8/31 Quiz 1 Practice Problems Flashcards
A 7-year old boy presented for a well child checkup. His height is 115 cm and weight is 27 kg. His BMI and weight characterization are:
20.4 kg/m2, obese
A 20-year old woman and her 52-year old mother both have BMI values of 24 and have identical heights. Even though they have similar levels of physical activity, the mother has a lower metabolic rate. This difference is most likely caused by the mother’s:
lower fat-free mass
The image below depicts the overall body composition of a healthy adult. Based on this information, which of the following 4 elements make up ~95% of the human body?
Carbon, Hydrogen, Nitrogen, Oxygen
When hospitalized patients need nutrition support, their resting energy expenditure (REE) is calculated to ensure that they are provided appropriate numbers of calories. For obese patients, the REE calculation should include an adjusted body weight instead of true body weight because:
Adipose is metabolically less active than lean tissue
A 53-year old woman with a BMI of 27 and a waist circumference of 89 cm is considered at high risk for hypertension, cardiovascular disease and type 2 diabetes. One reason is that waist circumference is correlated with:
Visceral (abdominal, surrounding the central organs) fat
Which of the following people would likely be in positive nitrogen balance?
23 year old pregnant woman.
The fat that we store in adipose tissue for future energy is primarily composed of:
triacylglycerol
Which of the following is NOT catabolized through various metabolic pathways and eventually the TCA (a.k.a. Kreb’s) cycle to ultimately produce ATP to fuel the body’s energy needs?
cholesterol (amino acids, fatty acids, glycogen are)
In a healthy adult, which fuel source has the smallest number of stored calories?
glycogen
- Embryonic stem cells are pluripotent. Discuss what pluripotent cells produce in the following situations:
a. When cultured in the laboratory
b. When transplanted to a morula stage embryo (in an animal model)
c. When transplanted to an animal embryo at a stage equivalent to 8 week human embryo
d. When transplanted to an adult animal
Discussion: Embryonic stem cells are pluripotent, which means they are capable of producing multiple adult tissues, but do not retain the ability to make a full organism. In the laboratory, pluripotent cells are able to produce progeny with characteristics of cells derived from of all three germ layers (endoderm, mesoderm and ectoderm). If transplanted to early (morula) stage embryos, they will participate in normal development. In mature tissues, including late stage embryos or adults, they produce tumors containing multiple types of differentiated cells (i.e. teratomas).
- Factors from sperm (likely including PLCζ) are required for the maternal nucleus of the zygote to complete meiosis II. Discuss why this is important and what would be the likely outcome for the embryo if meiosis II did not occur
Discussion: The primary function of meiosis is to reduce chromosome number to one half the normal complement. This is why meiotic divisions are often referred to as “reduction” divisions. If meiosis II does not occur in the egg-derived zygotic nucleus, the embryo will persist in a triploid state, with a preponderance of maternal chromosomes. Most cells are viable in the triploid state (i.e. triploidy does not interfere with normal cellular functions), but this condition is not tolerated during development, in part due to genomic imprinting. A preponderance of maternal chromosomes will create an imbalance between male and female imprinted genes. In humans, triploidy is nearly always fatal in early embryogenesis.
- What is the karyotype and potency of primordial germ cells? Discuss the role of these cells in development and what is the likely outcome for the embryo if they migrate to locations other than the genital ridge.
Discussion: Primordial germ cells (PGCs) are diploid cells that migrate through the hindgut to populate the genital ridge in early development. In normal development, they will give rise to oogonia or spermatogonia (depending on whether the gonadal ridge expresses SRY or not). These cells are often considered “totipotent” because they produce gametes that subsequently give rise to the next generation. However, on their own, PGCs are not able to produce an organized body plan (i.e. development only takes place after meiotic reduction and fertilization). In ectopic locations, these cells produce disordered masses containing multiple cell types, and are therefore more accurately viewed as pluripotent.
