Bio/Biochem from learning log Flashcards
The most common catabolic reaction in the human body is:
difference between catabolic and anabolic reactions
Metabolism of carbohydrate and fat spares protein tissue. All of teh following are true of fats except: A. fats may be used in cell structure- true; B. fats may be used as hormones- true; C. Fats are more efficient form of energy storage than proteins (true); D. Fats are less efficient form of energy storage than carbohydrates:
Fat is a more efficient form of energy storage than proteins (which obviously makes sense when you think about it)
Like cellulose, chitin is a polysaccharide that cannot be digested by animals. Chitin differs from cellulose in that it possesses an acetyl-amino group at the second carbon. What molecule is a reactant in the breaking of the beta-1,4 glycoside linkages of cellulose and chitin?
Hydrolysis (water as reactant) breaks beta-1,4 glycoside linkages of cellulose and chitin
Muscles cause movement at joints by:
decreasing in length, thereby bringing the muscle’s origin and insertion close together usually by moving the insertion; neurons cause contractions in muscles, not tendons, and the neural signals are not initiated by the muscle.
Studies have shown that atheletes have different proportions of muscle cell types according to the type of activity they perform. Which type of muscle cell would be best suited for a marathon runner? A. Slow oxidative fibers B. Fast oxidative fibers, C. Slow glycolytic fibers, D. Fast glycolytic fibers
Slow oxidative fibers fatigue slowly because they undergo aerobic respiration and while they do not contract quickly, they are the last type of muscles to fatigue. That combined with the large amount of myoglobin making this fiber type red, makes this type of muscle ideal for long-term running. Slow glycolyic fibers are not one of the three most common muscle types.
Which of the following statements regarding muscles that act as third class levers int eh body is LEAST likely true? A. They require more force to perform theh allotted work than i fno lever were present; B. The force of the muscle contraction is applied between the fulcrum and the load force; C. This arrangement increases the bulk of the body; D. This arrangement increases the precision of the movement.
Answer: C, All muscles are third class levers require more force to perform a certain amount of work than if no lever were present due to the fact that the force of the muscle acts between the fulcrum and the load force The need for greater force to perform smaller movements inc the body’s capacity for precision movement
Skeletal muscle contraction may assist in all of the following EXCEPT: A. movement of fluid through the body; B. body temperature regulation; C. posture; D. Peristalsis
Peristalsis reers to the muscle contraction which pushes partially digested food through the digestive tract; it is accomplished through teh action of smooth NOT SKELETAL muscle, all other options are correct
In PCR amplification, a primer is hybridized to teh end of a DNA fragment and acts as the initation site of replication for a specailized DNA polymerase. The DNA fragment to be amplified is shown below. Assuming that the primer attaches exactly to the end of the fragment, which of the following is most likely the primer? 5’-ATGNNNNGCT-3’
DNA polymerase replicates in the 5’ to 3’ directon, and that this replication occurs in teh opposite direction 3’ to 5’* of the template strand. In other words, the DNA polymerase can only read from 3’ to 5’ so it must sytart at the 3’ end of the template DNA fragment. The complement of the 3’ end of the DNA fragmetn in teh question stems is 5’ -AGC-3’, answer is D; other A and C are complementary to any portion of the depicted DNA srand so can be eliminated; B is reverse of complementary to teh tempalte strand if it was written in reverse, but at the 5’ end of the template, DNA polymerase could not work with this primer
Which of the following regarding resitrction enzymes such as EcoRI and HindIII, is most likely false? A. They are made by bacteria as a defense against viruses; B. EcoRI is synthesized in teh cytosol; C. they recognize sequences that rea hte same forward and backward; D. DNA cut by EcoRI easily joins to DNA cut by HindIII?
Restriction enzymes are proteins produced by bacteria that recognize and cut specific sequences of viral DNA, protecting the bacterium from teh virus, so choice A is true, chice B is true becuase these molecules are produced in cytosol since they are prokartyotic proteins; prokaryotes do not have membrane bound organelles, including the ER. *restriction enzymes are typically targeted to palindromic sequen es which read the same forward and backward, and strands of DNA cut by teh same restriction enzyme can be joined tgoether, the sme is not true of strands cut by different enzyme which will form noncomplementary ends due to tehir distinct target sequences. For this reason, answer D is false
Cystic fibrosis is an autosomal rescessive trait resulting from teh dysfunction of a sodium chloride transporter protein. Which of the following describes the best experimental approach to developing a mouse model for cystic fibrosis? A. take the human protein encoding the sodium chloride transporter and express it in mice. B. Identify the mouse homologue of the sodium chloride transporter and develop a knockout mouse. C. identify the mouse homologue of hte sodium chloride transporter and feed the mice food laced with RNAi for that protein. D. Identify the promoter for the mouse homologue of the sodium chloride transporter and change it to a weaker one.
