embryology and mitosis meiosis + science 3

  1. (haploid nuclei of sperm + ovum) --> diploid one cell zygote ---> 2 cell --> 4 cell -- > 8 cell -- > 16 cell (morula) --- blastocyst (with fluid filled cavity called blastoceol in middle) -- > gastrula (3 layers - ectoderm/ mesoderm/ endoderm)
    (haploid nuclei of sperm + ovum) --> diploid one cell zygote ---> 2 cell --> 4 cell -- > 8 cell -- > 16 cell (morula) --- blastocyst (with fluid filled cavity called blastoceol in middle) -- > gastrula (3 layers - ectoderm/ mesoderm/ endoderm)
  2. This gastrula layer goes on to produce these organs

    brain
    spinal cord
    skin
    hair
    nails
    sweat glands
    lining of mouth, anus, nostrils
    ectoderm
  3. This gastrula layer gives rise to these organs

    kidneys (excretion)
    blood vessels
    blood
    heart
    bone 
    cartilage
    adrenal cortex
    gonads
    musculoskeletal

    Lymph vessels and spleen
    mesoderm 

    means of getting around
  4. this gastrula layer gives rise to these organs

    interior lining of body including lining of 
    GI TRACT

    URINARY BLADDER

    URETHRA

    PANCREAS

    LIVER and gall bladder

    LUNGS

    (esophagus -stomach - intestines - colon)
    (Larynx, pharynx, trachea, thyroid, parathyroid)
    ENDODERM (GUULLP) - associate with stomach

    • GI
    • Urethra
    • Urinary bladder
    • liver and gall bladder
    • lungs
    • pancreas

    • (esophagus -stomach - intestines - colon)
    • (Larynx, pharynx, trachea, thyroid, parathyroid)
  5. In eukaryotes, the process of asexual cell division is known as mitosis. Mitosis takes place in four phases: prophase, metaphase, anaphase, and telophase. Prophase prepares the cell for mitosis: the DNA condenses such that distinct chromosomes become visible, as sister chromatids (or copies of a given chromosome) join at a region known as the centromere. The kinetochore assembles on the centromere, and is the site where microtubule fibers that extend from the centrosome and form the mitotic spindle attach to pull the sister chromatids apart in later stages of mitosis. Other microtubules known as asters extend from the centrosome to anchor it to the cell membrane. Additionally, the nuclear envelope and the nucleolus disappear, and the mitotic spindle forms. In metaphase, the chromosomes line up at the middle of the cell along an imaginary line that is known as the metaphase plate. In anaphase, the sister chromatids are separated and pulled to opposite sides of the cell by shortening of the microtubules attached to the kinetochores. Telophase can be thought of as the opposite of prophase, as a new nuclear envelope appears around each set of chromosomes and a nucleolus reappears within each of those nuclei. The process of mitosis is completed by cytokinesis.

     

    In contrast, meiosis is a form of cell division that is essential for sexual reproduction. It takes place in germ cells (also known as sex cells). Meiosis differs from mitosis in that it has two stages and results in the formation of four daughter cells, each of which has only one copy of each chromosome (haploid, n), in contrast to mitosis, which generates cells with two copies of each chromosome (diploid, 2n) that are essentially identical to their parent cell.

     

    In prophase I of meiosis, homologous chromosomes (i.e., the maternal and paternal copies of a given chromosome) pair up with each other in a process known as synapsis, forming tetrads. While paired up, homologous chromosomes may exchange genetic information in a process known as crossing over. The crossing-over points are known as chiasmata. This process results in recombinant DNA that is another source of variation in sexual reproduction, in addition to the variability inherent to the process. In metaphase I, homologous pairs, which take the form of tetrads, line up at the metaphase plate. The orientation of the homologous pairs is random in terms of which side of the metaphase plate the maternal or paternal copy of a given chromosome in a homologous pair winds up. In anaphase I, the homologous pairs are separated, and one member of each pair is pulled to each side of the cell. In meiosis II, which operates similarly to mitosis, the sister chromatids are split up into two haploid daughter cells.
    In eukaryotes, the process of asexual cell division is known as mitosis. Mitosis takes place in four phases: prophase, metaphase, anaphase, and telophase. Prophase prepares the cell for mitosis: the DNA condenses such that distinct chromosomes become visible, as sister chromatids (or copies of a given chromosome) join at a region known as the centromere. The kinetochore assembles on the centromere, and is the site where microtubule fibers that extend from the centrosome and form the mitotic spindle attach to pull the sister chromatids apart in later stages of mitosis. Other microtubules known as asters extend from the centrosome to anchor it to the cell membrane. Additionally, the nuclear envelope and the nucleolus disappear, and the mitotic spindle forms. In metaphase, the chromosomes line up at the middle of the cell along an imaginary line that is known as the metaphase plate. In anaphase, the sister chromatids are separated and pulled to opposite sides of the cell by shortening of the microtubules attached to the kinetochores. Telophase can be thought of as the opposite of prophase, as a new nuclear envelope appears around each set of chromosomes and a nucleolus reappears within each of those nuclei. The process of mitosis is completed by cytokinesis.