- Choriocarcinoma is a relatively rare metastatic cancer that has characteristics similar to placental/chorionic tissue. Review the characteristics of this cancer, including the typical karyotype of the tumor, and discuss the likely origin of this form or cancer.
Discussion: Choriocarcinomas are commonly believed to be malignant transformation of tissue derived from a molar pregnancy. Incomplete moles result from abnormal fertilization of an egg by two sperm, and have a triploid genome, with two copies of paternally imprinted chromosomes. Complete moles are diploid tumors containing only paternal chromosomes. They result either from fertilization of an egg that has lost its nucleus, followed by duplication of the paternal DNA, or fertilization of an egg that has lost its nucleus by two sperm. These tumors expand rapidly, produce hCG and are highly invasive (mirroring the normal properties of trophoblast tissue). Cancers derived from such tumors are very aggressive and have a poor prognosis. Having had a molar pregnancy is a risk factor for choriocarcinoma.
- Review the organization of fetal membranes and placenta in monozygotic and dizygotic twinning. In cases where twins have separate placentas, chorionic and amniotic cavities, can the zygocity of twins (monozygotic or dizygotic) be determined?
Discussion: The extent to which membranes and placenta are shared depends on two factors; when the splitting occurred (for monozygotic twins) and how closely the twins implant in the uterus. Twins sharing a single placenta may be either monozygotic or dizygotic, if the placenta fused early in pregnancy. Twins with a common amniotic sac are almost always monozygotic, although (rarely) the amniochorionic mebranes of dizygotic twins that are intimately in contact can fuse, resulting in a single amniotic sac for dizygotic twins.
- In the mature state, the uterine-facing side of the placenta is encased in a layer of cytotrophoblast and the fetus-facing side is lined by the chorionic plate. Discuss how this arrangement comes about.
Discussion: At early stages of development, the chorionic membrane consists of three layers (moving from the uterus, inward to the fetus): the syncytiotrophoblast, the cytotrophoblast and extraembryonic somatic mesoderm (see slide 16 from lecture). The entire chorion initially produces villi, and cytotrophoblast cells migrate out to produce a “shell” of cytotrophoblast between the uterine endometrium and the outer ends of the villi. This puts cytotrophoblast both on the surface and deeper in the placenta (see slide 22 from the lecture). The outer layer of cytotrophoblast protects the placenta and helps prevent shedding of fetal cells into maternal tissues. On the fetal side, the chorionic membrane ultimately fuses with the amniotic membrane (see slides 23, 24), resulting in a structure known as the chorioamniotic membrane or (in the region of the placenta) the placental plate, consisting of (from the placenta moving in towards the fetus): syncytiotrophoblast, the cytotrophoblast and extraembryonic somatic mesoderm (that includes fetal vessels), amniotic epithelium (derived from the epiblast).
identify the images of the chest for sagittal coronal and axial sections
Sagittal is side cut, coronal is front on cut, and axial is top down view.
After body folding has produced cavities internal to the body (the coelomic cavities), they are divided into separate cavities for the lungs, heart and abdominal organs. Review how the diaphragm is formed to separate the thoracic organs from the abdominal organs. In rare cases (
The diaphragm forms from the transverse septum fusing to two protrusions from the posterior body wall (the pluroperitoneal folds). Later, there is a muscular ingrowth from the posterior body wall. Incomplete fusion results in hernias, that allow abdominal contents to move into the plural cavity, suppressing normal lung growth and branching. Hernias are more common on the left side, because the liver stabilizes the diaphragm on the right side. Antenatal repair of diaphragmatic hernias greatly improves outcome by reducing lung hypoplasia at birth.
Sirenomelia is characterized by fusion of the lower extremities, deformities in the lumbar spine, bilateral renal agenesis and severe malformation of the caudal viscera. Discuss how these defects could be caused by a failure of gastrulation in the caudal region of the primitive streak. (This will require you to learn something about kidney and limb formation that have not yet been presented).
