Answer:
1. Functions:
a. Umbilical artery >> carries deoxygenated blood from the fetus to the placenta
b. Umbilical vein >> transports oxygenated blood from the placenta to the fetus
c. Ductus venosus >> allows oxygenated blood from the placenta to bypass the liver
d. Ductus arterious >> allows most of the blood from the right ventricle to bypass the fetus's non-functioning lungs
e. Foramen ovale >> oxygenated blood from the umbilical vein to bypass the pulmonary circulation
2. After the bird:
1. Umbilical artery >> medial umbilical ligament
2. Umbilical vein >> round ligament of the liver
3. Ductus venosus >> ligamentum venosum
4. Ductus arteriosus >> ligamentum arteriosum
5. Foramen ovale >> fossa ovalis
Explanation:
The umbilical artery is a paired artery localized in the abdominal and pelvic regions, which carries deoxygenated blood from the fetus to the placenta through the umbilical cord. The medial umbilical ligament is the obliterated part of the umbilical artery that arises from the internal iliac arteries. In utero, the umbilical arteries carry waste products back to the placenta, whereas the umbilical vein carries oxygenated blood from the placenta to the fetus. The round ligament of the liver (also known as ligamentum teres hepatis) is a remnant of the umbilical vein that exists in the embryonic stage, it connects the left lobe of the liver to the umbilicus. The ductus venosus is a slender shunt that allows oxygenated blood from the placenta to bypass the liver, it connects the intra-hepatic portion of the umbilical vein to the inferior vena cava. The ligamentum venosum is an extrahepatic, slender, and fibrous remnant of the fetal ductus venosus that travels between the left portal vein and the inferior vena cava. The ductus arteriosus is a fetal artery that connects the aorta to the pulmonary artery. The ligamentum arteriosum is a nonfunctional vestige of the ductus arteriosus, it is attached to the superior surface of the pulmonary trunk. The foramen ovale is an oval-shaped, small, opening in the wall (septum) between the two upper chambers of the heart. The fossa ovalis is a vestige stricture of the foramen ovale of the embryonic heart, which forms a depression in the right atrium of the heart.
In sweet peas, the genes for flower color and pollen grain shape are 11 cM apart. A pure-breeding purple flowering plant with round pollen grains is crossed to a pure-breeding red flowering plant with long pollen grains. The resulting F1 offspring are all purple flowering plants with long pollen grains. What percent of offspring from a test cross analysis of the F1 individuals would you expect to be purple flowering plants with long pollen grains
Answer:
The correct answer is - 44.5%.
Explanation:
In this question, there are two traits are given, flower color and pollen shape. The purple flowers and long pollen are wild-type (dominant). Red flowers and round pollen are mutant or recessive. The cross is between a pure breeding purple and round pollen plant with red and long pollen plant. The offspring are all purple and long pollen plants. However, the offspring would be all heterozygotes genotype for both traits.
Linkage can be calculated as -
Linkage distance = % of recombinant offsprings formed in a test cross.
parental types are purple and long pollen, red and round pollen.
So, the recombinant type would be purple and round pollen, red and long pollen.
11% of the offspring would be recombinant type as the distance is 11cM. Therefore, 89% of the progeny would be parental type. Which means this 89% of parental offsprings have equal proportions of both phenotypes of parents. So, the purple flowers and long pollen would be half of 89%.
that is 89/2 = 44.5%.
Igneous rocks turn into sediments through the process of ______.
A . Erosion
B.crystallization
C.melting
D. Deposition
Guinea pigs can produce different coat colors, white, brown, and mixed white/brown (agouti). The allele B is codominant with allele W (white coat). Cross an agouti-colored Guinea pig with a brown Guinea pig. What are the odds of the offspring?
Answer:
Brown
Explanation:because the agouti itself is a mixture that remains of a pale brown color and the melanin content in the animals is in the darker colors so it is more likely that the brown color is predominant to the agouti due to the content of melanin that it
The descendants of a cross between an agouti heterozygous (AB) colored Guinea pig and a brown (BB) colored Guinea pig are half heterozygous agouti, half homozygous brown. The correct option is A.