     

    In contrast, meiosis is a form of cell division that is essential for sexual reproduction. It takes place in germ cells (also known as sex cells). Meiosis differs from mitosis in that it has two stages and results in the formation of four daughter cells, each of which has only one copy of each chromosome (haploid, n), in contrast to mitosis, which generates cells with two copies of each chromosome (diploid, 2n) that are essentially identical to their parent cell.

     

    In prophase I of meiosis, homologous chromosomes (i.e., the maternal and paternal copies of a given chromosome) pair up with each other in a process known as synapsis, forming tetrads. While paired up, homologous chromosomes may exchange genetic information in a process known as crossing over. The crossing-over points are known as chiasmata. This process results in recombinant DNA that is another source of variation in sexual reproduction, in addition to the variability inherent to the process. In metaphase I, homologous pairs, which take the form of tetrads, line up at the metaphase plate. The orientation of the homologous pairs is random in terms of which side of the metaphase plate the maternal or paternal copy of a given chromosome in a homologous pair winds up. In anaphase I, the homologous pairs are separated, and one member of each pair is pulled to each side of the cell. In meiosis II, which operates similarly to mitosis, the sister chromatids are split up into two haploid daughter cells.
  6. During spermatogenesis, the main function of ______ cells is to nourish the developing sperm cells. These cells are located in the epithelial lining of the seminiferous tubules and are activated by FSH.




    _____ cells are adjacent to the seminiferous tubules in the testicles. In response to stimulation by LH, they produce testosterone and other androgens.




    _______ cells are catecholamine-secreting (dopamine NE and epinephrine) neuroendocrine cells of the adrenal medulla. Remember, the catecholamines include epinephrine and norepinephrine. 





    ______ cells are follicular cells closely associated with the developing female oocyte (egg). They function to convert thecal androgens to estradiol prior to ovulation. After ovulation, they give rise to the corpus luteum and begin producing high levels of progesterone.
    Sertoli

    Leydig

    Chromaffin

    Granulosa
  7. Spermatogonia divide through mitosis into two primary spermatocytes. Primary spermatocytes go through meiosis I and divide into two secondary spermatocytes. This is where the transition from diploid (2n) to haploid (n) happens. Secondary spermatocytes then go through meiosis II, forming spermatids. A total of four genetically unique spermatids are formed from each primary spermatocyte.

     

    Spermatids initially lack some of the most important features of the mature sperm cells that are released during ejaculation, and they gain those features in a process known as spermiogenesis. The main events of spermiogenesis are as follows: (1) formation of the acrosomal cap, which facilitates the ability of a sperm to fertilize an egg; (2) formation of a tail; and (3) loss of excess cytoplasm. Spermiogenesis results in non-mature spermatozoa that are incapable of independent movement, and are transferred to the epididymis to undergo maturation.

     

    Mature sperm cells are very compact, and contain roughly the bare minimum of structures necessary for their functionality. A mature sperm cell has a head, a mid-piece, and a tail. The head contains the cell’s DNA and is surrounded by the acrosomal cap. The mid-piece contains abundant mitochondria, which are necessary because sperm cells require quite a bit of energy throughout their life cycle, and the tail provides motility.
    Spermatogonia divide through mitosis into two primary spermatocytes. Primary spermatocytes go through meiosis I and divide into two secondary spermatocytes. This is where the transition from diploid (2n) to haploid (n) happens. Secondary spermatocytes then go through meiosis II, forming spermatids. A total of four genetically unique spermatids are formed from each primary spermatocyte.

     

    Spermatids initially lack some of the most important features of the mature sperm cells that are released during ejaculation, and they gain those features in a process known as spermiogenesis. The main events of spermiogenesis are as follows: (1) formation of the acrosomal cap, which facilitates the ability of a sperm to fertilize an egg; (2) formation of a tail; and (3) loss of excess cytoplasm. Spermiogenesis results in non-mature spermatozoa that are incapable of independent movement, and are transferred to the epididymis to undergo maturation.

     

    Mature sperm cells are very compact, and contain roughly the bare minimum of structures necessary for their functionality. A mature sperm cell has a head, a mid-piece, and a tail. The head contains the cell’s DNA and is surrounded by the acrosomal cap. The mid-piece contains abundant mitochondria, which are necessary because sperm cells require quite a bit of energy throughout their life cycle, and the tail provides motility.
  8. During ______, interaction occurs between chromatids from homologous chromosomes, not between sister chromatids. You can remember this because sister chromatids are genetically identical, so crossing over between them would have no effect.
    synapsis
  9. A fatty acid is composed of a carboxylic acid head and a tail end composed primarily of:

    A. hydrocarbon groups.

    B. phosphate groups.

    C. amino groups.

    D. sulfate groups.
    A - hydrocarbon groups.
  10. Several Salmonella species are facultative anaerobes. Assuming that other external conditions are controlled for, would the expected growth rate of a Salmonella colony be slower in the presence or absence of O2?
    In the absence of O2, because these conditions result in lower production of ATP, which can fuel binary fission.

