend to be smaller and mutely colored. It is important to note that sexual dimorphism only means that there is a distinguishable difference between the males and females of a given species. Although the female is larger than the male in Choice C, sexual dimorphism still exists.
Suppose a mutation occurs to the original DNA template strand shown below. What kind of mutation has occurred? (Use the Genetic Code Table as necessary) Original DNA Template Strand 3'--TACGTAAGTATTGT--5' Mutated DNA Strand 3'--TACGTAATTATTGT--5' Frameshift mutation. Nonsense mutation. Silent mutation. Insertion. None of the above.
Nonsense mutation. The mutation in the DNA strand in this problem is an example of a nonsense mutation. A nonsense mutation occurs when the mistake results in a premature stop codon, ultimately terminating the translation of an mRNA sequence. In this problem, the eighth base from the 3' end of the DNA strand is thymine instead of guanine. As indicated by the genetic code table, this mutation changes the codon from AGT (serine) to ATT (STOP). Original DNA Strand: 3'--TACGTAAGTATTGT--5' Original mRNA Strand: 5'--AUGCAUUCAUAACA--3' Amino Acid Sequence Synthesized: Methionine-Histidine-Serine (STOP) Mutated DNA Strand: 3'--TACGTAATTATTGT--5' Mutated mRNA Strand: 5'--AUGCAUUAAUAACA--3' Amino Acid Sequence Synthesized: Methionine-Histidine (STOP) Choice A is incorrect because a frameshift mutation typically results when an insertion or deletion of a nucleotide causes a "shifted reading frame". Choice C is incorrect because a silent mutation results from an alteration in the nucleotide sequence that does not alter the end amino acid sequence after translation. Choice D is incorrect because "insertion" is not the correct term to describe this mutation.
A universal blood donor has blood type A+. B-. AB. O-. O+.
O-. People with blood type O- are considered universal donors, capable of providing blood transfusions to recipients with a variety of blood types. The ABO blood type classification system categorizes blood type based on the antigens present on the surface of a person's red blood cells. Additionally, blood is categorized according to the presence or absence of an antigen called the RhD factor. People with blood type O- are considered universal donors because their red blood cells lack the A, B, and RhD antigens. Therefore, O- blood can be used in most transfusions because the immune system of the person receiving the blood will not recognize the incoming blood as foreign.
A developing primary oocyte typically completes meiosis I immediately prior to ovulation. After arresting in metaphase of meiosis II, the developing ovum cannot resume meiosis II until sperm penetration is initiated. Given this information, where would the first polar body produced during the development of an ovum initially form? Uterus. Body cavity. Ampulla of fallopian tube. Ovary. Cervix.
Ovary. If a single primary oocyte undergoes meiosis I and II, only 25% of the resulting cells will develop into a mature egg. During oogenesis, a primary oocyte undergoes meiosis I and II to produce a single ootid and three polar bodies. The ootid, which has the bulk of the cytoplasm, has the potential to develop into a mature egg but the polar bodies cannot. Since the question states that the meiosis I is completed immediately prior to ovulation the most logical conclusion is that the first polar body initially forms in the ovary. The ovary is indeed the initial location of the formation of the first polar body.
Sebaceous glands can be found in all of the following areas EXCEPT for one. Which one is the EXCEPTION? A. Eyelids B. Palms C. Scalp D. Underarms E. Earlobes
Palms Sebaceous glands are tiny glands in the skin that secrete an oil called sebum. They are located in the dermis of the skin and found nearly everywhere in human skin EXCEPT for the palms of the hands and soles of the feet. The body releases this oil to waterproof and protect the skin, but it also is a lipid and reduces friction. It doesn't make sense to make your hands and feet slippery, so we can infer this must be the correct answer.
What type of microscopy is best for viewing live, unstained cell specimens? A. Cryo-electron microscopy B. Scanning electron microscopy C. Phase contrast microscopy D. Transmission electron microscopy E. It is not possible to view live cells with a microscope
Phase contrast microscopy phase contrast and flourescent can be used to view live cells
All of the following are part of mRNA processing EXCEPT for one. Which one is the EXCEPTION? A. Phosphorylation of 3' end B. Addition of a 5' guanine cap C. Attachment of poly-A tail D. RNA splicing of exons E. Alternative splicing of mRNA
Phosphorylation of 3' end Answer Choices [B], [C], [D], and [E] are all included in mRNA processing. The 5' cap and poly-A tail are added to provide stability to the mRNA molecule. The poly-A tail is made of repeated adenine nucleotides attached to the 3' end of the mRNA. The mRNA must remove the introns, or non-coding regions, from itself using snRNPs (small nuclear ribonucleoproteins). Additionally, the mRNA can splice exons in different ways, effectively creating a different mRNA to be translated. This way, a single gene from DNA can produce multiple mRNAs, which can produce multiple proteins. Note: Prokaryotes have no introns for deletion, and therefore do not undergo splicing of exons and are not capable of alternative splicing. Additionally, their mRNA transcripts undergo little processing, as they have no nucleus and transcription is able to occur simultaneously with translation.
A causative agent of gangrene, Clostridium perfringens, is an obligately anaerobic, endospore producing, gram positive bacterium. What would be a logical treatment for a patient who stepped on a nail contaminated with C. perfringens? Moving the patient to a hot sauna. Moving the patient to a hot pool. Place the patient's leg in an oxygen rich hyperbaric chamber. Place the patient's leg in an oxygen rich hypobaric chamber. Place the patient's leg in an oxygen depleted hypobaric chamber.
