Exam 1 Flashcards
(40 cards)
- The genetic code is considered degenerate because
A. more than one codon can code for a single amino acid.
B. one codon can code for multiple amino acids.
C. more than one anticodon can bind to a given
codon.
D. only one anticodon can bind to a given codon.
E. None of the above
A.
The degeneracy of the genetic codes describes how it is not precise or specific. More than one codon can code for each of the different amino acids. For example, both UAU and UAC code for tyrosine. Since there are three bases in a codon, there are 64 possible codons and only 20 amino acids. Obviously, there will be some repeating of amino acids among the codons. Answer choice B is the opposite of answer choice A and therefore not true. Answer choice C is incorrect because it is essential that the anti-codon be the exact base pair complement of the codon to maintain protein integrity. Answer choice D is true, but not the definition of the degenerate genetic code.
- Which statement about gastrulation is false?
A. In the amphibian, the initial site of gastrulation occurs at the gray crescent.
B. As a consequence of gastrulation the embryo now consists of two germ layers: ectoderm and endoderm.
C. In the amphibian, the infolding through which blastula cells migrate during gastrulation is called the blastopore.
D. The primitive gut that results from gastrulation is called the archenteron.
E. Mesoderm develops during gastrulation.
B. As a consequence of gastrulation the embryo now consists of two germ layers: ectoderm and endoderm.
The in-pocketing of the blastula during gastrulation actually allows for the formation of the third cell layer, the mesoderm. All the other answer choices are true.
- Polar bodies are formed during
A. male mitosis. B. female mitosis. C. male meiosis. D. female meiosis. E. Two of the above
D.
Polar bodies are haploid, or monoploid, nonfunctional gamete-like cells that are formed during female meiosis. Recall that meiosis is a two-stage process. In the first stage, a diploid cell undergoes a reduction division to form two haploid cells. In the second division, each of those haploid or monoploid cells undergoes a second division to form two haploid or monoploid cells each. So there are a total of four haploid or monoploid cells formed from each original diploid germ cell. In the case of sperm cells, four functional haploid or monoploid gamete sperm cells are formed. In the case of egg cells, the first meiotic division involves an unequal division of cytoplasm, and that results in the formation of one large cell and one small cell. The large cell will go on to divide again; the smaller cell is known as the first polar body. The small cell polar body may then divide again to form two other polar bodies. The large cell undergoes a second meiotic division, again unequally dividing the cytoplasm, with the larger resulting cell being the final haploid or monoploid ovum, and the other small cell again becoming a polar body. So it is possible that during a female meiotic division that one large ovum or functional egg cell and three polar bodies may be formed. Polar bodies are not formed in mitosis, which is how all other cell divisions occur in the body; that is, equal cell divisions that do not form sex cells.
- Which statement about the lac operon is NOT correct?
A. There are 3 structural genes that code for functional proteins.
B. There is a gene that codes for a repressor protein.
C. The promoter is the binding site of RNA
polymerase.
D. The repressor protein binds to the operator, halting
gene expression.
E. The lac operon is found in leukocytes.
E.
The lac operon is a set of control and structural genes in E. coli that allow the digestion of lactose. There are three structural genes controlled by an operator found on another part of the genome. In the abscence of lactose a repressor protein is bound to the operator, preventing RNA polymerase from binding to the DNA, thus preventing translation of the structural genes. However, when lactose is present (and glucose is absent), the repressor is removed and RNA polymerase can attach to the promoter, and translation occurs. This is known as an inducible system.
- If 18O-labeled glucose is given to a rat, where will the label first appear?
A. Exhaled O2 B. Exhaled CO2 C. Exhaled H2O D. Plasma H2O E. Intracellular H2O
B.