    Facultative anaerobes can produce energy in the presence or absence of O2. In the presence of O2, the bacteria undergo aerobic respiration, which produces approximately 19 times as many ATP molecules per molecule of glucose as does anaerobic respiration.
  11. electron flow in electron transport chain

    NADH to complex 1 and Succinate to complex 2 (succinate dehydrogenase creates FADH2 which delivers electrons to complex 2)

    complex 1 and 2 send electrons to coenzyme q which sends electrons to complex 3 which sends to cytochrome c which sends to complex 4 which sends to oxygen to create _____
    H2O WATER
  12. Plane of symmetry on a compound means it will have a meso compound which would count as one stereoisomer not 2. So if a compound has 2 chiral carbons which would normally be 2^2 = 4 stereoisomers if the compound is meso because of a plane of symmetry one of the isomers is meso so its 1 instead of 2 so its 3 total stereoisomers
    Plane of symmetry on a compound means it will have a meso compound which would count as one stereoisomer not 2. So if a compound has 2 chiral carbons which would normally be 2^2 = 4 stereoisomers if the compound is meso because of a plane of symmetry one of the isomers is meso so its 1 instead of 2 so its 3 total stereoisomers
  13. an ox   red cat

    electrons flow from _____ to _____ to _____

    In an electrolytic cell set up as described in the passage, electrons flow into the electrolytic solution from?
    anode oxidation

    cathode reduction

    anode to cathode to solution

    the negatively-charged cathode, in order to allow reduction.
  14. more protons = smaller atomic radii

     Protons, which are positively charged, attract electrons. Calcium has the highest atomic number between it and K+ (i.e. the most protons), so its electrons will be pulled closest to the nucleus.



     As a general rule of thumb for isoelectronic species, cations have the smallest radii. Ar is not a cation.




     As a general rule of thumb for isoelectronic species, anions have the largest radii.
    more protons = smaller atomic radii

     Protons, which are positively charged, attract electrons. Calcium has the highest atomic number between it and K+ (i.e. the most protons), so its electrons will be pulled closest to the nucleus.



     As a general rule of thumb for isoelectronic species, cations have the smallest radii. Ar is not a cation.




     As a general rule of thumb for isoelectronic species, anions have the largest radii.
  15. Phosphorylation usually occurs on serine (S), threonine (T), tyrosine (Y), and histidine (H) residues in eukaryotic proteins.
    Phosphorylation usually occurs on serine (S), threonine (T), tyrosine (Y), and histidine (H) residues in eukaryotic proteins.
  16. A ______ blot measures levels of protein in a sample. Specific proteins can be found on the western blot using antibodies against that protein.

    The _____ blot is a technique used to analyze protein post-translational modifications, such as lipids, phosphoro-moieties, and glyco-conjugates.

    A _____ blot shows levels of specific RNA sequences using complementary probes.

    A ______ blot (named after Edwin Southern) shows levels of specific DNA sequences using complementary probes.

    RT-PCR vs PCR
    western

    eastern

    northern

    Southern- The purpose of Southern blotting is to look for specific nucleotide sequences in the DNA being tested and/or to identify the size of the DNA fragments that contain the sequence of interest. The diagram below shows the logic of Southern blotting:

    RT-PCR- Detect RNA expression and PCR detect DNA sequence
  17. If researchers failed to take into account the effect of air resistance on the pitch, how would it impact their measurements of the efficiency of energy transfer from the arm to the baseball? 

    A. It would be lower than the actual efficiency, as there was a higher initial velocity at release.
    If researchers failed to take into account the effect of air resistance on the pitch, how would it impact their measurements of the efficiency of energy transfer from the arm to the baseball? 

    A. It would be lower than the actual efficiency, as there was a higher initial velocity at release.
  18. NaCN 

    strong or weak base?
    weak base
  19. CaHPO4

    Name and why?
    Dibasic calcium phosphate because HPO4^2- can accept 2 hydrogens so its dibasic phosphate and the Ca2+ making it even proves it

    PO4^3-
  20. minerals are not vitamins

     Vitamins are generally organic compounds (carbon) that are not naturally produced by the human body and are required to maintain good health. Minerals are also essential elements, such as calcium, phosphorus, potassium, sulfur, and magnesium, as well a number of trace elements, such as iron, cobalt, zinc, and selenium, which are required to maintain good health.

    minerals inorganic (no carbon except co2)
    minerals are not vitamins

     Vitamins are generally organic compounds that are not naturally produced by the human body and are required to maintain good health. Minerals are also essential elements, such as calcium, phosphorus, potassium, sulfur, and magnesium, as well a number of trace elements, such as iron, cobalt, zinc, and selenium, which are required to maintain good health.

    minerals inorganic
  21. In comparison to the cohesive forces between water molecules of the protein solution droplet, how can the strength of interaction between water and oil molecules at an oil-water interface be characterized?
    Weaker, because they are forces created by induced polarity in nonpolar molecules 

    The Van der Waals forces that exist between molecules of water and of oil are predominately of the induced dipole-dipole type. Here, a small, temporary dipole is induced in molecules of oil by the permanent dipole of water, resulting in a weak attraction between the molecules.



    This is a weaker interaction than the hydrogen bonding that predominates in the interaction between water molecules.
  22. Amphiphiles, which include _____ are chemical compounds that possess both hydrophilic and hydrophobic components.