What exactly is codominance?Codominance is a genetic effect that happens whenever the alleles are expressed in heterozygotes.
In this particular instance, allele B co-dominates with allele W, resulting in heterozygous (WB) colored Guinea pigs like one of the parents.
Thus, the correct option is A.
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The missing options of the question are:
A. 1/2 heterozygous brown, 1/2 homozygous agouti.B. 1/2 homozygous white, 1/2 heterozygous brown.C. 1/4 homozygous brown.D. 1/2 heterozygous agouti, 1/4 white.Fill in the bottom line of DNA with the complementary base pair:
Answer: ACCTGATCGTAGCT
Explanation:
Keep in mind that in DNA replication, the matching base pairs are A and T (adenine and thymine) and C and G (Cytosine and Guanine). This means that for DNA complementary strands, A's can ONLY bind with T's and C's can only bind with G's , and vice versa. So, your answer for the picture you have given is ACCTGATCGTAGCT
In the cell bio lab, we use company manufactured gels, however you can make you own polyacrylamide gels. List all of the ingredients found in an SDS-PAGE gel. Which ingredients are responsible for polymerizing the solution?
Answer:
The list of ingredients are used in this method are as follow-
1. Acrylamide - mostly 30%
2. Bisacrylamide - mostly 0.8%
3. 1.5M Tris-Hcl pH 6.8 for stacking gel and 8.8 for separating gel.
4. 10% SDS
5. 10% APS
6. TEMED
7. Double distilled water
Polymerization:
Acrylamide is a polymer that is used to dissolved in buffer and bisacrylamide is the cross linking agen.
Two types of gels are prepared for separation of protein. Stacking gel is usually 4 or 6% and separating gel is usually 10 to 20%. Lower concentration of stacking gel will allow the migration of all the protein same because of larger pore size. It will stack or line up the proteins for its next level of separation.
Higher concentration of separating gel will provide resistance to the movement of protein because of the low pore size and therefore linear bands can be achieved neatly.
is mucous an enzyme or harmone
Answer:
It's an Enzyme.
Explanation:
It is a viscous colloid containing inorganic salts, antimicrobial enzymes (such as lysozymes), immunoglobulins (especially IgA), and glycoproteins such as lactoferrin and mucins, which are produced by goblet cells in the mucous membranes and submucosal glands.
Which organelle is labeled A?
Wheres the cell membrane located?
Answer:
in cells the cell membrain is located in the middel but this can vary in diffrent cells Explanation:
The endogenous cholesterol synthesis by the liver that we need to function normally it can lead to
plaque formation within the blood vessels.
Оа.
False
b. True
Answer:
It should be true
Explanation:
online health class and quick double check
In an experiment, a small dialysis bag is filled with a 20% salt solution. It is placed in a
beaker filled with a 40% salt solution. Assuming that water can pass through the
small pores of the dialysis bag, whereas the large salt molecules cannot, what will
happen to the size of the dialysis bag?
Answer:
The bag will get smaller
Explanation:
Osmosis can be defined as the process of diffusion or movement of water molecules across a semipermeable membrane from an area of higher concentration of water (i.e., lower concentration solution) to a lower concentration of water. For example, in cells, there are specialized pores called 'aquaporins' which are membrane proteins that form channels to transport water molecules by facilitated diffusion. In this case, the outside of the bag has a higher salt concentration than inside, thereby water molecules will move by facilitated diffusion through pores from inside to outside the bag.
What is homeostasis? What are some of the things my body does to keep it at the right temperature?
Answer:
Blood flow to your skin increases to speed up heat loss into your surroundings, and you might also start sweating so the evaporation of sweat from your skin can help you cool off.As heat is lost to the environment, the body temperature returns to normal.