When radioactively-labeled glucose is given to a rat, it will enter the respiration reactions beginning with glycolysis and proceeding through the Krebs cycle and the electron transport chain. The glucose will be broken down into pyruvic acid during glycolysis. However, at that point, no CO2 is formed. As pyruvic acid forms acetyl coenzyme A, CO2 is released, and this CO2 may contain the labeled oxygen. Later, as acetyl coenzyme A enters the Krebs cycle and the Krebs cycle reactions occur, more CO2 is released. This CO2 may also contain
the labeled oxygen. Answer choice A is incorrect because we generally do not exhale oxygen. The only oxygen exhaled is that which is not absorbed into the alveoli. Answer choices C, D, and E are all incorrect. The water that is formed as a result of respiration is formed from the electrons that are sent down the electron transport chain in the mitochondria, combined with hydrogen ions and molecular oxygen, which is taken in through the lungs. This molecular oxygen is the final electron acceptor of the chain. Therefore the oxygen in respiratory waste product water comes from the oxygen that is inhaled into the alveoli. It does not come from oxygen that is part of the glucose molecule that is expired.
- An individual with type AB antigen on his red blood cells, in an emergency
A. may receive a transfusion of type O blood.
B. may receive a transfusion of type A blood.
C. may receive a transfusion of type B blood.
D. All of the above
E. None of the above
D.
People with the blood type AB are known as the universal recipients. They have both A and B blood group antigens on the surface of their red blood cells. Therefore, their blood will have no antibodies to any of the blood antigens and they can receive blood from type A, B, and O donors.
- Which of the following statements about oxidative phosphorylation is NOT correct?
A. It occurs on the inner membrane of the mitochondrion.
B. It involves O2 as the final electron acceptor.
C. It produces two ATPs for the FADH2.
D. It can occur under anaerobic conditions.
E. It involves a cytochrome electron transport chain.
D.
Oxidative phosphorylation occurs in the cristae of the mitochondria. NADH and FADH2 donate their electrons to a series of cytochrome molecules and form an electron gradient, which forms a proton gradient. This proton gradient drives a proton pump that is coupled to an enzyme, which produces ATP. The donated electrons are ultimately picked up by O2 and form H2O, one of the metabolic wastes of cellular respiration. Each FADH2 from the Krebs cycle results in two ATP, while each NADH results in three ATP. However, the NADH from glycolysis only produce two ATP due to the energy needed to get their electrons into the mitochondria from the cytoplasm.
- Glycogen
A. is a storage polymer of glucose.
B. is found in both animals and plants.
C. is degraded by a process called glycogenesis.
D. is synthesized by a process called glycogenolysis. is an unbranched molecule.
A.
Glycogen is a storage polysaccharide sometimes known as animal starch. It is made usually in the liver as a result of high glucose concentrations in the blood. Answer choice B is incorrect because plants do not produce glycogen, rather they produce starch. Answer choice C is incorrect because glycogen is degraded by a process called glycogenolysis, and answer choice D is incorrect because glycogen is synthesized by a process called glycogenesis. Answer choice E is incorrect because glycogen is also a highly branched molecule.
- Which of the following is a characteristic of tRNA?
A. It is a long filament of RNA.
B. It is produced in the nucleolus.
C. It has a poly-A tail.
D. It has some short double-stranded segments.
E. It is the template for protein synthesis.
D.
tRNA is a smaller type of RNA, which functions as a carrier of amino acid molecules. Like mRNA, tRNA is coded for by DNA, but, unlike mRNA, it is a comparatively short ribonucleotide polymer instead of a long filament. Although it is largely single-stranded, there are short double-stranded segments in tRNA where the nucleotide chain loops back upon itself. mRNA is actually the template for protein synthesis and has a poly-A tail, which plays a role in the degradation of mRNA. rRNA is produced in the region of the nucleus known as the nucleolus.
- The process in which a cell engulfs large particulate is called
A. pinocytosis. B. exocytosis. C. cytokinesis. D. phagocytosis. E. osmosis.
D.
Phagocytosis is the process of engulfing large matter such as a bacterium.
Answer choice A is incorrect because pinocytosis is the process of taking in small amounts of liquid
Answer choice B is incorrect because exocytosis is the process of releasing proteins from the cell.