    Phospholipids are the most important example of an amphiphilic compound for the MCAT. Phospholipids are triacylglyceride derivatives with two nonpolar fatty acid “tail” regions and a hydrophilic phosphate-containing “head.” The predominant structural component of the cell membrane is the lipid bilayer,

    NO2, SO3-, etc are charged or polar so they are hydrophillic 

    straight chain hydrocarbons are hydrophobic
    soaps, detergents, and biological molecules such as lipoproteins and bile acids,

    Phospholipids are the most important example of an amphiphilic compound for the MCAT. Phospholipids are triacylglyceride derivatives with two nonpolar fatty acid “tail” regions and a hydrophilic phosphate-containing “head.” The predominant structural component of the cell membrane is the lipid bilayer,

    NO2, SO3-, etc are charged or polar so they are hydrophillic 

    straight chain hydrocarbons are hydrophobic
  23. Detergents, which can solubilize impurities and denature protein structure, are another important class of amphipathic molecules. In a technique known as sodium dodecyl sulfate-polyacrylamide gel electrophoresis (SDS-PAGE), the strong anionic detergent SDS is used to denature native proteins into their unfolded polypeptide states. SDS imparts most proteins with an even distribution of charge per unit mass, and denatures secondary and tertiary structure (except for disulfide bonds). SDS-PAGE is a useful technique because it allows proteins to be separated only based on mass. More specifically, the electrophoretic mobility displayed by these proteins will be a linear function of the logarithms of their molecular weight.
    Detergents, which can solubilize impurities and denature protein structure, are another important class of amphipathic molecules. In a technique known as sodium dodecyl sulfate-polyacrylamide gel electrophoresis (SDS-PAGE), the strong anionic detergent SDS is used to denature native proteins into their unfolded polypeptide states. SDS imparts most proteins with an even distribution of charge per unit mass, and denatures secondary and tertiary structure (except for disulfide bonds). SDS-PAGE is a useful technique because it allows proteins to be separated only based on mass. More specifically, the electrophoretic mobility displayed by these proteins will be a linear function of the logarithms of their molecular weight.
  24. A scientist uses an ultrasound device mounted to a vehicle to measure fluid flow underground. The device makes use of the Doppler effect to track fluid movement in the water table. Which of the following scenarios is most likely to produce a readable Doppler shift?

    I. The fluid is flowing at a velocity twice that of the sound-emitting device, in the same direction as the device is moving.

    II. The fluid is flowing at the same velocity and in the same direction as the sound-emitting device is moving.

    III. The fluid is not moving at all.
    I. The fluid is flowing at a velocity twice that of the sound-emitting device, in the same direction as the device is moving.

    ONLY

    The Doppler effect is used to analyze moving objects or fluids. A Doppler shift will be registered only if the fluid is moving relative to the source of the sound (the device). More specifically, at least some component of the fluid's velocity must exist in the same directional plane as the wave's velocity (otherwise, the device will register the fluid as not moving at all), and this component must be different from the velocity of the sound source.

    f observed = (f actual  ) (v +/- vo)/ (v -/+ vs)

    v = 3 x 10^8

    use top sign if its moving toward. so vo + is observer moving toward and vs - if its moving toward
  25. The azimuthal quantum number corresponds with _______

    principal quantum number corresponds with ______
    Approximate geometric shape of the orbital

    The potential energy of the electron. Approximate radial size of an electron cloud.
  26. The problem with ______ (far-sightedness) is the image is formed behind the retina rather than on the retina. A ______ lens will _____ light rays before they can pass the retina and focus light closer to the retina.

    _______, also known as near-sightedness, is a condition of the eye where the light that comes in does not directly focus on the retina but in front of it, causing the image that one sees when looking at a distant object to be out of focus. You need a _____ lens for this.
    presbyopia/ Hyperopia,

    converging

    converge

    Myopia

    diverging
  27. Light inside the thin glass tube of a laproscopic surgical device strikes the edge of the glass tube and is entirely reflected back into the tube, with none of the light exiting to the surrounding medium. Which of the following must be true?
    θincident ≥ θcritical

    For a light ray to totally internally reflect, rather than exit and refract (bend), the light ray must strike the edge of the glass tube at an angle equal or greater than the critical angle.


    An important special case occurs when light moves into a medium with a smaller index of refraction (that is, when n2 > n1). A classic example of this is when light is moving from water to air. As this happens, the angle θ with the normal will increase—in other words, the ray of light will bend further away from the normal. As the angle of the incident ray (θ1) increases, there will come a point where the angle of the refracted ray (θ2) reaches 90°. This is known as the critical angle. If we increase the angle beyond the critical angle, the light can no longer refract at all. Instead, all the light rays are reflected within the original medium. This is known as total internal reflection.
  28. Besides directly killing bacteria, how does pasteurization with hot water treat IPB populations?
    By decreasing the solubility of oxygen, preventing further bacterial growth 

    Oxygen, like all gases, is more soluble in cold water than warm water. Thus, the process of heating water (pasteurization) would lower the solubility of oxygen and reduce the amount available to IPB.
  29. In locations with very low dissolved O2 concentrations in drinking water, the observed Ksp of ferrous (II) hydroxide will be:
    the same.

     Equilibrium constants, including Ksp, are not affected by concentration. These constants are only altered by changing temperature, as shown below for a variety of salts
  30. _____ bond stretching typically exhibits a frequency range of 1650-1750 cm-1.