Place the patient's leg in an oxygen rich hyperbaric chamber. A logical treatment for a patient who stepped on a nail contaminated with C. perfringens would be to place their leg in an oxygen rich hyperbaric chamber. Gangrene is a condition characterized by the death of tissues (necrosis) due to lack of blood flow (ischemia). The lack of blood flow causes the oxygen content of the gangrenous tissues to decrease significantly. Anaerobic bacteria, such as those of the genus Clostridium, are able to proliferate under such oxygen depleted conditions and produce toxins that can promote further cell destruction. This question is essentially testing your understanding of the terms "obligate anaerobe" and the roots "hyper" and "hypo". Obligate anaerobes are microorganisms that not only are unable to use oxygen during growth, but are actually killed by its presence. With an understanding of this definition Choice C and Choice D are good candidates for the correct answer choice. Choice C is correct because hyperbaric (greater than atmospheric pressure) chambers would increase the likelihood of hemoglobin oxygen saturation and penetration into ischemic tissues. Choice D is incorrect because hypobaric (less than atmospheric pressure) chambers would decrease the likelihood of hemoglobin oxygen saturation and penetration into ischemic tissues. Choice A and Choice B unreasonable choices because slightly warmer conditions would not have an effect on any endospore producing microorganism. Endospores are resilient exterior, encapsulating structures that allow cells to resist harsh environmental conditions due to fluctuations in temperature, pH, and moisture. Note: This question is a good example of the verbal application based biology questions commonly found on the official. You will not be expected to know the nuances of gangrene for your exam, however, you should be able to deduce logical conclusions from information provided in the question stem.
A bilateral animal lacking a coelom or pseudocoelom would belong to which phylum? A. Porifera B. Annelida C. Cnidaria D. Platyhelminthes E. Nematoda
Platyhelminthes Platyhelminthes are the flatworms and are known to be acoelomate and bilateral. Some of the flatworms include planarians (planarians, planar meaning flat) and tapeworms (tape is flat). Tapeworms do not have a complete digestive tract because they just absorb pre-digested food from the stomach of their hosts. Porifera are the sponges. You can remember this because sponges are porous and "porifera" sounds similar. They are also classified with the parazoa, meaning even though they are multi-cellular, the cells are not organized and no organs develop. Rotifera are rotifers, again both names sound very similar. Rotifers are especially cool because they are microscopic animals. They have a pseudocoelom and are bilateral. Cnidaria are radial animals and include hydrozoans, jellyfish, sea anemones, and corals. Annelida are segmented worms, such as leeches, earthworms, and polychaete worms. They have bilateral symmetry and a coelom.
All of the following are true of red blood cells EXCEPT one. What is the EXCEPTION? A. The kidneys influence the rate of red blood cell production B. Components of hemoglobin can be excreted from the body through the gastrointestinal tract C. Red blood cells are the most abundant type of blood cell D. Red blood cells are rigid in order to prevent damage to the cell E. More than 95% of oxygen traveling through the blood is traveling within red blood cells
Red blood cells are rigid in order to prevent damage to the cell important feature of RBC is their flexibility to allow them to fit inside capillaries that are more narrow than the diameter of the RBC
what is the portion of the human kidney that the collecting duct empties into called? Renal cortex. Ureter. Renal pelvis. Renal medulla. Bladder.
Renal pelvis The arrow in the diagram is indicating the location of the renal pelvis. The renal pelvis is the funnel-shaped, proximal end of a ureter. Filtrate processed in the kidneys is collected in the renal pelvis prior to being released into the ureter. It is important to realize that occasionally official exams will present biology questions in the form of a picture or illustration where the examinee is expected to identify a structure within the image.
Acne is caused by infections of blocked, oil secreting, glands in human skin. Which gland is likely the site of an acne infection? Sebaceous gland. Apocrine gland. Eccrine gland. Mammary gland. Adrenal gland.
Sebaceous gland. Sebaceous glands are oil-secreting glands located over the majority of the surface of the human body, with the exception of the soles of your feet and the palms of your hands. The obstruction of a sebaceous gland can cause the oil secretions, also known as sebum, to accumulate and become infected by bacteria (Propionibacterium acnes), ultimately resulting in acne. However, overall, sebum benefits our bodies by serving a protective function in the innate immune system. The fatty acids in sebum maintain the pH of human skin at a relatively acidic level (pH = 3~5), ultimately discouraging most microbial growth. Although apocrine glands (B) secrete some lipids, proteins, and water, they are not primarily the sites of an acne infection. Eccrine glands (C) primarily secrete water and other substances such as salts, and metabolic wastes. Mammary glands (D) primarily produce milk, not oil. Adrenal glands (E) do not secrete oil resulting in acne.
Protons are removed from the stroma during the light reactions of photosynthesis. Assuming that the dark reactions operate most efficiently while the light reactions are occurring, what relative pH would you conjecture the enzyme rubisco functions most optimally at? Much lower than normal pH. Slightly lower than normal pH. The same pH. Slightly higher than normal pH. None of the above.
Slightly higher than normal pH. This question could be answered through rote memory of the optimal pH of the enzyme rubisco. However, a more reasonable approach would be to deduce the answer from fundamental biology knowledge combined with information provided in the question. In order to answer this question correctly you need to understand that rubisco is a key enzyme in the dark reactions of photosynthesis and that these reactions occur in the stroma. Secondly, you need to know that removing protons from an enclosed space will increase its corresponding pH. Next, the question states that the dark reactions operate more efficiently in conjunction with the proton transporting action of the light reactions. If the dark reactions "like" the protons being removed from the stroma then the most logical conclusion is that rubisco functions optimally in a higher, or more basic, pH range. This is indeed true because the optimal pH for rubisco is roughly equal to 8. Your fundamental biology knowledge of photosynthesis combined with information from the question logically leads to the correct answer. Note: this question is a good example of a verbal application based style question commonly found on the official.
Which of the following is most correctly associated with the respiration physiology of some insects? Stomata. Spiracles and tracheal tubes. Malpighian tubules. Gills. Insects do not respire.
Spiracles and tracheal tubes. Most insects respire by use of a network of trachea and spiracles. Stomata are the openings where gas exchange occurs in the leaves of plants. Malpighian tubules are found in many insects, but serve excretory functions rather than respiratory. Gills are structures found in organisms such as fish, sharks, and some mollusks.
A football player tears an anterior cruciate ligament (ACL) located in his knee during a game. What classification of joint suffered the injury? Fibrous. Cartilaginous. Synovial. Suture. None of the above.
Synovial. The three main classifications of joints that you need to be aware of for the official are described as fibrous, cartilaginous, or synovial. Fibrous joints are, not surprisingly, connected by fibrous tissue and lack a joint cavity. An example of a fibrous joint is the sutures that connect the bones of the cranium. Cartilaginous joints do not have a joint cavity and are adjoined by cartilage. An example of a cartilaginous joint is the location where a rib attaches to the sternum. Synovial joints contain a true joint cavity filled with synovial fluid and typically have free range of motion. The human knee and shoulder are the most common examples of this classification of joint. Note: This question is a good example of the verbal application based biology questions commonly found on the official.