Answer choice C is incorrect because cytokinesis is the division of the cytoplasm during mitosis
Answer choice E is incorrect because osmosis is the movement of water from an area of lower solute to an area of higher solute concentration.
- Which organelle is chiefly responsible for digestive breakdown of the cell during autolysis?
A. Pinocytotic vesicle B. Golgi body C. Ribosome D. Mitochondria E. Lysosome
E.
The lysosome is a small, membrane enclosed vesicle containing digestive enzymes at a very low pH. These enzymes are sometimes used to digest worn-out organelles within the cell. In some cells, such as phagocytic cells, they are used to help break down ingested bacteria and foreign material. Autolysis is when the lysosomes release there contents into the cytoplasmof a cell, causing cell death. Autolysis is a method by which a cell can undergo programmed death, called apoptosis. An example of apoptosis is seen during the molding of the human hand during fetal development. Answer choice A is incorrect; a pinocytotic vesicle is a small vesicle formed by the cell membrane in order to take into the cell a small particle or drop of liquid. Answer choice B is incorrect; a Golgi body, normally located at the end of a section of endoplasmic reticulum, has the function of packaging and modifying proteins. The Golgi apparatus may surround these proteins and vesicles for export out of the cell, and may also add or delete sections of these proteins that have been formed. Answer choice C is incorrect; ribosomes produce proteins under the direction of messenger RNAs using amino acids, which are present in the cytoplasm. Answer choice D is incorrect; mitochondria are the site of cell respiration and all energy-producing processes within the cell. They are also known as the “powerhouses” of the cell, as they produce ATPs.
- All of the following about the Krebs cycle are true EXCEPT
A. the Krebs cycle occurs in the matrix of the mitochondrion.
B. the Krebs cycle is linked to glycolysis by pyruvate. C. the Krebs cycle is the single greatest direct source
of ATP in the cell.
D. citrate is an intermediate in the Krebs cycle. E. the Krebs cycle produces nucleotides such as
NADH and FADH2.
C.
The single greatest direct source of ATP in the cell is the electron transport chain. The Krebs cycle does occur in the matrix of the mitochondria and oxidative phosphorylation (the electron transport chain) occurs in the inner membrane. The final product of glycolysis is pyruvate, which is converted into acetyl coA for use in the Krebs cycle. Citrate is an intermediate in this cycle (it is the first molecule formed; another term for the Krebs cycle is the Citric Acid cycle). The Krebs cycle only forms two ATP directly, all of the other ATPs that are formed are produced by oxidative phosphorylation (the nucleotides NADH and FADH2 donate their electrons to the electron transport chain).
- Which statement about cyclic AMP (cAMP) is NOT true?
A. cAMP is formed from ATP.
B. The enzyme that catalyzes the formation of cAMP
is adenylate cyclase.
C. The enzyme that catalyzes cAMP formation is
generally located in the cytoplasm.
D. Membrane receptors are capable of activating the
enzyme that forms cAMP.
E. cAMP is regarded as a second messenger, since it
can trigger a cascade of intracellular reactions after a hormone binds to the cell membrane.
C.
cAMP is a second messenger triggered after a receptor binds a ligand. Ligands (such as hormones and neurotransmitters) will bind their membrane receptors, activating it. Through a G-protein intermediate the enzyme adenylate cyclase will be activated, and will covert ATP into cAMP. Adenylate cyclase is attached to the inner layer of the phospholiped bilayer; it is not located in the cytoplasm. cAMP is responsible for carrying the chemical stimulus into the cytoplam and triggering a response, and so is called a 2o messenger.
- The ectodermal germ layers give rise to
A. nails, blood vessels, and epidermis. B. adrenal cortex and epidermis. C. neurons and epidermis. D. kidneys, blood vessels, and heart. E. tooth enamel, blood vessels, and epidermis.
C.