Imagine you found S. aureus to be resistant to Penicillin by Kirby Bauer analysis, but susceptible to Penicillin treatment in liquid culture (in other words, a MIC was determined). Which of the following are possible explanations for this inconsistency?
a. The concentration of Penicillin was higher in the Penicillin antibiotic disk than that used in liquid culture treatment.
b. The Penicillin disks used for Kirby Bauer analysis were expired/no longer active.
c. The concentration of bacteria was lower on the Mueller Hinton plate for Kirby Bauer analysis than that used in liquid culture treatment
Which of the following describes a predator?
A a fish that is killed and eaten.
B a bear that kills and eats fish
C a worm that lives inside a bear
D a bear that has a worms in its gut.
Answer:
b
Explanation:
a predator is a bear that kills and eats fish
Answer:
B. bear that kills and eat fish
Which level of organization includes all the other levels of organization?
answer : biosphere
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Answer:
The highest level of organization for living things is the biosphere; it encompasses all other levels. The biological levels of organization of living things arranged from the simplest to most complex are: organelle, cells, tissues, organs, organ systems, organisms, populations, communities, ecosystem, and biosphere.
Answer:
biosphere
The highest level of organization for living things is the biosphere; it encompasses all other levels. The biological levels of organization of living things arranged from the simplest to most complex are: organelle, cells, tissues, organs, organ systems, organisms, populations, communities, ecosystem, and biosphere.
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Why does each trophic level have a smaller amount of organisms or biomass than the one below it? A. Because higher trophic levels have lower survival rates for offspring. B. Because each trophic level has larger organisms than the one below it. C. Because the higher trophic levels have to hunt the lower trophic levels, so they- use most of the energy that they gain from eating. D. Because each trophic level is only absorbing 10% of the previous, so there is not enough energy to sustain larger numbers.
Each trophic level has a smaller amount of organisms or biomass than the one below it because it is only absorbing 10% of the energy from the lower level, so there is not enough energy to sustain larger numbers.
A trophic level refers to a level in the food chain consisting of the same or biologically similar organisms. Examples of trophic levels include producer, consumer, and decomposer.Organisms in the lower trophic level serve as food for organisms in the next trophic level of the food chain.Only about 10% of the energy derivable from one trophic level gets to the next trophic level, the remaining is lost as heat during metabolic processes.Hence, the number of organisms progressively decreases as we climb higher in the food chain. The limit to the amount of energy transferred as we move higher means that the number of individuals that can be sustained will also be limited.The correct option would, therefore, be D.
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Bacteria break down the nitrates and release oxygen. True or false? Short answer
Answer:
True
Short Answer-Denitrifying bacteria transform nitrate in extremely wet soils and swampy grounds where there is very little oxygen, i.e. the conditions are anaerobic. The bacteria get the oxygen they need for respiration from the breakdown of nitrates.
Explanation:
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From a chemical view, how is an amino acid is being recognized by its specific aminoacyl tRNA synthetase?
Answer:
An aminoacyl-tRNA synthetase (aaRS or ARS), also called tRNA-ligase, is an enzyme that attaches the appropriate amino acid onto its corresponding tRNA. It does so by catalyzing the transesterification of a specific cognate amino acid or its precursor to one of all its compatible cognate tRNAs to form an aminoacyl-tRNA.
Chlorophylls are a. seen best in carrots, beets, and red ferns. b. pigments that absorb only blue and green wavelengths. c. the main pigments that absorb the wavelengths most efficient at driving photosynthesis. d. All of the answers are correct.
Answer:
....kghvkhbiblyhlugkufv
Explanation:.....
Answer:A
Explanation:
I think A because they are all plants
How do endophytes affect grazing animals?
A. Endophytes pull nutrients from the soil, making plants more nutritious for grazing animals.
B. Endophytes that produce toxins can be poisonous to grazing animals.
C. Endophytes fix nitrogen from the air, making plants more nutritious for grazing animals.
D. Endophytes are poisonous to plants, depriving grazing animals of forage.
Answer:
B. Endophytes that produce toxins can be poisonous to grazing animals.
How many federal agencies are responsible for managing land resources in the United States?