Which of the following is false regarding the benthic zone of aquatic biomes? The benthic zone can be part of the aphotic zone. The benthic zone cannot be part of the photic zone. The benthic zone is home to organism that feed on detritus. Temperatures are typically cooler in benthic zones. Benthic zones can be found in both freshwater and marine environments.
The benthic zone cannot be part of the photic zone. Choice B is false because the benthic zone can indeed be part of the photic zone of an aquatic biome. The benthic zone refers to the lowest strata of water in an aquatic biome. Choice A is true because the benthic zone can be part of either the photic or aphotic zones of an aquatic biome. Like wise, Choice E is true, because the benthic zone can apply to both freshwater and marine environments. Choice C and Choice D are true because benthic water is typically cooler and home to many organisms that feed on decaying organic matter that falls from the water above.
Bicoid is an example of a maternal effect inheritance gene in D. melanogaster. Which of the following situations is absolutely necessary for the lethal expression of the bicoid phenotype? The father of the fly is homozygous for the mutant gene. The mother of the fly is homozygous for the mutant gene. The fly is homozygous for the mutant gene. The mother and father are heterozygous for the mutant gene. The fly is heterozygous for the mutant gene.
The mother of the fly is homozygous for the mutant gene. The bicoid gene in D. melanogaster is important because it is an example of a maternal effect inheritance gene. The expression of the bicoid phenotype is only dependent on the genotype of the mother. If a mother is homozygous for the mutant genotype than her offspring will lethally express the trait.
Where does the gray crescent appear relative to the sperm entry point in amphibian eggs? The same side as the point of sperm entry. The opposite side as the point of sperm entry. The gray crescent does not form in frog eggs. The sperm entry point does not effect the formation of the gray crescent. Amphibians do not have eggs.
The opposite side as the point of sperm entry. The gray crescent is a region of darker colored cytoplasm that forms opposite to the point of sperm entry in some amphibian eggs. The gray crescent is indicative of the dorsal side of the embryo.
Which of the following correctly defines the term epitope? A phagocytic cell found in lymph nodes that destroys invading bacteria. A stage of apoptosis. A synonym for the major histocompatibility complex (MHC complex). A mutated T cell that cannot distinguish self from non-self. The part of a macromolecule which a B cell or T cell recognizes as foreign. Next
The part of a macromolecule which a B cell or T cell recognizes as foreign. an epitope or antigenic determinant is the region or a macromolecule on the surgace of an invading pathogen that the host immune system recognises as foreign. MHC are a group of cell surace proteins that bind to epitopes and present the epitope to immune cells. epitopes are the :imporant part: of an antigen that is uniqu and recognizable by the immune cells. this process epitope is on an antigen presenting celll (APC) only identical twins have the exact same MHC. MHC are important in organ transplants as the transplanted organ cells will be identifies as nonself and would be destroyed. for this reason an organ reciever must be immunosupressed.
All of the following statements are true about receptor tyrosine kinases EXCEPT one. Which one is the EXCEPTION? A. The protein passes through the membrane seven times B. They often bind to growth factors C. They often form dimers after extracellular signaling D. Phosphatases turn the receptor off E. The ligand binding site is extracellular
The protein passes through the membrane seven times
Which of the following mammalian extraembryonic membranes functions to produce developing red blood cells? Yolk sac. Allantois. Chorion. Amnion. Epidermis.
Yolk sac. One of the major functions of the yolk sac in mammalian embryos is to aid in the formation of developing red blood cells. In contrast to the yolk sac in birds, the mammalian yolk sac does not actually contain any yolk. Choice B is incorrect because the allantois functions in gas, nutrient, and waste exchange by developing into part of the umbilical chord. Similarly to the allantois, the chorion functions in gas exchange. Choice D, amnion, is incorrect because it does not function in the production of red blood cells, rather this extraembryonic layer is the fluid filled cavity that helps to protect and cushion a developing embryo from trauma. Choice E, epidermis, is not an extraembryonic layer and is therefore incorrect. the yolk is INTRambryonic (part of the embryo), the yold is not to be confused with the yolk sac. which is an EXTRAEMBRYONIC structure. in bids repties fish and monotrmes, the yolk sac encloses the yolk, the yolk provides nourishment to the cells in mammals the yolk sac completely ourishes the fetus until the placenta completely develops and takes over. also helps with the formation of basic blood vessels and bloood cells.
According to the endosymbiosis theory of evolution, the innermost layer of the plasma membrane of a chloroplast most closely resembles that of a fungi. a protist. a eukaryote. a cyanobacterium. an archaea.
a cyanobacterium. Endosymbiotic theory explains the origin of chloroplasts and mitochondria. Since the presence of membrane bound organelles, such as mitochondria and chloroplasts, is one of the defining characteristics of eukaryotic cells, the theory also provides insight into the origin of eukaryotic cells. According to endosymbiotic theory, chloroplasts arose from a mutually beneficial relationship that was formed due to the engulfment of a cyanobacterium by an ancient prokaryote. The relationship was beneficial to the host cell because the photosynthetic cyanobacterium was capable of harvesting energy from the sun and thus provided some of this energy to the host prokaryote. Likewise, the engulfed cyanobacterium may have benefited from the relationship by residing in a safer internal environment. Mitochondria are believed to have been derived through a similar mechanism to chloroplasts, however due to the engulfment of a proteobacterium as opposed to a cyanobacterium. Why would anyone ever believe this occurred? One of the major sources of evidence in support of this theory is that the innermost plasma membrane of chloroplasts and mitochondria most closely resembles that of a bacterium. Other major evidence in support of the endosymbiosis theory is that mitochondria and chloroplasts divide independently, by a mechanism similar to binary fission, from the host cell in which they reside. Additionally, DNA in chloroplasts and mitochondria resembles the circular DNA found in bacteria, contrasting greatly with the linear DNA found in most eukaryotes. Lastly, chloroplasts and mitochondria have 70S ribosomes similar to bacteria, while eukaryotes have 80S ribosomes. All of this evidence strongly suggests that the origin of chloroplasts and mitochondria, and thus eukaryotic cells (which are implicitly defined by the presence of membrane bound organelles), are of bacterial origin.