      _____ bond stretching in a carboxylic acid typically exhibits a frequency range of 2500-3300 cm-1.



     _____ bond stretching typically exhibits a frequency range of 2100-2260 cm-1.



    ______  bond stretching typically exhibits a frequency range of 1000-1320 cm-1.

     Diatomic molecules (e.g. O2, N2, and Br2) do not return IR signals because no net change in the dipole moment occurs.
    C=O

    O–H

    C≡C

    C–O

     Diatomic molecules (e.g. O2, N2, and Br2) do not return IR signals because no net change in the dipole moment occurs.
  31. If a hydrogen NMR spectrum were obtained for a double-stranded amidine dimer, would the amide N-H proton resonances experience any shielding effects?
    Yes, they would exhibit downfield shifts due to deshielding effects. 

     If hydrogen bonds are present, the electron density around hydrogen atoms bonded to nitrogen atoms will shrink, as these electron clouds would tend to surround nitrogen more than hydrogen
  32. ______ isomers have the same molecular formula but differ in their connectivity, as exhibited by L1 and L2.




    ______ are chiral mirror images of each other. They have the same atomic connectivity. 



    _______, or stereoisomers that are not enantiomes, also have the same connectivity. Differ in one or more chiral carbons but not all.




    _______ isomers are another form of stereoisomer. A classic example is the "cis" and "trans" configurations around a double bond, which does not apply here.
    Constitutional

     Enantiomers

    Diastereomers

    Geometric
  33. Structural isomers describe the different ways that the atoms in a compound can be connected. The three main subcategories of structural isomers are chain isomers, functional isomers, and positional isomers. Chain isomers have different arrangements of the carbon ‘skeleton.’ Functional isomers are isomers where the molecular formula remains the same, but the type of functional group in the atom is changed. For example, a compound with an oxygen atom in addition to several carbon atoms and the corresponding number of hydrogens could be an alcohol with an –OH group, or an ether with a C–O–C group. Positional isomers have a given functional group in different locations (e.g., 1-pentanol vs. 2-pentanol).




    Stereoisomers involve different ways that substituents can be positioned. Cis-trans isomerism on the MCAT most frequently involves alkene bonds. Rotation about pi bonds is restricted, meaning that two different groups of atoms attached to each carbon of the C=C bond can be arranged in different ways to give different molecules. These atoms or groups can be given priorities, with atoms with higher atomic numbers given higher priorities. If the highest-priority groups for each carbon are on the same side of the molecule, that molecule is denoted as the cis or Z isomer. If they are on opposite sites, the isomer is trans or the E isomer. Cis-trans isomers generally have similar chemical but different physical properties.




    Optical isomers come in pairs and typically contain one or more chiral centers (a C atom bonded to four unique substituents). These substituents are arranged differently around the chiral carbon, in such a way that the molecule cannot be rotated to make the two arrangements match. If all of the chiral centers have different configurations, the isomers are non-superimposable mirror images (similar to your hands), known as enantiomers. Enantiomers have identical chemical and physical properties, but they differ in their rotation of plane-polarized light (opposite directions) as well as the products they yield when reacted with another chiral reagent. If the isomers have multiple chiral centers but differ only at some, they are known as diastereomers. Since diastereomers have very similar chemical, but less similar physical properties, the MCAT can drop subtle clues about how molecules were isolated and/or separated as indicators of whether compounds are enantiomers or diastereomers.




    Chiral carbons can be described as R or S. The groups around the carbon are given priorities, with the lowest-priority group oriented away from the observer. If the priority of the remaining groups decreases (1 → 2 → 3) in a counterclockwise direction, it is an S chiral center, while if priority decreases (1 → 2 → 3) as you move in a clockwise direction, it is an R chiral center.
    Structural isomers describe the different ways that the atoms in a compound can be connected. The three main subcategories of structural isomers are chain isomers, functional isomers, and positional isomers. Chain isomers have different arrangements of the carbon ‘skeleton.’ Functional isomers are isomers where the molecular formula remains the same, but the type of functional group in the atom is changed. For example, a compound with an oxygen atom in addition to several carbon atoms and the corresponding number of hydrogens could be an alcohol with an –OH group, or an ether with a C–O–C group. Positional isomers have a given functional group in different locations (e.g., 1-pentanol vs. 2-pentanol).




    Stereoisomers involve different ways that substituents can be positioned. Cis-trans isomerism on the MCAT most frequently involves alkene bonds. Rotation about pi bonds is restricted, meaning that two different groups of atoms attached to each carbon of the C=C bond can be arranged in different ways to give different molecules. These atoms or groups can be given priorities, with atoms with higher atomic numbers given higher priorities. If the highest-priority groups for each carbon are on the same side of the molecule, that molecule is denoted as the cis or Z isomer. If they are on opposite sites, the isomer is trans or the E isomer. Cis-trans isomers generally have similar chemical but different physical properties.




    Optical isomers come in pairs and typically contain one or more chiral centers (a C atom bonded to four unique substituents). These substituents are arranged differently around the chiral carbon, in such a way that the molecule cannot be rotated to make the two arrangements match. If all of the chiral centers have different configurations, the isomers are non-superimposable mirror images (similar to your hands), known as enantiomers. Enantiomers have identical chemical and physical properties, but they differ in their rotation of plane-polarized light (opposite directions) as well as the products they yield when reacted with another chiral reagent. If the isomers have multiple chiral centers but differ only at some, they are known as diastereomers. Since diastereomers have very similar chemical, but less similar physical properties, the MCAT can drop subtle clues about how molecules were isolated and/or separated as indicators of whether compounds are enantiomers or diastereomers.




    Chiral carbons can be described as R or S. The groups around the carbon are given priorities, with the lowest-priority group oriented away from the observer. If the priority of the remaining groups decreases (1 → 2 → 3) in a counterclockwise direction, it is an S chiral center, while if priority decreases (1 → 2 → 3) as you move in a clockwise direction, it is an R chiral center.
  34. A pure sample of (R)-limonene has a specific rotation of +125.6. If a mixture of (R)-limonene and (S)-limonene has a specific rotation of +62.8, what are the percentages of the R and S enantiomers in this mixture?

    75% R, 25% S



    Since the enantiomers in this case must be present in unequal proportions, a weighted average of the components should be used. In this case, the weighted average is given; therefore, we can set up an equation. If we designate x as the percentage of R and 1 - x as the percentage of S, we can solve for x using this equation:


    (125.6)(x) + (-125.6)(1-x) = 62.8
    125.6x - 125.6 + 125.6x = 62.8
    251.2x = 188.4
    x = 0.75
    1 - x = 0.25 

    Therefore, R makes up 75% of the solution, while S comprises the remaining 25%.
  35. Which of the following acids is expected to generate the strongest hydrogen bonding?

    A. HI

    B. HF

    C. HCl

    D. HBr
    B. HF

    Hydrogen bonding is a unique type of intermolecular force that occurs when H is bonded to very electronegative elements like F, O or N. Fluorine is the most electronegative element on the periodic table. Of the answer choices, only hydrogen fluoride is capable of hydrogen bonding.
  36. _______ (solid to liquid),
    ________ (liquid to gas),
    _______ (solid to gas),
    _______ (gas to liquid),
    _______ (liquid to solid), 
    _______ (gas to solid).

    3 endothermic phases ____

    3 exothermic phases _____
    melting or fusion

    evaporation

    sublimation

    condensation

    freezing


    deposition

    • melting, sublimation, evaporation 
    • deposition, condensation, freezing
  37. ______ processes involve the breaking of bonds or intermolecular interactions, which require a source of heat. In contrast, ______ processes typically involve bond formation or an increase in intermolecular force strength.

    But for ATP its different. 

    Forming ATP (ADP + Pi --> ATP) = (exo or endo)


    Breaking ATP ( ATP --- > ADP + Pi) endo or exo?
    Endothermic

    exothermic

    endothermic

    exothermic
  38. pKa + pKb = 14

    pH + pOH = 14

    Ka x Kb = 10^-14

    pKa = -logKa
    pKb = -logKb

    pH = -log[H+]
    pOH = -log[OH-]

    example: find pH if given [OH-] = 3.6 X 10^-11 M

    pOH = -log (OH-)
    pOH = -log (3.6 x 10^-11)
    pOH = 10.64
    14 - 10.64 = pH
    3.36 = pH

    Ka = 1.8 x 10^-5
    pKa = 4.82

    H+ = 1 X 10^-3
    H+ = 3 X 10^-4
    pH = -log H+
    pH = -log (1 x 10^-3)
    pH = 3
    -------
    pH = -log H+
    pH = -log (3 x 10^-4)
    pH = 3.5

    [H+] = 10^-pH
    [OH-] = 10^-pOH

    kw = [H+][OH-] = 10^-14
    pKa + pKb = 14

    pH + pOH = 14

    Ka x Kb = 10^-14

    • pKa = -logKa
    • pKb = -logKb

    • pH = -log[H+]
    • pOH = -log[OH-]

    example: find pH if given [OH-] = 3.6 X 10^-11 M

    • pOH = -log (OH-)
    • pOH = -log (3.6 x 10^-11)
    • pOH = 10.64
    • 14 - 10.64 = pH
    • 3.36 = pH

    • Ka = 1.8 x 10^-5
    • pKa = 4.82

    • H+ = 1 X 10^-3
    • H+ = 3 X 10^-4
    • pH = -log H+
    • pH = -log (1 x 10^-3)
    • pH = 3
    • -------
    • pH = -log H+
    • pH = -log (3 x 10^-4)
    • pH = 3.5

    • [H+] = 10^-pH
    • [OH-] = 10^-pOH

    kw = [H+][OH-] = 10^-14
  39. A patient presents in the emergency department having ingested a large quantity of tolbutamide. Intravenous administration which of the following compounds is most likely to increase the rate of urinary excretion of the drug?


    NaHCO3




    for weakly acidic drugs, the uncharged state is capable of diffusion through membranes much more than the charged form. Thus, to prevent reabsorption, we must maximize the charged form of tolbutamide. We can do this to a weak acid by deprotonating the drug (rendering it negatively charged) via administration of a base. This can will increase blood and urinary pH (i.e. decreasing [H+]) and increase the fraction of ionized drug present. Urinary alkalization can be accomplished by administration of a basic salt, like NaHCO3 (the salt of the conjugate base of carbonic acid).




     NaCl and KCl are salts that are very close to neutral (neither acidic nor basic). Because Cl- is the conjugate base of a strong acid (HCl), it will only be able to act as a base when pH is extremely low.




     ClO4- is also the conjugate base of a strong acid (HClO4), so it will also fail to act as a base under biological conditions. Additionally, NH4+ is acidic. Because of this, NH4ClO4 is acidic overall, not basic.
  40. The equivalence point of the titration of tolbutamide with NaOH was reached by adding 50 mL of NaOH. Which of the following correctly describes the solution during this process?  tolbutamide (C12H18N2O3S; pKa = 5.3)

    A. The solution had a pH less than 7.1 at the equivalence point.

    B. After addition of 25 mL of NaOH, the pH of the solution was greater than 5.

    C. The concentration of the charged form of tolbutamide was greater than neutral tolbutamide during the titration.

    D. The concentration of the charged form of tolbutamide was less than neutral tolbutamide during the titration.
    B. After addition of 25 mL of NaOH, the pH of the solution was greater than 5.



     If 50 mL of NaOH represents the volume of titrant required to reach the equivalence point, then 25 mL is the volume added at the half equivalence point. At this point, one half of the original tolbutamide present will have been converted to its conjugate base, and their concentrations will be equal (eliminate choices C and D). According to the Henderson-Hasselbach equation, when the values of protonated acid and conjugate base are equal, pH = pKa + log 1 = pKa + 0 = pKa. At the half-equivalence point, pH of solution equals the pKa of the analyte. According to paragraph 3, tolbutamide’s pKa is 5.3.





    A: The titration of a strong base and a weak acid results in the formation of a basic salt, the hydrolysis of which causes the pH at the equivalence point to be greater than 7.




    C, D: The relative concentrations of the charged versus neutral forms of tolbutamide changed throughout the titration.
  41. acid base strength

    low pKa =
    low pKb =
    high Ka =
    high Kb =

    The rates of diffusion of four drugs were tested: acetazolamide (pKa = 7.2), sulfadiazine (pKa = 6.5), warfarin (pKa = 5.0), and cephalexin (pKa = 3.6). Which drug will have the strongest conjugate base?

    A. Acetazolamide

    B. Sulfadiazine 

    C. Warfarin 

    D. Cephalexin
    • low pKa = strong acid
    • low pKb = strong base
    • high Ka = strong acid
    • high Kb = strong base

    A. Acetazolamide
  42. Ten moles of the monoprotic, weakly acidic medication aspirin were added to water to make one liter of solution. If the pH of the resulting solution was 5.9, what is the approximate Kb for the non-diffusible form of aspirin?

    A. 0.1

    B. 0.01 

    C. 0.001

    D. 1 



     Since we are given pH in the question stem, we will not be able to find Kb immediately. Instead, we need to calculate Ka and solve for Kb from that value. The Ka for the dissociation of a generic acid HA can be written as Ka = [H+][A-]/[HA], where all concentrations are measured at equilibrium. In the solution of aspirin described, the initial concentration of drug is 10 M. Since only a small amount of this weak acid will dissociate, this value is a good approximation for our final equilibrium [HA].




    Next, we must find the proton concentration. Remember, [H+] = 10-pH. Here, the pH of the solution is 5.9, so [H+] = 10-5.9 M ~ 10-6 M. Since each HA molecule dissociates into equal parts [H+] and [A-], our value for [A-] must be 10-6 M as well. Returning to the Ka expression, Ka = [(10-6 M )(10-6 M)] / (10 - 10-6 M).




    Remember, we can estimate that [HA] = 10 M, even though its true final value is 10 M - 10-6 M!
    [(10-6 M )( 10-6 M)] / (10 M) = 10-12 M / 10 M = 10-13 M


    In water at 25°C, Ka•Kb = 10-14. Given this, Kb = 10-14/Ka = 10 -14/10-13 = 10-1 = 0.1.
    [H+] = 10^-pH
  43. Closed system: Exchange heat but not mass.

    Open system: Exchange heat and mass.

    Isolated system: Exchange neither heat nor mass.

    What system is earth?

    What system is a human being?
    Earth satisfies both conditions of a closed system, namely that there is energy exchange but not mass exchange with the surroundings.

    It is an open system by the strict thermodynamic definition because it exchanges energy with its surroundings (heat and work), and material in the form of gases, food, waste products, water.
  44. Why does the energy flux for each system in Table 1 always sum to 0?
    The energy being absorbed by the system is equal to the energy being released by the system.
  45. What is the primary process responsible for the loss of latent heat and entropy from the ocean at the air-sea interface in Table 1? 

    A. Precipitation 

    B. Condensation 

    C. Evaporation 

    D. Melting
    C is correct. A negative latent heat implies that the phase change that is occurring is consuming energy. Therefore, the phase change that is occurring is melting, evaporation or sublimation. The fact that this process is occurring at the air sea interface means that the phase change must involve the gas phase. That fact rules out melting. Since sublimation is not an option, will go with evaporation. It makes sense that with evaporation, the ocean is losing entropy at the expense of the atmosphere which is gaining gas particles.
  46. SN1 reactions proceed faster in protic solvents. SN2 reactions prefer polar aprotic solvents.

    So SN1 works good for propanol
    SN1 reactions proceed faster in protic solvents. SN2 reactions prefer polar aprotic solvents.

    So SN1 works good for propanol
  47. as frequency increases stiffness _____

    as frequency increases mass _____
    increases

    decreases
  48. _______ myelinate nerves in the central nervous system, which increases nerve conduction velocity.