The ectodermal germ layers give rise to the epidermis of the skin and also the nervous system. The endodermis, or endodermal germ layer, gives rise to the lining of the digestive system, its associated glands and organs (such as the liver and pancreas), and the lungs. Most of the other organs and systems of the body are mesodermal, including the excretory system, the reproductive system, the muscular and skeletal systems, and the circulatory system. In general, if you cannot identify a particular tissue as being specifically endodermal or ectodermal, there’s a very good chance that it is one of the many mesodermal-derived tissues. Choices A, B, D, and E all contain tissues or structures that are mesodermal in origin.
- Which statement about the cell plasma membrane is NOT correct?
A. It serves as a selectively permeable barrier to the external environment.
B. It serves as a mediator between the internal and external environments.
C. In eukaryotes it contains the cytochrome chain of oxidative phosphorylation.
D. It contains phospholipids as a structural component.
E. It contains proteins that in some cases span the membrane.
C.
The plasma membrane separates the cellular contents from the environment and is responsible for the permeability of the membrane and what is allowed in and out. The fluid mosaic model of the plasma membrane states that it is a bilayer of phospholipid interspersed with proteins acting as receptors, pores, and channels. The pores and channels cross the entire membrane. The cytochrome chain is actually located in the cristae (the inner membrane) of the mitochondria.
- Unequal division of the cytoplasm occurs in
A. production of sperm cells. B. production of egg cells. C. mitosis of an epidermal cell. D. binary fission in bacteria. E. None of the above
B.
Unequal cytoplasm division occurs when egg cells are produced during the meiotic process of oogenesis. All meiotic divisions are two stages. In the first stage of egg production, also known as oogenesis, the precursor diploid cell produces two daughter cells, but one of the daughter cells receives almost the entire amount of cytoplasm, while the other becomes a nonfunctioning polar body. In the second division of oogenesis, the large daughter cell divides again, and once again one of the new daughter cells receives almost all of the cytoplasm,and the other becomes a small nonfunctional polar body. The original first polar body may also divide to form two nonfunctional polar bodies. The final result is a potential four haploid or monoploid cells, but only one of them—the one that received a greater amount of cytoplasm during each meiotic division—becomes a functional egg cell. Answer choice A is incorrect; during spermatogenesis, one diploid precursor cell forms four functional haploid, or monoploid, sperm cells. In this case, both divisions are equal, and all sperm cells are equal in amount of cytoplasm. Answer choice C is incorrect; during mitosis of epidermal cells cytoplasm is distributed equally. Likewise, answer choice D is incorrect; in the binary fission of bacteria, in which bacterial cells are dividing as a means of reproduction, cytoplasmic division will be equal.
- Green (G) is dominant over yellow (g) in peas, and smooth peas (S) are dominant over wrinkled peas (s). Which cross must produce all green, smooth peas?
A. GgSs x GgSs B. Ggss x GGSs C. GgSS x ggSS D. GgSs x GGSS E. None of the above
D.
Both green and smooth are dominant phenotypes. We want to produce only green smooth peas, so we want only dominant phenotype offspring. Therefore we must avoid any crossing that may result in the mating or combining of two recessive alleles. In choice A, crossing Gg and Gg could result in approximately 1/4 yellow offspring. Similarly, in choice B, ss crossed with Ss would produce approximately half wrinkled phenotype offspring. In choice C, Gg crossed with gg is likely to produce approximately half yellow offspring. Choice D is correct in that one of the parents is a double dominant, in that all offspring will have the dominant phenotype, regardless of the genotype of the other parent.
- Which statement regarding protein synthesis is false?
A. tRNA molecules shuttle amino acids that are incorporated into the protein.
B. tRNA molecules have the amino acid bound to the 3’ end of the molecule.
C. The process does not require energy.
D. Ribosomal RNA is needed for proper binding of
the mRNA message.
E. The message is read from the 5’ end to the 3’ end.
C.
Protein synthesis does require energy.
Answer choice A is true because tRNA brings the amino acid to the ribosome where it interacts with the mRNA with the proper sequence.