what is an antigen a protein comprised of heavy and light chains a foreign macromolecule with a distinct configuration that does not belong to the host molecule and that triggers an immune response. proteins that cause agglutination in the blood a and b all of the above
a foreign macromolecule with a distinct configuration that does not belong to the host molecule and that triggers an immune response. an antigen represents an antibody generating foreign macromoleule. antigens include bacteria, fungi, viruses, protezoa, and parasites such as worms.
false? B lymphocytes and T lymphocytes populate lymph nodes natural killer cells are not phagocytotic the largest phagocytic cells are the macrophages which may be found in the lymph nodes about 10% of WBC are compromised of neutrophils all true
about 10% of WBC are compromised of neutrophils lymph nodes contain B cells T cells and macrophages. these large macrophages contain fibril like pseudopodia that can attatch to a bacterium. macrophages are the larges phagocytic cells and will phagocytize bacteria. fixed macrophages are very numerious in lymphatic organs such as the spleen and the lymph nodes. neutrophils have an average lifespan of a few days and and tend to self destruct as foreign invaders are destroyed. neutrophis make up 60% of the WBC never let monkeys eat bananas
Platyhelminthes are acoloemates. pseudocoelomates. coelomates. bacteria. archaea.
acoloemates. Platyhelminthes are considered acoelomates because these organisms lack a true coelome. The lack of a true coelome is one of most important defining characteristics of this phylum. Coelomes are advantageous to some organisms because they can help to separate and protect organs and aid in locomotion.
which statement is true pine trees, spruces, and fir trees are examples of gymnosperms the dominant stage of life cycle in mosses is the gametophyte oaks maples and grasses are examples of angiosperms alternation of generations can be exhibited in the life cycle of land plants all correct
all correct angiosperms include fruits, flowering plants, maple, oaks, grass, PEAS.
the term motor unit refers to an entire muscle a single muscle fiber all the muscle fibers innervated by one nerve fiber all the motor nerves in one muscle all the sliding filaments of actin and myosin in one muscle fiber
all the muscle fibers innervated by one nerve fiber
what is true about HIV the hallmark is progressive depletion of CD4+ t cell pop (helper T cells) quite early in the infection CD4+ memory t cell destruction occurs in large nubers some strains of HIV are more virulent than others hiv can lie dormant in cells for many years, possibly decades all true
all true
which statements are representative of the second law of thermodynamics systems tend to move towards greater disorder energy transfers are always accompanied by some amount of loss energy is required for highly organized systems for most systems heat represents loss of energy all true
all true in most energy systems, energy is partly converted to heat. there is an unstoppable move in the direction of disorder.
which statement is true the nucleolus is the site of ribsome assembly lipid synthesis occurs in the smooth enodplasmic retuculum in euk organelles such as lysosomes nuclei and mito differ in size and sedimentation rates during centrifugation nuclei will sediment more readily than mito or bibosomes all true
all true during centrifugation large organlelles such as nuclei can be sedimented at forces of 1000X the force of graity lighter organelles such as mito remain in the supernatant. ribosomes are made in the nucleolus whicle the SER is invilved in lipid synthesis and drug detox
which is correct angiosperms include monocots and dictos cambium inside tissue cant develop into xylem and phloem gymnosperms are flowering plants nontracheophytes contain vascular tissue
angiosperms include monocots and dictos
In ecology, the competitive exclusion principle asserts that two different species cannot occupy the same niche at the same place and time. asserts that two individuals of the same species cannot coexist in the same area, if competing for the same mate. asserts that competition for resources occurs between all species within an ecosystem. only applies to marine ecosystems. asserts that the most unfit member of any competitive population will likely be excluded from the other individuals.
asserts that two different species cannot occupy the same niche at the same place and time. In ecology, the competitive exclusion principle states that two different species cannot occupy the same niche at the same place and time. In other words, two different species that are competing for the same limiting resources cannot coexist in the same place at the same time. Choice B is incorrect because the competitive exclusion principle primarily refers to resources for survival such as food, sunlight, or habitable space. Access to viable mates doesn't necessarily fall into the category of resources for survival and is therefore incorrect. Choice C may be true in certain conditions; however, it does not accurately define the competitive exclusion principle. Choice D is completely incorrect because the competitive exclusion principle applies to all ecosystems and is not solely limited to marine environments. Choice E is incorrect and has no relevance to the competitive exclusion principle.
what is not true about a vaccine many vacccines are made from killed or weakened pathogens vaccines may be from toxic but inactivated by products of dangerous organisms such as bacteria vaccines represent strategies used to reduce the risk of illness vaccines are given for diseases such as hepatitis and can be made from viral fragments attenuated vacccines contain a low concentration of the pathogen and will illicit an immune response
attenuated vacccines contain a low concentration of the pathogen and will illicit an immune response an attenuated vaccine contains a version of the living microbe that has been weakened and modifies in the lab. however the ability to induce protectice immunity remains as virulencce is lost. WILL NOT illicit an immune response.
which white blood cell listed below is least numerous monocyte neutrophil lymphocyte basophil eosinophil
basophil NEVER LET MONKEYS EAT BANANAS
A breed of drosophila can have a purple body (P) which is dominant to orange body (p). They also can either have vision (V), which is dominant to blindness (v). These are all autosomal genes. In an experiment, true breeding purple bodied, blind males (PPvv) are crossed with true breeding orange bodied flies with vision (ppVV) One of the F1 progeny males (PpVv) was test crossed, and the offspring were as follows: Purple, blind: 332 Purple, vision: 94 Orange, Vision: 390 Orange, blind: 85 Total # of offspring: 901 What is the recombination frequency in map units (m.u)? A. 80 m.u. B. 10 m.u. C. 20 m.u. D. 40 m.u. E. 51 m.u.
c. 20mu 94+85=179 total offspring 900 179/900 = 20. -if there are disproportionately more of 2 categories of offspring and disproportionally less of the other two categories then this means the genes are linked and do not assort with complete independence. -look at the two categories that have the lowest numbers. these are the ones with gene recombination. -add these two numbers together and divide by the total # of offspring to calculate the m.u. cM or % crossover.
In bacteria, DNA polymerase I produces new DNA in the 3' to 5' direction of the strand being synthesized. produces new DNA in the 5' to 3' direction of the template strand. synthesizes RNA. directly synthesizes ketone bodies. can only synthesize new DNA by adding nucleotides to an existing strand.
can only synthesize new DNA by adding nucleotides to an existing strand. Bacterial DNA polymerase I, as is true with all DNA polymerases, can only synthesize new DNA by adding nucleotides to an existing strand. DNA polymerase I requires a free hydroxyl group at the 3' position of the sugar backbone in order to stimulate the addition of new nucleotides. Choice A and Choice B are incorrect because DNA polymerase I synthesizes new DNA by adding nucleotides in the 5' to 3' direction of the strand being synthesized. Due to the antiparallel arrangement of a DNA double helix, this is analogous to saying that DNA is synthesized in the 3' to 5' direction of the template strand. Choice C is incorrect because RNA polymerases synthesize RNA, not DNA polymerases. Choice D is erroneous.
a stable complex of carbon monoxide and hemoglobin in RBC is termed carboxyhemoglobin carbaminohemoglobin oxyhemoglobin deoyhemoglobin 2,3-DPG`
carboxyhemoglobin carboxyhemoglobin is HB-CO. carbon monoxide is a toxic gas ad it can bind to hemoglobin. O2 works as the antidote if one is poisoned. however, even small amounts may lead to neurological damage and even death, CO can also bind to myglobin. CO binds to these 100000X more than oxygen. if CO2 is carries on hemoglobin we have carbaminohemoglobin. 2,3_DPG or 2,3BPG is a intermediate in glycolysis and made in the RBC which helps to unload O2 from hemoglobin
A terrestrial biome that can be located along the California coastline and is characterized by wet winters, dry summers, and scattered vegetation is known as a temperate deciduous region. temperate coniferous region. taiga. chaparral. desert.
chaparral. A terrestrial biome that is located along the California coastline and is characterized by wet winters, dry summers, and scattered vegetation is known as a chaparral. It is important to note that chaparral biomes are located in other regions around the world including: the Mediterranean, southwestern Australia, and Chile.
Which ratio below is most likely to greatly increase within a spermatid as it matures into a spermatozoa? cytoplasm/mitochondria. ribosomes/cytoplasm. chromatin/cytoplasm. mitochondria/centrioles. cytoplasm/centrioles.
chromatin/cytoplasm. The ratio of chromatin/cytoplasm is most likely to greatly increase as a developing spermatid matures into a spermatozoan. Spermatogenesis is the process by which spermatogonia develop into mature spermatozoa. Spermiogenesis refers to the later stages of spermatogenesis during which a spermatid develops into a spermatozoa. During spermiogenesis, the spermatid sheds much of its cytoplasm, ribosomes, and other temporarily inessential cytoplasmic components into a residual body which is then degraded by Sertoli cells. In contrast, mitochondria, centrioles, and chromatin are retained and form a highly specialized developing spermatozoan which is efficiently suited for propelling genetic material. Because the amount of chromatin in the developing spermatozoan remains relatively constant, and the amount of cytoplasm decreases greatly, the ratio of chromatin/cytoplasm within the developing spermatozoan is likely to increase substantially. The ratio of cytoplasm/mitochondria (A) and cytoplasm/centrioles (E) are likely to decrease during spermiogenesis because the amount of cytoplasm is likely to decrease while the amount of mitochondria and centrioles is likely to remain relatively constant. The ratio of ribosomes/cytoplasm (B) is likely to decrease because the ribosomes and much of the cytoplasm are shed during spermiogenesis. The ratio of mitochondria/centrioles (D) is likely to remain relatively constant, and therefore not increase, because the amount of mitochondria and centrioles are remains relatively constant. However, one of the centrioles does eventually elongate to form the flagella. Note: The question is representative of the verbal application based style questions commonly found on the official.
An individual who is color-blind most likely has some deficiency in which part of the eye? A. Cornea B. Lens C. Cones D. Pupil E. Rods
cones cones detect color concentrated in fovea of retina rods detect low level light and detail not sensitive to color
In regards to the concept of inclusive fitness, Hamilton's rule recognizes the following variables: benefit to the altruist, cost to the recipient, coefficient of relatedness. cost to the altruist, benefit to the recipient, and coefficient of relatedness. cost to the altruist, benefit to the recipient, distribution coefficient. benefit to the altruist and cost to the recipient only. cost to the altruist and benefit to the recipient only.
cost to the altruist, benefit to the recipient, and coefficient of relatedness. Hamilton's rule is dependent on several variables and attempts to quantify the relative likelihood of a seemingly altruistic action by an individual. The rule depends on the fitness cost to the altruist, the possible benefit or likelihood of improving the beneficiaries fitness, and how closely related the two individuals are (coefficient of relatedness). According to Hamilton's rule, an action that is not costly to the altruist and greatly beneficial to a closely related beneficiary increases the likelihood of an altruistic act.
after a night on the town and heavy drinking which os the following would most likely acccount for the increased urine production decrease in aldosterone decrease in vasopressin decrease in blood ph increase in ADH prozimal tubule reabsorbing more water
decrease in vaso pressin vasopressin as also called ADH is involved in the bodys ability to retain water, thus reducing urine output. casopressin can also raise Bp due to vasocontriction alcohol will black the creation of vasopressin thus the urge to urinate occurs. this peptide hormone is made in the hypothalamus and sotres and reelased by the posterior pituitary gland
The shaft of the long bone is called the: A. metaphysis. B. periosteum. C. epiphysis. D. diaphysis. E. medullary cavity.
diaphysis The diaphysis is the name of the shaft of a long bone, which is made of compact bone. The epiphysis is the rounded end of a long bone. The metaphysis is the wider portion of a long bone situated near the epiphysis, this is where the majority of growth happens during puberty. The periosteum is a membrane that covers the surface of all bones, except at the joints of long bones. It provides nourishment and sensation to the bones. The medullary cavity is the central cavity of bone shafts. It usually contains red bone marrow, which produces blood cells, and yellow bone marrow, which consists of adipose tissue.