    ______ myelinate nerves in the peripheral nervous system, which increases nerve conduction velocity.
    Oligodendrocytes

    Schwann cells
  49. capacitance = C/V (Charge in coulombs / voltage)

    The capacitance of a nerve membrane can be increased by:

    A. decreasing the width of the membrane.

    B. decreasing the surface area of the membrane.

    C. decreasing the charge stored across the membrane.

    D. increasing the voltage difference across the membrane.
    A is correct. Remember that capacitance is coulombs per volt. So decreasing the charge (coulombs) would decrease the capacitance. That lets us eliminate choice C. Increasing the voltage would also decrease the capacitance, letting us eliminate choice D. For the MCAT, you should also remember that capacitance is directly proportional to area but inversely proportional to the distance between the two sides of the capacitor. Thus, decreasing the width of the membrane (choice A) would actually increase capacitance.
  50. specific gravity = density of object / density of water

    density of water 1000 kg/m^3

    1m^3 = 1000L
    specific gravity = density of object / density of water

    density of water 1000 kg/m^3

    1m^3 = 1000L
  51. The pI of any amino acid can be calculated as the average of the two most relevant pKa values.

    If the amino acid is basic indicated by two pKa that have a basic pH (more than 7) and just one with an acidic then to get pI you need to get the average of the two highest pKa

    If the amino acid is acidic indicated by two pKa that have a acidic pH (less than 7) and just one with a basic pH then to get pI you need to get the average of the two smallest pKa to get pI
    The pI of any amino acid can be calculated as the average of the two most relevant pKa values.

    If the amino acid is basic indicated by two pKa that have a basic pH and just one with an acidic then to get pI you need to get the average of the two highest pKa

    If the amino acid is acidic indicated by two pKa that have a acidic pH and just one with a basic pH then to get pI you need to get the average of the two smallest pKa to get pI
  52. CO2 + H2O <----> H2CO3 <--> HCO3- + H+

    HCO3- (Bicarbonate)

    H2CO3 (Carbonic acid)

    For example, hyperventilation (rapid shallow breathing) results in excess CO2 being expelled from the blood, causing the pH to rise.

    As H+ increases HCO3- decreases and CO2/H2O increase.
    CO2 + H2O <----> H2CO3 <--> HCO3- + H+

    HCO3- (Bicarbonate)

    H2CO3 (Carbonic acid)

    For example, hyperventilation (rapid shallow breathing) results in excess CO2 being expelled from the blood, causing the pH to rise.

    As H+ increases HCO3- decreases and CO2/H2O increase.
  53. Impulse = Force x time
    Impulse = Force x time
  54. Thin-layer chromotagraphy (TLC)

    3 causes for TLC plate streak
    TLC, streaking can be caused by overloading the spot with sample

     by using a sample that is too concentrated.




    •  If a TLC sample is impure, its multiple components will travel different distances along the plate, which could lead to a streaked appearance.
    • --------------------------
    • Chromatography is used to separate, identify, and purify the components of a mixture.

    Although there are many chromatography techniques, they share the principle that the molecules in a mixture are applied onto a stationary phase (usually a solid), while a fluid known as the mobile phase (generally a solvent chosen to match the target molecules, e.g. polar or nonpolar) containing the molecules of interest travels through the stationary phase. Molecules of interest in the mobile phase will interact with the stationary phase with different levels of intensity. Molecules that interact more strongly with the stationary phase will take longer to pass through it, whereas molecules that interact more weakly with the stationary phase will pass through it more quickly. Common factors that shape these interactions include molecular characteristics related to adsorption, polarity- or charge-based affinity for the stationary/mobile phase, and differences in molecular weight. A chiral stationary phase can also be used to separate stereoisomers based on the principle that the various enantiomers of a compound may interact differently with such a stationary phase.
  55. 1 atm = 760 mmHg = 760 Torr = 10^5 Pa (N/m2)

    TK = TC + 273

    Volume can be expressed as any unit of distance cubed or as any standard unit of volume, but the most common units are mL, L, or cm3, where 1 L = 103 mL = 103 cm3.

     PV = nRT
    1 atm = 760 mmHg = 760 Torr = 10^5 Pa (N/m2)

    TK = TC + 273

    Volume can be expressed as any unit of distance cubed or as any standard unit of volume, but the most common units are mL, L, or cm3, where 1 L = 103 mL = 103 cm3.

     PV = nRT
  56. How much does the rate of effusion of dichloromethane at 200 K change when the temperature is changed to 800 K? Assume all other conditions are identical. 


    800k to 200k
    square root of rate 2 over rate 1
    square root of 800/ 200
    square root of 4 = 2
    How much does the rate of effusion of dichloromethane at 200 K change when the temperature is changed to 800 K? Assume all other conditions are identical. 


    • 800k to 200k
    • square root of rate 2 over rate 1
    • square root of 800/ 200
    • square root of 4 = 2
  57. _______ is the amount of energy needed for the ______ (melting) of 1 unit of mass. The substance with the ____ latent heat of fusion requires the least amount of energy per unit mass for melting.
    Latent heat of fusion

    fusion

    lowest
Author
JAM41MAN
ID
340926
Card Set
embryology and mitosis meiosis + science 3
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em
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