B. because question stem indicates cystic fibrosis is caused by a loss of function mutation in a specific transporter protein. A model should involve the complete loss of function of the protein or its model homologue. Both RNA interference and substituting a weaker promoter will cause a partial loss of function, these methods would not be the best methods for making a model. Choice A might initially sound appealing, it does not invlve any practical loss of function. Expressing human protein in mic emight cause it to stop functioning normally, but the mouse homogloue of the cholride transporter would not be affected, meaning the mice would be essentially wild type. A much more effective method for generating a modle would be to generate a knockout for the homologous mouse gene. B is the best answer.
Pancreatic beta cells were isolated from a mouse after a fast. The same mouse was then fed, followed by another round of beta cell collection. If samples from both groups of cells were run on teh same DNA microarray, what color would be seen on the region of the chip containing probes for the insulin gene? A. green, B. yellow, C. red, D. No color
A. green, the lecture states that a gene that is up-regulated after some treatment will appear green on a gene chip, wherease a gene that is down-regulated will appear red. After eating, insulin levels generally rise, insulin is necessary to help remove nutrients like glucose from bloodstream for intracellular use and storage), likely corresponding with a rise in expression of the gene that codes for insulin. The spot on the chip specific to insulin mrNA, then should appear green rather than red. Yellow spot would indicate a gene that was neither up-regulated nor down -regulated. there is no reason to expect no labelign would occur.
Developmental biologists have developed inducible knockout mice for certain genes. In these animals, an enzyme responsible for excising the gene of interest is only active after treatment with certain pharmaceuticals. This development would be most useful for the study of which of the followign genes? A. The LacZ gene, B. A telomerase gene that is active in many cell types during old age; C. A dopamine transpoter gene that becomes active during puberty in certain brain regions; D. a Hox family gene necessary for proper embryonic neurulation.
One of hte main disadvantages of standard knockout models is that they cannot be use dto study genes that are essential for surviving during development. Knocking out these genes is lethal to the embryonic mouse, makign breeding (and continuation of the knockout line) impossible. for this reason, a Hox gene ncessary for neurulation could only be studied using conditional knockouts or a similar method liek RNA intereference. This makes choice d the best answer. Although both the genes named in choices B and C could be studied using a conditional knocout, they are also accessible via traditional knockout technology, so condiitonal knockout technology would not be most helpful for thier study. LacZ prokaryotic gene that is not expressed nroamlyl in mice, so choice a can be eliminated.
If gluconeogenesis is occuring in the liver, which metabolic pathway will be happening in a red blood cell?
pentose phosphate pathway is use to generate NADPH which will be used to reduce glutathione in red blood cells. In teh fastign state, red blood cells rely upon liver gluconeogenesis for their glucose supply. They do not have hte enzymes necessary to produce glucose themselves. red blood cells do not store glycogen, so they cannot have glycogenoltuic pathway active. Teh citrc acid cycle occurs in the mitochondria which red blood cells do not possess.
Glucose-6-phosphate is the enzyme that catalyzes the last step of gluconeogenesis, teh formation of glcuose from glucose 6 phosphate. If a child were born without a functional version of this enzyme, which of following problems would be expected?
Glucose 6 phosphate catalyzes the conversion of glucse 6 phosphate to glucose. glcuose 6 phosphate cannot easily cross cell membranes becuase of its cahrged phosphate group. withot glucose6phosphate there would be a buildup of glcuose 6 phosphate at the site of glucogenesis: THE LIVER! Glucogenesis occurs during the fastign state. at this time, te liver is the primary supplier of glucose. If it cannot export glcuose, blood glcuose levels will be low. Choice A and B can be eliminated because loss of this enzyme would not impact/affect glycolysis or glycogenesis. Chocie C is not the best ansewr becuase muscle tissue is not a major source of blood glcuose even when this enzyme is functional. =answer=dec blood glucose levels during the fasting state
When hormone-sensitive lipase is active during the fasting state, which of the following processes will not be active in a muscle cell? A. glycolysis, B. gluconeogenesis, c. citric acid cycle, d. electron transport chain
Glucogenesis occurs in teh liver, not in muscle.
A cell that is drastically HYPERTONIC to its environment will most likely experience which of the following?
Bursting! When cell in hypertonic environment, it has a higher solute concentration than the environment. this results in an influx of water and potential bursting. Choice B shriveling is unlikely because shriveling occurs when the environment is hypertonic. the tonicity of the cell is nto associated with cell division or presence of pathogens, so mitosis and infection are irrelevant
Transmission of action potentials in the heart requires a large scale movement of ions. which of the followign intercerllular connections would most likely be enriched in cardiac muscle cells? A. tight junctions; B. desmosomes; C. intercellular bonds; D. gap junctions
gap junctions! Tight junctions are impermeable connections that do not allow ions to flow; makes them unlikely to be found in a tissue that requires the movement of ions. Desmosomes specalize in cell-to-cell adesion, so choice B is possible but not necessarily related to the ion movement in teh question. Intercellular bonds can be ruled out not a major type of junction. Gap junctions best answer because these gap junctions allow the ion flow that occurs in cardiac muscle cells.
Streptococcus pyogenes is a common cause of pharyngitis, or inflammation in the throat. How might these bacteria be distinguished in culture? A. They will stain pink, indicating they are gram-positive; B. they will have a rod-like apperance; C. they will stain purple, indicating they are gram-negative; D. they will show a growth pattern resembling beads on a chain.