which is true regarding lakes and ponds during summer, the deeper waters are high in nutrients and the upper layers high in O2 during summer, the deeper waters are high in O2 and nutrients during summer, the deeper waters are high in O2 and the upper layers high in nutrients during summer, winter, and fall oxygen is always highest at the surface during summer, winter, and fall oxygen is always lowest at the top
during summer, the deeper waters are high in nutrients and the upper layers high in O2 during the summer warm water is at the surface and is well aerated because of oxygen production by plants and mixing by the wind. the bottom layer will be high in nutrients and in the spring and autum ocygenated water from the surace does to the bottom and nutrients from the bottom goes to the top this process is known as a turnover and is vital for the survival an growth of organisms at all levels
ebola. what is incorrect humans pigs and monkeys can carry the ebola virus ebola spreads through direct contact with infected bodily fluids and tissues ebola targets mainly RBC which cause the patient to bleed uncontrollably people who die from ebola have multiple organ failures, low BP, and shock.
ebola targets mainly RBC which cause the patient to bleed uncontrollably ebola infects WBC such as monocytes and macrophagffes along with fibroblass dendritic cells and endothelial cells which release proteins that can cause clots in the blood stream. ebola blocks interferon release/
Which embryonic germ layer(s) eventually develop(s) into the adrenal gland? Ectoderm. Mesoderm. Endoderm. Ectoderm and mesoderm. Endoderm and mesoderm.
ectoderm and mesoderm The adrenal gland is derived from two different embryonic germ layers. The adrenal cortex is derived from the mesoderm, while the adrenal medulla is derived from the ectoderm. Note: This question can also be categorized as a developmental biology question.
during the respiratory cycle, the intrapleural pressure of the lungs is most negative during the: A. end of exhalation. B. end of inhalation. C. beginning of exhalation. D. beginning of inhalation. E. peak relaxation of the diaphragm.
end of inhalation. During inhalation our diaphragm contracts and therefore the volume in our lungs increases. This increase in volume leads to a decrease in pressure, which causes air to flow into our lungs. At the very end of inhalation is the point where we cannot breathe in anymore because our intrapleural pressure reaches its maximum negative value. During exhalation we relax our diaphragm, the volume of the lungs decreases, and air is pushed out of our lungs.
A first degree burn, such as a mild sunburn, damages the epidermis. dermis. hypodermis. dermis and hypodermis. epidermis, dermis, and hypodermis.
epidermis. First-degree burns, such as normal sunburns, only affect the epidermal layer of skin. Second-degree burns typically damage both the epidermal and dermal layers of the skin. Third-degree burns damage every skin layer.
birth control pills are most closely associated with which two hormones? testosterone and estrogen estrogen and progestin progesterone and entergastrone FSH and LH insulin and glucagon
estrogen and progestin birth control pills are designed to decrease the body into thinking that a pregnancy has occurred. estrogen and progestin (a synthetic progesterone-like hormone) are most commonly used. a mini pill that only contains progestin is quite common too. progesterone will decrease FSH and LH thus pregnancy will not occur. An enterogastrone is any hormone secreted by the mucosa of the duodenum in the lower gastrointestinal tract in response to dietary lipids that inhibits the caudal (or "forward, analward") motion of the contents of chyme. Examples include: secretin. cholecystokinin. gastric inhibitory peptide.
The massive influx of nitrogenous fertilizer runoff from farmland into aquatic ecosystems can cause destructive algae blooms. This process is called carbon fixation. heterotrophism. autotrophism. eutrophication. oxidation-reduction.
eutrophication. Eutrophication occurs when an influx of nutrients, such as nitrogen, destroys aquatic ecosystems by creating anaerobic water due to algae blooms. This phenomenon is a major problem in bodies of water such as the Gulf of Mexico where nitrogen rich runoff from farmland is creating anaerobic seawater and ultimately killing marine life.
The nervous system acronym EPSP stands for\ inhibitory postsynaptic potential. excitatory postsynaptic potential. inhibitory presynaptic potential. excitatory presynaptic potential. excitatory presynaptic propagation.
excitatory postsynaptic potential. EPSP stands for excitatory postsynaptic potential. EPSP's can be caused by the influx of ions into a neuron, resulting in the depolarization of the membrane. If the EPSP depolarization is significant enough to reach the threshold of the neuron then an action potential will be initiated. In contrast, inhibitory postsynaptic potentials (IPSP's) can be caused by the influx of ions which prevent the depolarization of the membrane.
which statement is incorrect external fertilization often results in many zygotes, but few survive internal fertilization usually produces fewer zygotes, but many survive external fertilization involves cooperative before leading to copulation external fertilization occurs mostly in wet environments all true
external fertilization involves cooperative before leading to copulation in external fertilization eggs shed by the female are fertilized by the male in the environment. external fertilization is seen in most fish and amphibians. internal fertilization involved sperm being deposited into the female reproductive tract. think copulation. internal fertilization offers protection from predators, whereas in external fertilization, predators greatly reduce the chance of surviving into adulthood.
What is a likely symptom if a person does not produce adequate thyroid hormone? A. Fast heart beat B. Sweating C. Weight loss D. Feeling cold E. Accelerated growth in a developing child
feeling cold hypothyroidism would result in lower metabolic rate and less metabolic hear production.
___________ are unique among landplants they have 2 independent living structures within their life cycle the sporophte and the gametophyte angiosperms ferns moss liverworts gymnosperms
ferns they contain two sepereate living structures called the sporophyte and the gametophyte. they are not dependent on eachother.
which of these cell types can most easily change their character in response to chemical signals megakaryocyte erythrocyte fibroblast adipose osteoclast
fibroblast fibroblasts seem to be the least specialized in the connective tissue family. the fibroblast is involved in collagen and extracellular matrix synthesis. these cells are derived from primitive mesenchyme, fibroblasts show a remarkable ability to differentiate into other cell types. fibroblasts can differentiate into adipose, muscle, bones and cartilage cells. mesenchyme- a loosely organized, mainly mesodermal embryonic tissue that develops into connective and skeletal tissues, including blood and lymph.
Which heterotrophic group of organisms can be microscopic and contains chitin in their cell walls? A. Protista B. Fungi C. Echinodermata D. Archaea E. Dinoflagellates
fungi Heterotrophic means the organism must seek its energy from sources other than itself. Fungi can be microscopic or macroscopic, such as a mushroom. The distinguishing feature is that fungi contain chitin in their cell walls, which is a nitrogen containing polysaccharide. Protista are the "odd" organisms that can be animal-like, fungus-like, unicellular, or multicellular. Echinodermata include sea stars and sea urchins. Archaea are prokaryotes that are more closely related to eukaryotes than bacteria. Dinoflagellates are a type of protist.
There are 50 black marbles and 50 white marbles in a bag. A student randomly grabs 2 marbles out of the bag, both of which happen to be black. The student erroneously concludes that the other 98 marbles remaining in the bag are also black. This student would most likely have difficulty understanding gene flow. genetic drift. natural selection. mutation. inbreeding.
genetic drift. The student would most likely have difficulty understanding the concept of genetic drift. Genetic drift is an evolutionary mechanism that occurs when a small, non-representative sample of a population evolves independently due to less genetic variation within the group. Genetic drift is to evolution as sampling error is to statistics. Because the boy initially only grabbed two marbles from the bag, he erroneously concluded that a similar frequency of marble colors would persist in future attempts. Since the boy did not take a statistically significant number of marbles from the bag, his assumptions were incorrect. Note: This question is an excellent example of a verbal application based biology question on the official. Although the question is rather difficult, it truly tests whether or not a student understands genetic drift through a novel example that they have likely never seen before.
which is false about liver albumin synthesis bile production destruction of worn-out old RBC's converts nitrogenous waste into urea glucose storage
glucose storage all of these are important facts about the liver except that GLYCOGEN is stored in the liver not glucose. the hormone insulin, secreted by beta cells of the pancreas decrease blood sugar levels and allow it to be converted into glycogen which is mostly stored in the liver, but is also stored in muscle cell.s albumin is used to transfer hormones and fatty acids. they are proteins that are water soluble. bile is made in the liver and stores in the gallbladder. helps emulsify fats. its release is by the signal of CCK or secretin in the duodenum
During isoelectric focusing, the pH of the gel used is always neutral. always 1. held at a consistent acidic level. held at a consistent basic level. held at a gradient.
held at a gradient. During isoelectric focusing, the pH of the gel used is held at a gradient. The gradient is necessary to separate different molecules based on their respective charges at various pH levels. Negatively charged molecules move towards to the positively charged end of the gel, while positively charged molecules move towards the negative end of the gel. As is the case with a protein, the amino acid residues gain or lose hydrogen atoms as they pass through the pH gradient. Eventually, the protein movement will remain relatively static upon reaching their isoelectric point, where the net charge of the molecule is zero.
Which of the following represents a structural motif that is seen in regulatory proteins that bind to DNA? A. Leucine-finger motif B. Helix-turn-helix motif C. Alpha-zipper motif D. Leucine-tail motif E. Beta-turn motif
helix-turn-helix the four common structural motifs as seen in regulatory proteins that bind to nucleic acids are -zinc finger motif -leucine zipper motif -helix turn helix motif -helix loop helix motif
Memory is primarily a function of the hippocampus. hypothalamus. cerebellum. medulla oblongata. pons.
hippocampus. The hippocampus primarily serves the body through memory functions.
Which part of the brain monitors the body's temperature? A. Hypothalamus B. Pons C. Medulla D. Cerebral cortex E. Cerebellum
hypothalamus The hypothalamus controls the body's temperature, hunger, thirst, sleep, and circadian cycles. One of its most important functions is that it links the nervous system to the endocrine system through the pituitary gland. The pons relays signals from the forebrain to the cerebellum. The medulla controls autonomic functions, such as respiration, digestion, and the pumping of the heart. The cerebral cortex is a complex portion of the brain that is involved in higher-level thinking and sensory perception. Lastly, the cerebellum is involved in motor control, such as balance and coordination.
During oxidative phosphorylation (but before chemiosmosis), the pH of the mitochondrial matrix: A. increases as ATP is generated B. decreases as oxygen enters the mitochondrion C. increases as the electron transport chain proceeds D. is always lower than the pH of the intermembrane space E. decreases when water is formed.
increases as the electron transport chain proceeds During oxidative respiration, electrons flow through the electron transport chain (ETC) until they react with oxygen to form water. During this time, protons are pulled into the intermembrane space, or OUT of the matrix. If protons are being removed from the matrix, then the pH must INCREASE (remember, a higher pH indicates more base, and a lower pH indicates more acid). Choice [A]: As ATP is generated, protons flow back into the matrix, meaning the pH of the matrix decreases since it is more acidic. Choice [B]: If oxygen is present in the mitochondrion, then the ETC will continue to proceed, H+ ions will continue to flow out of the matrix, and the pH of the matrix will increase, not decrease. Choice [D]: The pH of the matrix is often HIGHER than the intermembrane space because protons are being pumped out of the matrix and into the intermembrane space. Choice [E]: If water is being formed then the ETC must be operating, meaning the pH of the matrix must be INCREASING since protons are being pumped out of the matrix.
which statement is false passive diffusion involves the passage of a solute across a membrane from high to low concentraiton active transport is transport across a membrane against a concentration gradient from low to high conc. carried molecules are needed for facilitated diffusion and active transport ions diffuse across membranes down their concentration gradients all true
ions diffuse across membranes down their concentration gradients all cells have voltages across their plasma membranes and act like a battery. two forces act to drive ion diffusion: a chemical force due to the ions concetration gradient and the electrical force. together this is called the electrochemical gradient IONS CANT F*CKING PASS
The three dimensional structure of a protein is primarily dictated by its amino acid sequence. the number of nitrogen containing amino acid residues. the number of glycine residues. the number of proline residues. the time required to translate the mRNA into a polypeptide.
its amino acid sequence. The three dimensional structure of a protein is primarily dictated by its amino acid sequence. The relative location of various amino acid residues within a protein establishes the chemical interactions that occur between the atoms of one or more polypeptides.
The concentric rings that surround the haversian canals found in a cross section of compact bone are called lamellae. lacunae. osteons. canaliculi. Volkmann canals.
lamellae For the official you should be aware of several important features of compact bone. In preparation for your exam make sure you understand the function and are able to identify lamellae, lacunae, osteons, canaliculi, Haversion canals, and Volkmann canals. Lamellae are the concentric ring structures that surround the haversian canals in compact bone. Lacunae (B) are the small cavities within the lamellae that typically contain osteocytes. Osteons (C) are the circular structures in compact bone that contain the haversian canals, surrounding lamellae, as well as other structures such as the Volkmann canals. Canaliculi (D) are the small crevices that connect lacunae in compact bone. The filopodia of adjacent osteocytes extend through the canaliculi where they join one another via gap junctions. The gap junctions allow osterocytes to exchange nutrients and signaling molecules. Volkmann canals (E) are the pathways by which blood vessels can pass from the periosteum to the osteons within a particular section of compact bone.
Louis Pasteur disproved the theory of spontaneous generation with the use of swan neck flasks. His experiment ultimately demonstrated that life cannot be created from non-life. a cultured organism can cause disease. anaerobic organisms exist. mutation is a valid evolutionary mechanism. bacterial colonies will continue to grow as long as resources are available.
life cannot be created from non-life. Louis Pasteur's swan neck flask experiment disproved spontaneous generation by demonstrating that life cannot be created from non-life. In other words, by supporting the theory of spontaneous generation, scientists and non-scientists alike, incorrectly believed that life could be created from non living matter such as rocks or sunshine.
Artificially prepared vesicles made of a lipid bilayer are called miscelles. Capsule. liposomes. ribozymes. platelets.
liposomes Liposomes are artificially prepared vesicles made of a lipid bilayer. By separating the external environment from internal, liposomes provide a chemically controlled environment more conducive to independent metabolism. Choice A is incorrect because micelles, although similar to liposomes, are lipid membranes that are characterized by a hydrophobic interior and hydrophilic exterior. Choice B is incorrect because a capsule is usually composed of polysaccharides and covers the exterior of bacteria. Choice D is incorrect because a ribozyme is a term used to describe RNA that serves dual functions as a component of protein synthesis as well as a catalyst. Choice E is incorrect because platelets are involved in an immune system response and does not reflect the quotation.
which of the following would NOT cause a deviation in the hardy weinberg equilibrium? mutation small pop size meiosis natural selection founder and bottle neck effects
meiosis genetic equilibrium would be psosible if a pop was very large no net mutations random mating occured isolation from other pops and no natural selection occured. these are needed to maintain equilibrium. bottlenecking will reduce overall genetic variability while the founder effect inroduces new genes. meiosis will not change the hardy weinberg equiibrium because the gene frequency is NOT altered for the genes to eeep foing back into the gene pool
Which of the following options correctly sequences the humoral immune response? use antibodies, mature b cells, plasma cells, naive b cells and place in order
naive b cells mature b cells plasma cells antibodies The humoral immune response can be summarized by naïve B cells -> mature B cells -> plasma cells -> antibodies. After naïve B cells become mature in the bone marrow, they move to secondary lymphoid organs and await activation from helper T cells. After activation, mature B cells begin differentiating into plasma cells. In turn, plasma cells produce antibodies which are capable of binding to invading antigens.
All of the following are components of the nitrogen cycle, EXCEPT one. Which one is the EXCEPTION? A. Denitrifying bacteria regenerate atmospheric nitrogen, N2 B. Nitrogen accounts for around three quarters of the Earth's atmosphere C. Dead, decaying plants and animals contribute as a nitrogen source as they decompose D. Nitrifying bacteria convert nitrates, NO3- in the soil into ammonium, NH4+ E. Nitrogen fixing bacteria exist in root nodules of legumes
nitrogen source as they decompose D. Nitrifying bacteria convert nitrates, NO3- in the soil into ammonium, NH4+
any change in the nucleotide seq of DNA is called a mutation. in sickle cell anemia a single AA subsitution occurs. a glutamic acid replaced by a valine. which statement is incorrect? the mutation occurs on the beta chain of hemoglobin the electrophoretic patttern could be different between normal hemoglobin HBA and sickle cell anemia patients with sickle cell anemia are homozygous for the sickle gene sickled cells are very fragile and rupture easily. capillaries may become blocked by the long abnormally formed cells causing pain and interfering with normal function offspring with sickle cell traits are homozyhgous for the abnormal gene and have a resistance to malaria
offspring with sickle cell traits are homozyhgous for the abnormal gene and have a resistance to malaria HBA is more negative than HBS since it contianed a glutamin acid residue. patients with sickle cell disease are homozygous for the sickle gene while patients with the sickle cell TRAIT are heterozygous and have both kinds of hemoglobin in approx. equal amounts.
he primary force that drives the reclamation of blood from the interstitial space at the venous end of a capillary is blood pressure. muscle movement. voluntary control by the brain. osmotic pressure. active transport.
osmotic pressure. There are two major forces that dictate flow into or out of capillaries: blood pressure (capillary hydrostatic pressure) and osmotic pressure. Blood pressure is the force the capillary wall exerts on blood. Osmotic pressure opposes blood pressure and results from the tendency of water to move inside capillaries due to the abundance of solute particles in the blood. At the arteriole end of capillaries, blood pressure tends to be greater than osmotic pressure. As a result, blood flows out of the capillary and into the interstitial space. At the venous end of a capillary, osmotic pressure is usually greater than blood pressure. As a result, blood is reclaimed from the interstitial space and flows into the capillary.
an ultra marathon runner may run 100 miles in a snigle race.what will most likely occur from the stress being placed on his bones osteoclastic activity would increase, anad calcium phosphate is removed from the bone osteoblastic activity would increase, anad calcium phosphate is released into the matrix osteoblastic activity would decrease and collagen fibers are produced osteoblastic acivity would increase and calcium chloride is released into the matrix all of the above may occur
osteoblastic activity would increase, anad calcium phosphate is released into the matrix upon stressing a bone, bone forming cells called osteoblass deposit collagen and release calcium phosphate to strenghten the bone. the mineral hydroxy apetite is produced.
cri du chat syndrome is due to deletion of the short arm of chromosome 5 and is associated with an array of congenital problems such as abnormal development of facial structures. gastroinestinal malformations and infants that make crying sounds that sound like meowing. this syndrome is due to monosomy trisomy partial trisomy partial monosomy inversion
partial monosomy partial monsomy occurs when only part of one chromosome is lost. in partial trisomy the addidtion of only a portion of another chromosome is seenl