The defects of sirenomelia are all consistent with a deficiency in mesoderm production. Limbs form largely from lateral plate mesoderm, kidneys from intermediate mesoderm and the spinal vertebrae from somites. When insufficient mesoderm is produced during gastrulation and/or this mesoderm fails to migrate to lateral positions, production of these later structures is reduced and/or collapsed towards the midline, resulting in fusion or hypoplasia of mesodermally derived structures
The drug cyclopamine was discovered by researchers in Logan, UT in 1968, based on the increase in cyclopic sheep born to ewes that had grazed on the plant, Veratrum californicum (see image). Based on what you know about holoprosencephaly, discuss the likely biologic function of cyclopamine.
Holoprosencephaly is most frequently caused by disruption of the Sonic Hedgehog (Shh) pathway (see lecture slide 15). Shh plays important roles in patterning the midline of the body, including the nervous system. When Shh signaling is reduced, lateral structures contract towards the midline, resulting in midline fusion of normally bilateral structures. The teratogenic agent in V. californicum (later named cyclopamine) inhibits Shh signaling, by direct binding to a component of the Shh signaling pathway, and is widely used in developmental biology research for this purpose.
you measure the flux of two different drugs across a cell membrane and obtain a graph of diffusion rate vs. concentration for both of the athat is simply linear, and directly related to the concentration gradient, 1 is steeper than 2. What can we conclude
the membrane permeability of drug 1 iss much greater than drug 2 and they are both simple diffusion.
the osmolarity of the intracellular fluid of a heart cell is 300mOsm/liter. if the osmolarity of the intersitiual fluid suddenly decreaseds to 200 Osm/liter which of the following is the most likely to occure
water concentration in the interstitual fluid is greater than in the intracellular fluid thus water will diffuse into the cell and the cell volume will increase.
which of the following characteristics are shared by simple and facilitated diffusion of glucose?
Occurs down an electrochemical gradient
accidently infused with a large volumes of solution that cuases lysis of red blood cells. what could account for this? normal plasma and intracellular osmolarity is 300 mOsm.
look for a hypotonic solution to be added. remember ions break up and multiply the conc. to get the mOsm.
what would occur if block the Na-K pump
the intracellular Sodium would increase, and the potassium would decrease. It will also decrease the efflux of any pumps reliant on the Na gradient.
Which of the following best describes the changes in cell volume that will occur when red blood cells (previously equilibrated in a 280-milliosmolar solution of NaCl) are placed in a solution of 140 millimolar NaCl containing 20 millimolar urea, a relatively large but permeant molecule?
Cells shrink initially, then swell over time and lyse
Cells shrink transiently and return to their original volume over time
C) Cells swell and lyseD) Cells swell transiently and return to their original volume over time
E) No change in cell volume will occur
B. A solution of 140 millimolar NaCl has an osmolarity of 280 milliosmoles, which is iso-osmotic relative to “normal” intracellular osmolarity. If red blood cells were placed in 140 millimolar NaCl alone, there would be no change in cell volume because intracellular and extracellular osmolarities are equal. The presence of 20 millimolar urea, however, increases the solution’s osmolarity and makes it hypertonic relative to the intracellular solution. Water will initially move out of the cell, but because the plasma membrane is permeable to urea, urea will diffuse into the cell and equilibrate across the plasma membrane. As a result, water will re-enter the cell, and the cell will return to its original volume.
. Assume you are studying a cell (nerve, heart, skeletal, etc.) with an intracellular K+ concentration of 130 mM and an interstitial K+ concentration of 5 mM. Also assume the cell membrane contains only potassium ion channels that are only permeable to potassium ions. Which of the following is correct regarding the resting membrane potential (Vm) and the potassium equilibrium potential (VK).