Answer choice B is true because tRNA molecules do have the amino acid bound to their 3’ end. The mRNA is read from 5’ to 3’ as the ribosome moves along the message. Answer choices D and E are both true.
- If a male with blood type A marries a female with blood type B, which of the following types is impossible for a first generation child?
A. Type B B. Type A C. Type O D. Type AB E. All types are possible.
E.
A and B blood groups are co-dominant over blood antigen O. Therefore if a man heterozygous for blood type A (AO) married a woman heterozygous for blood type B (BO), they could possibly have children with the genotypes AO, BO, OO, or AB. Therefore, all blood types are possible in this mating.
- Which of the following associations of germinal tissues and developed tissues is incorrect?
A. Mesoderm : heart B. Ectoderm : nervous system C. Endoderm : intestinal tract D. Mesoderm : pancreas E. Ectoderm : epidermis of skin
D.
The ectoderm develops into the skin, the nervous system, and the eyes while the mesoderm develops into the musculoskeletal system and the internal organs. The endoderm develops into the digestive tract and is associated with organs such as the pancreas and liver, the respiratory tract, and the bladder lining. The pancreas would therefore develop from the endoderm.
- All of the following organelles are membrane-bound EXCEPT the
A. rough endoplasmic reticulum. B. Golgi apparatus. C. nucleus. D. mitochondria. E. ribosome.
E.
The ribosome, found in both prokaryotes and eukaryotes, can be found in the cytoplasm or bound to the endoplasmic reticulum. It is involved in polypeptide synthesis. Membrane-bound organelles are found only in eukaryotes, and include the ER, Golgi apparatus, mitochondria, nuclei, and lysosomes.
- The basis for the pairing of the two strands of DNA in the double helix is
A. covalent bonding. B. ionic bonding. C. hydrogen bonding. D. hydrophobic interactions. E. hydrophilic interactions.
C.
DNA is a double-stranded helix composed of the purines adenine and guanine, and the pyrimidines cytosine and thymidine. Adenine binds with thymidine while guanine binds with cytosine via weak hydrogen bonds. These weak bonds enable the helices to separate easily to facilitate DNA replication.
Answer choice A is incorrect because covalent bonding is characterized by shared electron pairs while answer choice B is incorrect because ionic bonding is characterized by electron transfer. These are both strong forms of intermolecular bonds. Answer choice D is incorrect because hydrophobic interactions are attractive forces between non-polar molecules while answer choice E is incorrect because hydrophilic interactions are attractive forces between polar molecules.
- C6H12O6+O2→CO2+H2O
Which organelle is this process completed in?
A. in the cytoplasm. B. in the area of the cell membrane. C. in the nucleus. D. in the mitochondria. E. in the area around the ribosomes.
D. in the mitochondria
Respiration begins in the cytoplasm but is completed in the mitochondria. The first step of respiration is glycolysis, which occurs in the cytoplasm. Pyruvate from this reaction is converted to acetyl CoA, which enters the Krebs cycle in the mitochondria. NADH and FADH2 from the Krebs cycle enter the electron transport chain
in the cristae, the inner membrane of the mitochondria.
- Cells that are involved in active transport, such as cells of the intestinal epithelium, utilize large quantities of ATP. In such cells there are
A. high levels of adenylate cyclase activity. B. many polyribosomes. C. many mitochondria. D. high levels of DNA synthesis. E. many lysosomes.
C.
A cell that is involved in active transport, such as the epithelial cells of the intestine, will require large amounts of ATP. If a cell utilizes large amounts of ATP, it must have many mitochondria. Answer choice A is incorrect because high levels of adenylate cyclase activity is found in cells that are the target cells for hormones. Answer choice B is incorrect because many polyribosomes are found in cells that have a high level of protein synthesis. Answer choice D is incorrect because high levels of DNA synthesis are found in cells that undergo rapid reproduction and mitosis. Answer choice E is incorrect because many lysosomes would be found in phagocytic cells enabling them to digest the foreign material they’ve endocytosed.