Answer:
How many federal agencies are responsible for managing land resources? Most of the current land use concerns relate to federal lands controlled by four federal agencies: Bureau of Land Management, U.S. Forest Service, National Park Service, and Fish and Wildlife Service.
Explanation:
Answer:
The answer is 4 on edge
Explanation:
15. Assuming the fluorescence threshold is set to be 1. Which gene has the highest gene expression rate? What is the Ct value? Gene with highest expression = ______________ Ct value = _______________
Answer:
The correct answer is - Gene 1 and 14.5 (or 14).
Explanation:
The cycle threshold is denoted as Ct value. It is the cross point at the threshold line meets with fluorescence signal reaction curve. In the given RTpcr experiment, the threshold line is fluorescence intensity 1 thus, Ct value for each gene will be the point at which each curve meets the fluorescence intensity line at 1. Ct value of is inversely proportional to amount of nucleic acid in the given sample.
Here, the Ct for gene 1 is the lowest, so the amount of DNA would be highest.
Gene with highest expression= Gene 1
Ct value= 14.5
in some plants pink flowers are dominant over yellow flowers if Mario crosses two hybrids what happens
Which of the following factors would LIMIT carrying capacity?
A. Far from factories or roads
B. A large nearby river
C. Abundant food
D. A small space to live in
Answer:
D. A small space to live in
Explanation:
Having a small space to live in would directly limit carrying capacity.
Since there is a small space, only a few organisms would be able to live in that space.
This means that carrying capacity is limited, because the space does not have enough room to sustain more people.
So, the correct answer is D. A small space to live in
Please help
ASAP 10 points
Answer:
I would guess B, systematic sampling. I might be wrong though.
Explanation:
Linda received the flu vaccine this year.
Which of the following statements is true?
Linda's T-cells will produce memory cells against the virus.
Linda's body will produce antigens against the flu virus.
Linda's immune system will produce lifelong immunity against any form of the flu.
Linda's immune system will produce antibodies against the flu virus.
Answer:
Linda's immune system will produce antibodies against the flu virus
Answer:
Linda's immune system will produce antibodies against the flu virus
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A DNA fragment is introduced into the lacZ gene of a plasmid, which also contains an ampicillin resistance gene. What is the appearance of bacteria transformed with this plasmid if they are spread on plates containing ampicillin and X-gal
Answer: White colonies that are resistant to ampicillin.
Explanation:
The lac operon is an operon required for the metabolism of lactose in enteric bacteria such as Escherichia coli . It has three structural genes, a promoter, an operator and a regulator, all regulated by the availability of glucose and lactose. The lac repressor, a protein, senses lactose and blocks transcription of this operon. It acts as a repressor when lactose is present. A catabolite-activating protein (CAP), on the other hand, acts as a glucose sensor. The bacterium should express the lac operon only when lactose is available and glucose is not available. Thus, genes can always be transcribed, except when the Lac repressor protein is bound to the operon region, for which it has a high affinity (i.e. in the absence of lactose), where the Lac repressor protein maintains its high affinity for the operon region, preventing RNA polymerase from transcribing the structural genes. Thus, the system remains closed with consequent energy savings for the bacterium. In the presence of lactose, it binds to the Lac repressor protein and generates a conformational change that decreases its affinity for the operator region. Thus, the operator region is left free, RNA polymerase can freely transcribe the structural genes and the synthesized β-galactosidase (an enzyme) can degrade lactose to glucose plus galactose for energy.
Thus, the lac z gene encodes the enzyme β-galactosidase, which catalyzes the hydrolysis reaction of lactose to glucose and galactose. In gene cloning experiments, a compound called X-gal is used as an indicator of cells expressing the β-galactosidase enzyme. X-gal is hydrolyzed by the enzyme to galactose and another compound that is oxidized giving an insoluble blue compound. Thus, if X-gal and a β-galactosidase inducer are dissolved in the medium of a culture plate where the transformed bacteria is found, colonies grown on the plate that possess a functional lac z gene (either because they were not transformed by the plasmid, or if they were but the plasmid does not have the cloned fragment or gene that disrupts the lac z gene) can be clearly distinguished by their blue coloration. If they have another gene inserted interrupting the lac z gene, they will not be able to produce the enzyme that degrades X-gal, resulting in white colonies since X-gal is not degraded giving that characteristic blue color.