D. question asking how one might DISTINGUISH bacteria from other bacteria in culture. Hint is the name ends with “coccus” indicating that these are cocci, or spheres. Answer can be derived by POE. A is misleading since purple strain is associated with positive Gram stain. B is inaccurate since bacilli are rod-shaped. C is misleading since a pink stain is associated with a negative Gram stain. D is correct since this is true of hte bacteria and is also the source of theri name. The hint is that they resemble beads, indicating they are spherical.
Which of he following must be present for a population to be considered a Hardy-Weinberg equilibirum? A. natural selection; B.large population size; C. non-random mating; d. emigration
B. large population size; Hardy-weinberg equilibrium describes a population in which NO EVOLUTION is occuring. Natural selection and non-random mating contribute to evolution by the selection and propagation of certain genes at greater frequencies than others. Emigration contributes to evolution by the introduction of new genetic information into a population. B is the best answer because a large population size is necessary to protect the population from evolution caused by GENETIC DRIFT
Which of the following does NOT correctly describe a characteristic that is true of eukaryotic cells but not prokaryotic cells? A. Eu cells have a membrane-bound nucleus; b. Eu DNA is coiled with histone proteins; C. EU cells contain ribosomes; D. Eu flagella are composed of microtubules.
C. Eu cells contain ribosomes; A and B obviously correct. Prok. do not have any membrane bound organelles not only no membrane bound nucleus and mitochondria! Prokarytoic DNA naked and lacks histone proteins. EU flagella are composed of microtubules, but prok. cells are made of flagellin** Both EU and Prok. cells contain ribosomes but prok. ribosomes are smaller. Flagella of bacteria are made of from the protein flagellin. Microtubules are a main building block of other organelles: tail of sperm cell, spindle apparatus, cilia of the fallopian tubes!
Which of the following virus types require that an RNA dependent RNA polymerase be carried within the virus for replication? I. Retrovirus, II. Positive sense virus, III. Negative sense virus. Options: A. I only, B. III only, C. I and II only, D. II and III only
B. Negative sense virus only. Postive sense viruses have coding RNA that can immediately be transcribed by the host ribosomes. they do NOT REQUIRE RNA- dependent RNA polyemrse (RdRP) b/c these viruses can simply carry teh gene for it and have hte host translate new RdRP. This makes option II positive sense virus a weak answer so not C or D, answer is negative sense viruses. Negative sense viruses contain the complement of the coding strand. Since human cells do not normally make RNA from other RNA, there is no way for these viruses to begin replication. This makes option III a strong choice and make choice B the best answer. Retroviruses are virsues that use an enzyme called reverse transcriptase to insert themselves into the host genome. Knowledge of their polarity is not required since it is only an option by itself and can be elimiated with option II.
Most viruses that infect animals: A. enter the host cell via endocytosis; B. do not require a receptor protein to recognize the host cell; C. leave their capsid outside the host cell; D. can reproduce independentl of a host cell
A. most animal viruses enter host cells via receptor-mediated endocytosis. Becasue of this, a receptor is required to recognize the sepcific appropriate host cell. Unlike bacteriophages, animal viruses do not leave capsids outside host cells. Finally, no virus can reproduce independently, as a defining characteristic of viruses is reproduction within a host cell.
Which of the following structures is (are) found in bacteria? I. A cell wall containing peptidoglycan; II. a plasma membrane lacking cholesterol; III. Ribosomes
D. All three are correct, I, II, III. Eubacteria have cell walls containing peptidoglycan. Bacterial cell membranes do not contain cholesterol. Finally, ribosomes are present in all cells, Eu or Prok.
Prior to infecting bacteria, bacteriophage must: A. reproduce makign copies of the phage chromosome; B. integrate its genome into the bacterial chromosome; C. penetrate the bacterial cell wall completely; D. attach to a receptor on the bacterial cell membrane
What occurs first is receptor attachment. Answer D is correct. Bacteriophages can only infect those bacteria who have specific receptor** that the virus can attach to and use to enter the cell body. Viruses do not reproduce until they enter the cell. Bacterial cell wall penetration is the first step of infection, right after attachnment to the receptor. Integration of the viral genome doesnto occur until after infection.
DNA from phage resistant bacteria is extracted and placed on agar with phage-sensitive E. coli. After incubation it is determined that these E. coli are now also resistant to phage attack. The most likely mechanism for their acquisition of resistance is: A. transduction, B. sexual reproduction, C. transformation, D. conjugation
C. Transformation is the process of picking up stray DNA frm environment and incorporating it. DNA already on agar so originally sensitive bacteria most likely underwent transformation and incorporated the resistance gene. Transduction is transfer of genes via a viral vector, and while that could potentially occur here, transfomraiton would definitely have to occur first. Sexual reproduction and conjugation involve transfer and swapping of genes btw multiple bacteria, and the resistance gene are on the extracted DNA in environment, not in a bacterium that could conjugate or do sexual reproduction