A. Vm will be –85 mV but VK will be -75mV.
B. Both Vm and VK will be –60 mV.
C. Both Vm and VK will be –85 mV.
D. Vm will be –100 mV but VK will be -85mV.
Answer C. . If the membrane is only permeable to potassium then Vm will be equal to the potassium equilibrium potential (VK). Nearly all cells in the body have a membrane potential (Vm), inside negative. Resting membrane potential is negative because the resting cell membrane is highly permeable to K (ie, K channels are open) and there are impermeable anions inside the cell. K ions diffuse out of the cell down their concentration gradient leaving behind the impermeable anion. This creates a negative internal potential. In this setting, the ratio of K concentrations (K outside/ K inside) is the major determinant of membrane potential and can be calculated from the Nernst equation (below) which you should remember. If the membrane is only permeable to potassium then Vm will be equal to the potassium equilibrium potential (VK). However, most cells have a small sodium leak which makes the actual Vm less negative than VK. You can think of VK as the predicted Vm if there is no sodium leak. However, in most cells the Nernst equation gives a pretty good approximation of the actual Vm.
You are monitoring resting membrane potential (Vm) in a cell and notice that it has gone from -90mV to -70mV. Which of the following will cause this effect?
A. Intracellular potassium concentration, [K+]i, has increased.
B. The interstitial concentration of potassium, [K+]o, has decreased.
C. The cell has been exposed to a drug that increases the resting inward leak of Na ions.
D. The cell has been exposed to a drug that stimulates the Na-K pump.
Answer C. The inward Na leak that occurs in resting cells will make Vm less negative (ie, will depolarize Vm). Choice A is incorrect because a rise in [K+]i will make Vm more negative, ie. Vm will hyperpolarize. Another way to think about this is that Vm is the internal negativity required to maintain the K concentration gradient across the membrane (high inside, low outside). Thus the higher the intracellular K concentration the more negative Vm needs to be. Similarly, choice B is incorrect since a decrease in [K+]o will make Vm more negative. Choice D is incorrect since stimulation of the Na-K pump will increase [K+]i and thus make Vm more neagative.
- The action potential and ionic currents from a cell are shown in the figure. Choose the correct answer as related to this figure.
A. An increase in in the current labeled A with epinephrine will trigger greater release of calcium from the sarcoplasmic reticulum (SR).
B. Application of a drug that selectively blocks calcium currents will block the current labeled C.
C. The action potential and currents shown in the figure are from a skeletal muscle cell.
D. Application of a drug that selectively blocks sodium current will block the current labeled A.
Answer A. These signals are from a ventriculer myocyte. The inward current labeled A is the calcium current (ICa). It does two important things: 1) helps maintain the long duration of the action potential, 2) Ca influx via ICa triggers a large release from sarcoplasmic reticulum (SR) which produces the Ca transient. Epinephrine and norepinephrine increase ICa causing greater SR Ca release. Choice B is incorrect because this current is a potassium current (IK1) and blocking ICa will not affect IK1. Just looking at the shape and long duration of the action potential (AP) indicates this is not a skeletal muscle cell (choice C). APs in skeletal muscle look like those in nerve and are very brief in duration. D is incorrect because it is a calcium current and not a sodium current.
. You are recording action potentials from two sites, separated by 1 cm, on the outer surface (epicardium) of the left ventricle. You find that it takes 10 msec for action potentials to propagate between the sites. What is the action potential conduction velocity in meters/sec and which of the following would cause it to decrease?
A. The conduction velocity is 10 meters/sec and could be decreased by decreasing sodium current.
B. The conduction velocity is 1 meter/sec and could be decreased by increasing sodium current.
C. The conduction velocity is 1 meter/sec and could be decreased shortening action potential duration.
D. The conduction velocity is 1 meter/sec and could be decreased by closure of gap junctions.
Answer D. Action potential conduction velocity is distance/time. In this case the wavefront travels 0.01 meters in 0.01 se giving a conduction velocity of 1 meter/sec. Closure of cardiac gap junctions reduces the local circuit current flowing out in front of the action potential and will slow conduction velocity. Choince A is incorrect because velocity is 1 meter/sec and not 10 meter/sec. Choice B is incorrect because increasing sodium current will also increase conduction velocity. Choice C is incorrect because AP duration will not affect conduction velocity.