The white, non-transforming colonies are eliminated by adding an antibiotic to the medium for which the plasmid provides resistance (in this case ampicillin), so that we can select the recombinant colonies that carry the vector with our sequence, simply by their color.
So, if bacteria are transformed with a plasmid (with ampicillin resistance) cloned with a gene that interrupts the lac z gene, the bacteria will be white because they do not synthesize the enzyme that degrades X-gal and will be resistant to ampicillin.
Una de las aplicaciones más importa del electromagnetismo son
Answer:
Las principales aplicaciones del electromagnetismo se emplean en: La electricidad. El magnetismo. La conductividad eléctrica y superconductividad.
Explanation:
If Sarah had skin vesicles on the anterolateral region of her neck, which peripheral nerve is infected by the virus? To which peripheral nerve plexus does this nerve belong?
Answer:
The correct answer is -
- transverse cutaneous nerve
- cervical plexus
Explanation:
transverse cutaneous nerve is infected by the virus that causes skin vesicles on the anterolateral region of her neck. It belongs to the cervical peripheral nerve plexus.
Name three ways that humans benefit from the use of seedless plants
Answer:
They support life by being the first vegetation to spring up on harsh terrain where soil is scarce. Even when they perish, seedless plants give back to nature. Certain seedless plants like moss and liverworts actually leave behind a layer of fertile soil for other plants when they perish.
Explanation:
Seedless plants have historically played a role in human life through uses as tools, fuel, and medicine. Dried peat moss, Sphagnum, is commonly used as fuel in some parts of Europe and is considered a renewable resource.
A 17-year-old student has experienced reversible, periodic attacks of chest tightness with coughing, wheezing, and hyperpnea. She states that expiration is more difficult than inspiration. She is most comfortable sitting forward with arms leaning on some support. X-rays revealed mild overinflation of the chest. Results from laboratory and pulmonary function tests are as follows:
• Frequency 20 breaths/min
• Vital capacity (VC) 2.9 L
• FEV1.0 1.4 L
• FEV1.0/FVC 56%
• Functional residual capacity (FRC) 3.89 L
• Total lung capacity (TLC) 6.82 L
• PaO2 70 mm Hg
• PaCO2 26 mm Hg
• Pulse 108 beats/min
• BP 120/76 mm Hg
Intermittent use of a bronchial smooth muscle dilator (1:1000 epinephrine by nebulizer) for several days caused marked improvement, resulting in the following laboratory and pulmonary function tests:
• VC 4.15 L
• FEV1.0 3.1 L
• FEV1.0/FVC >75%
• FRC 3.7 L
• TLC 5.96L
• PaO2 89 mm Hg
• PaCO2 38 mm Hg
• Pulse 129 beats/min BP 122/78 mm Hg
1. What is the disorder of this 17-year-old student?
2. Is this primarily a restrictive or an obstructive disorder? Why?
3. Write the formula for determining residual volume (RV).
4. Determine the residual volume (RV) before and after the use of the bronchodilator.
a. RV before using the bronchodilator:b. RV after using the bronchodilator:
5. Why is expiration more difficult than inspiration in this person?
6. What does the change in pulmonary function after the bronchodilator therapy indicate?
7. Why does the bronchodilator exaggerate the tachycardia?
8. What causes the hypoxemia and the hypocalcemia in this person?
9. A beta2-adrenergic agent was prescribed for further use because it has less cardio stimulatory (beta1) effect. Based on your knowledge of beta1 and beta2 receptors, why is this a good suggestion?
10. An anticholinergic agent was also suggested as a possible nebulizer agent. How might this helps the breathing problem?
Answer:
Frecuencia 20 respiraciones / min
Explanation:
