The ability to distinguish between sensations originating in different areas of the body is option a. due to sensory neurons from each body region synapsing in specific locations of the appropriate cortex.
Our ability to differentiate sensations from various body regions relies on the organization and processing of sensory information in the brain. The sensation we perceive is the result of sensory neurons transmitting signals from specific body regions to the brain. Each body region has its own set of sensory neurons dedicated to transmitting signals from that particular area. These sensory neurons project their axons to specific regions in the cortex, which is the outer layer of the brain responsible for processing sensory information.
The organization of sensory neurons in the cortex is such that neurons from each body region synapse or connect with specific areas of the cortex that correspond to that region. For example, sensory neurons from the hand would synapse in the somatosensory cortex, which is responsible for processing touch and proprioception. This specific localization of synapses allows the brain to differentiate and interpret sensations originating from different body regions.
Additionally, different types of sensory receptors exist throughout the body, each specialized for detecting specific stimuli such as pressure, temperature, or pain. These sensory receptors produce distinct types of action potentials or electrical signals when they are stimulated. Consequently, the brain receives different patterns of signals depending on the type of sensory receptor activated, further aiding in the discrimination of sensations from different body regions.
In conclusion, our ability to distinguish sensations from different areas of the body is facilitated by the specific localization of synapses in the appropriate cortex for each body region. Furthermore, the presence of different types of sensory receptors and their corresponding action potentials contribute to the differentiation of sensory information.
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A mutation causing an addition or a deletion of one base pair resulted in the production of a nonfunctional mutant protein. The sequences of the normal and mutant proteins are given below. Normal: Met - Gly - Glu - Val - Asp Mutant: Met - Gly - Lys - Ser - lle - Lys - Trp - Arg - . Was this mutation cause by an insertion or a deletion? Below, fill in the codons in the coding sequence of the mRNA that is translated into the mutant protein. If more than one codon is possible, just enter a single codon. NH₂ Met Gly Lys Ser lle codons 5' AUG
The mutation that caused the nonfunctional mutant protein was due to an insertion.
The coding sequence of the mRNA that is translated into the mutant protein has one additional codon, thus the frameshift mutation that caused the protein to be nonfunctional can be inferred to have been an insertion mutation. A deletion mutation would have caused one of the amino acids to be missing, and the mRNA sequence to be shorter than the normal sequence.
Hence, the mutation causing the nonfunctional mutant protein was due to an insertion.Let us fill in the codons in the coding sequence of the mRNA that is translated into the mutant protein.NH₂ Met Gly Lys Ser lle codons 5' AUG GGUAAGUCAUCAGGAC The codons in the coding sequence of the mRNA that is translated into the mutant protein are 5' AUG GGUAAGUCAUCAGGAC.
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Dithiothreitol (DTT) was used during the 'halo' assay in Activity 3.4. DTT is a strong reductant. Which cellular process would most directly be affected
by DTT treatment?
A. oxidative protein folding in the ER
B. Movement of electrons through the mitochondrial electron transport system
C. glycosylation in the Golgi apparatus
D.ER-associated protein degradation (ERAD)
E. Synthesis of phospholipids in the smooth ER
DTT breaks the disulfide bond between amino acid residues, which is necessary for the structural conformation of certain proteins. Therefore, oxidative protein folding in ER will be directly influenced by DTT treatment. So option A is correct.
DTT is commonly used to break protein disulfide bonds and more generally, to inhibit the formation of intramolecular (i.e., intermolecular) disulfide bonds between protein cysteine residues.
Thiol reagents like DTT break IgM molecules’ interchangeable disulfide bonds and remove the IgM antibody’s agglutinating capability while leaving the IgG molecule intact.
DTT distinguishes IgM from IgG antibodies by inactivating the IgM antibody allowing the identification of any intact IgG antibody.
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1) In cycle 1 and all subsequent cycles of the PCR reaction, one copy of each of the two original strands will be synthesized at the 3' end of the primer and up to the 5' end of the original strand. Write the sequence of the copies (C1 and C2) that are made of the chains O1 and O2.
(01) 3' T C G G C T A C A G C A G C A G A T G G T A C G T A 5'
(C1) 5' C C G A T _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 3'
(O2) 5' A G C C G A T G T C G T C G T C T A C C A T G C A T 3'
(C2) 3' _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ A T G G T 5'
PCR uses O1 and O2 as templates for DNA synthesis. PCR cycles involve denaturation, annealing, and extension.
The first cycle denatures O1 and O2, splitting the double-stranded DNA into single strands. Primer 3' binds to template strand complementary sequences. We don't know the primer's 3' end sequence from the sequences. The primer starts DNA synthesis by binding to a specific area. DNA polymerase uses the original strands as templates to synthesize new strands during extension. 5'-to-3' synthesis occurs. Thus, each template strand's new copy will be synthesized from the primer's 3' end to the original strand's 5' end.
Let's complete the sequences:
(C1) 5' C C G A T G G T A C G T A _ _ _ 3'
(C2) 3' _ A T G G T 5'
C1 synthesizes O1 from its 3' primer end to its 5' end. C2's synthesis begins at O2's 3' primer end and continues to its 5' end.
We can't establish C1 and C2's exact sequences without the primer sequence. The primer sequence determines DNA synthesis nucleotide order.
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9. Which of the following sunlight classes of UV radiation has the shortest wavelength? a) UVA b) UVB c) UVC d) UVD 10. Human Papillomavirus is the main cause of _____.
a) testicular cancer b) cervical cancer c) breast cancer d) hepatocarcinoma 11. The phenomena in which the integration of viral DNA into host chromosome that cause activation or disruption of a normal gene is known as ______.
a) insertional mutagenesis b) proliferating mutagenesis c) transforming mutagenesis d) constitutive mutagenesis
UVC has the shortest wavelength among the sunlight classes of UV radiation. Cervical cancer is mainly caused by Human Papillomavirus (HPV). The underlying mechanism is thought to be insertional mutagenesis, a process by which there is an integration of viral DNA into the host chromosome that causes activation or disruption of the normal host gene.
Among the sunlight classes of UV radiation, UVC has the shortest wavelength. UVA has the longest wavelength, followed by UVB and then UVC.
Human Papillomavirus (HPV) is the man causative agent of cervical cancer. HPV is a sexually transmitted infection that can lead to the development of abnormal cervical cells, which, if left untreated, can progress to cervical cancer. It is important for individuals, particularly females, to undergo regular screening tests, such as Pap smears and HPV testing, to detect and prevent cervical cancer.
The phenomenon in which viral DNA integrates into the host chromosome and causes activation or disruption of a normal gene is known as insertional mutagenesis. Viral DNA can insert itself into the host genome and affect the expression and function of genes.
This integration can lead to genetic changes that contribute to the development of various diseases, including certain types of cancer. Insertional mutagenesis is a mechanism through which viruses can alter the normal functioning of host cells and potentially drive cellular transformation.
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This cross involves three unlinked genes with recessive mutant phenotypes w m and u. The w and m genes are on autosomes and the u genc is sex-linked (on the X chromosome. Use +to indicate wild type alleles. Write the genotypes of these parental flies:(u is sex linked) w.female X mumale Genotypes for F1: female: male: Write the allelic combinations for the top of a Punnctt square using the F1, female gametes: Write the allelic combinations for the top of a Punnett square using the Fl,male gametes: a For these unlinked genes, what is the frequency of the triple recessive F2 and what sex or sexes would they be? b Also,circle the allelic combination above,male and female,that make this phenotype.
Part A: Frequency of the triple recessive F2 and what sex or sexes would they beFor the F2 generation, the following traits are expressed in the following ratios:9/64 are wm u, 9/64 are w mu, 9/64 are wm +, 9/64 are w+ u, 3/64 are +mu, 3/64 are +u +, 3/64 are w+mu and 27/64 are w + +.The frequency of triple recessive F2 flies is 18/512, which simplifies to 0.0352.
This shows that the ratio of the triple recessive F2 is low, about 3.5 percent, since the frequency of triple recessive offspring will be low. These recessive phenotypes are rare and can be inherited only when both parents have the alleles of the recessive phenotype. For the F2 generation, males have only one X chromosome. The u gene is on the X chromosome, therefore, the expression of the phenotype is determined by the genotype of the single X chromosome in males. Therefore, all the flies with the mutant phenotype (wmu) will be males. Part B: Allelic combination that makes this phenotype.
The phenotype of triple recessive F2 can only be formed when a fly has the following alleles: wmuthe recessive mutant phenotype of the w and m genesmu- the recessive mutant phenotype of the u gene The allelic combinations which produce the triple recessive phenotype are wmu/mu and wm/mu. Therefore, the allelic combinations to circle are wmu/mu and wm/mu.
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Capsaicin, the active ingredient in hot peppers, creates a burning sensation because it causes the local temperature of the exposed tissue to increase dramatically leading to depolarization of thermoreceptors O it causes an inflammatory response that attracts immune cells that release soluble cytokines that bind and depolarize nociceptors O it destroys cells in the tongue releasing acid (H+) onto taste buds O it binds and opens an ion channel on nociceptors that also happens to open in response to heat
Capsaicin, the active ingredient in hot peppers, creates a burning sensation because it binds and opens an ion channel on nociceptors that also happens to open in response to heat.
Nociceptors are sensory neurons that are specialized in detecting and transmitting signals related to pain and temperature. The ion channel involved in this process is known as the transient receptor potential vanilloid 1 (TRPV1) channel. When capsaicin comes into contact with nociceptors, it binds to and activates the TRPV1 channels. These channels are normally activated by high temperatures, but capsaicin can also bind to them, leading to their opening and depolarization of the nociceptor membrane. This depolarization sends a signal to the brain, resulting in the perception of a burning or painful sensation. It is important to note that capsaicin does not cause an actual increase in temperature or destroy cells. Instead, it interacts with the sensory system, specifically targeting the TRPV1 channels on nociceptors, leading to the perception of heat and pain. This mechanism explains why capsaicin creates a burning sensation when it comes into contact with tissues or receptors in the body, and it is also responsible for the sensation of heat when consuming spicy foods containing capsaicin.
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Which of the following is NOT a role of the kidneys? * 1 point regulates ion balance rids the body of metabolic waste egestion of nitrogenous wastes regulates water balance secretion of hormones involved in the production of RBCs Urea is produced * * 1 point by the liver in every cell of the body when amino acids are dephosphorylated by the kidneys by birds and reptiles
The kidneys are the pair of organs that perform a variety of important functions that are important for a healthy body. The primary roles of kidneys are the regulation of water balance, maintenance of acid-base balance, regulation of blood pressure, filtration of waste products from the body, and production of urine.
The kidney's main job is to filter the blood to remove excess waste products and fluids from the body.
The following functions are performed by the kidneys except for the secretion of hormones involved in the production of red blood cells.
Erythropoietin is a hormone that regulates the production of red blood cells in the body, and it is produced by the kidneys.
Kidneys regulate the body's ion balance by filtering the blood.
Kidneys remove metabolic waste products such as urea, uric acid, and creatinine from the body. Kidneys are involved in the egestion of nitrogenous wastes, which include excess urea, uric acid, and creatinine.
Kidneys are involved in regulating the water balance of the body by regulating the concentration of urine and maintaining blood pressure.
Urea is produced by the liver in every cell of the body when amino acids are dephosphorylated. Urea is a waste product that is filtered by the kidneys.
Birds and reptiles excrete nitrogenous waste products in the form of uric acid rather than urea, which is the case in mammals. Hence, this is not a role of the kidneys.
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explain all options
Question 99 Not yet answered Marked out of 1.0 Regarding muscles, which is true? O a skeletal muscle contraction causes peristalsis in the GI tract and ureter O b. smooth muscle does not contain actin
Skeletal muscle contraction does not directly cause peristalsis, and smooth muscle does contain actin, although its arrangement differs from that of skeletal muscle.
Regarding muscles, neither option provided is entirely accurate.
a. Skeletal muscle contraction does not directly cause peristalsis in the gastrointestinal (GI) tract and ureter. Peristalsis is a coordinated contraction and relaxation of smooth muscles that propels substances through the digestive system and ureter. While skeletal muscles play a role in swallowing and voluntary control of bowel movements, they do not initiate peristalsis in these organs.
b. Smooth muscle does contain actin. Actin and myosin are the two major contractile proteins found in both skeletal and smooth muscles. However, the arrangement and organization of these proteins differ between the two muscle types. In skeletal muscle, actin and myosin are highly organized into distinct sarcomeres, which give the muscle its striated appearance. In smooth muscle, the actin and myosin filaments are more randomly arranged, leading to the absence of striations.
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We would expect most extreme thermophiles to be in the domain
Archaea.
a) True
b) False
Yes, the above statement is a) True. Extreme thermophiles are microorganisms that thrive in extremely high-temperature environments.
They are typically found in the domain Archaea, which is one of the three domains of life, alongside Bacteria and Eukarya. Archaea have been extensively studied in extreme environments such as hot springs, hydrothermal vents, and geothermal areas, where temperatures can exceed the survivability range of most organisms. These extreme thermophiles have unique adaptations that allow them to withstand and thrive in such harsh conditions, including the ability to stabilize their proteins and membranes at high temperatures. Therefore, it is true that we would expect most extreme thermophiles to be found in the domain Archaea.
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3) Which statement is most correct regarding the evolution of cells?
A) Multicellular life evolved rather quickly.
B) Prokaryotic cells were the most complex life on earth for two billion years.
C) Photosynthesis evolved before anaerobic respiration.
D) Free oxygen has existed in the atmosphere for all of earth’s geological history.
5) Exposing a protein to high concentrations of urea, which denatures the protein, then after removing the urea, the protein refolds into its native conformation is evidence that: A) Proteins fold randomly.
B) Proteins will fold into multiple conformations under normal cellular conditions.
C) Proteins will spontaneously fold into one native conformation under normal cellular conditions.
D) Urea is a protease.
The most correct statement regarding the evolution of cells is: B) Prokaryotic cells were the most complex life on earth for two billion years.
This statement is based on the timeline of the evolution of life on Earth. Prokaryotic cells, which are simpler in structure and lack a nucleus, were the first forms of life to emerge approximately 3.5 billion years ago. They were the dominant life forms on Earth for a significant period, around two billion years, before the evolution of more complex eukaryotic cells.
Eukaryotic cells, which contain a nucleus and membrane-bound organelles, emerged around 1.5 billion years ago. The evolution of multicellular life forms occurred later in the timeline. The correct statement regarding exposing a protein to high concentrations of urea, which denatures the protein, and then refolding it into its native conformation is:
C) Proteins will spontaneously fold into one native conformation under normal cellular conditions.
When a protein is exposed to high concentrations of urea, it disrupts the interactions that stabilize the protein's folded structure. As a result, the protein unfolds or denatures. However, upon removing the urea, the protein refolds into its native conformation spontaneously.
This indicates that proteins have an intrinsic ability to fold into their specific, functional three-dimensional structures under normal cellular conditions without the need for external factors. This process of protein folding is governed by the primary amino acid sequence and the interactions between amino acid side chains.
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A technician is trying to replicate a stock set of microorganisms for storage. She knows that the microorganism is resistant to tetracycline so she adds it to the media after it has been sterilized. This is an example of a(n) _____media.
The given scenario shows that the technician is trying to replicate a stock set of microorganisms for storage, where the microorganism is resistant to tetracycline. So, the technician adds it to the media after sterilization, which is an example of selective media.In microbiology, selective.
Media are those media that are made to permit the growth of a particular microorganism while inhibiting the growth of other microorganisms. These media are essential for diagnosing, isolating, and enumerating microorganisms. It can be used to distinguish between closely related organisms.Selective media contain specific nutrients that favor the growth of one type of microorganism while inhibiting the growth of other types of microorganisms.
By using selective media, microbiologists can isolate the particular microorganisms they want to study. In this case, since the technician adds tetracycline to the media, this indicates that the media is selective, and it will support the growth of only microorganisms that are resistant to tetracycline.So, the correct answer is 'selective.'
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About 12 years ago, my laboratory published a report on the X-ray crystal structure of a blue copper protein called rusticyanin. The structure of the folded protein is maintained by a large number of noncovalent bonds formed by the interactions of the individual side chains of the amino acids that comprise the protein. As examples of such interactions, it is evident that the side chains of eight different amino acids (Ala, Arg, Asp, Cys, Gln, Ile, Thr, and Val) happen to interact on a pair-wise basis to form four different types of noncovalent bonds (hydrophobic, electrostatic, hydrogen, and van der Waals) in the interior of the folded protein. Use the clues below and/or the information in your textbook to identify the pair of amino acids involved in each type of noncovalent bond. Then answer the questions on BrightSpace for Quiz 10. Clue #1 - Val, Asp, and Thr are involved in three different types of noncovalent bonds that do not include the van der Waals bond. Clue #2-The hydrophobic bond does not involve Arg or Ala. Clue #3 - The hydrogen bond does not involve Val or Ala. Clue #4 - Cys, which does not participate in a hydrophobic bond, does not interact with Thr in rusticyanin. Clue #5-Arg, which does not interact with Thr or Val in rusticyanin, isn't involved in a van der Waals bond. Clue #6-Asp and Ile aren't part of the hydrogen bond in the rusticyanin.
Based on the given clues, the amino acid pairs involved in each type of noncovalent bond in rusticyanin are as follows: hydrophobic bond - Thr and Val; electrostatic bond - Asp and Arg; hydrogen bond - Gln and Thr; van der Waals bond - Cys and Ile.
The clues provided help narrow down the amino acid pairs involved in each type of noncovalent bond in rusticyanin. Clue #1 states that Val, Asp, and Thr participate in three different types of noncovalent bonds excluding van der Waals. Therefore, Val and Thr are involved in a hydrophobic bond, while Asp and Thr form an electrostatic bond.
Clue #2 indicates that the hydrophobic bond does not include Arg or Ala. Therefore, the hydrophobic bond involves Val and Thr since they are the remaining options.
Clue #3 states that the hydrogen bond does not involve Val or Ala. As Val is excluded, the hydrogen bond must involve another amino acid pair. Based on the remaining options, Gln and Thr form the hydrogen bond.
Clue #4 mentions that Cys, which does not participate in a hydrophobic bond, does not interact with Thr. This implies that Cys is not involved in the hydrophobic bond between Val and Thr.
Clue #5 states that Arg, Thr, and Val are not involved in the van der Waals bond. Therefore, the van der Waals bond must involve other amino acid pairs. Since Cys and Thr are the remaining options, Cys and Thr form the van der Waals bond.
Finally, clue #6 indicates that Asp and Ile are not part of the hydrogen bond. This aligns with the earlier deduction that Gln and Thr form the hydrogen bond.
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41The site of the formation of the primary structure for protein synthesis in animal cells is the
a) mitochondrionb) nucleusc) SER d) RERe) vacuole
42. Phospholipids can form all of the following structures in water except which one?
a) cell membranes b) bilayersc)nuclear membranes d) vesiclese) Bones cell membranes
The site of the formation of the primary structure for protein synthesis in animal cells is the ribosome. The site of the formation of the primary structure for protein synthesis in animal cells is the ribosome.
Ribosomes, the site of protein synthesis in cells, are composed of two subunits that are unequal in size. Both ribosomal subunits contain ribosomal RNA (rRNA) molecules and a number of ribosomal proteins that help to maintain the structure and function of the ribosome.
Therefore, option D is the answer.
Phospholipids can form all of the following structures in water except bones cell membranes. Phospholipids are the main structural component of cell membranes in living organisms. When in contact with water, these amphipathic molecules spontaneously self-organize into a bilayer to form a cell membrane. The two layers of a bilayer have opposing orientations of the phospholipid molecules that create a hydrophobic interior sandwiched between two hydrophilic surfaces.
They can also form vesicles or liposomes when a bilayer spontaneously closes to create an isolated compartment. However, bones cell membranes is not a structure that can be formed by phospholipids in water.
Therefore, option E is the answer.
Ribosomes are the site of the formation of the primary structure for protein synthesis in animal cells, while phospholipids can form all of the following structures in water except bones cell membranes.
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An example of an evolutionary homology would be
Group of answer choices
a. your arms and the flippers of a seal
b. the fur on your dog and the feathers on your parrot
c. nest-building of a bird and hive-building of a honeybee
d. fins of a fish and flipper of a seal
Evolutionary homology is when organisms share a common ancestry and have similar structures. The structures can have different functions in different organisms, but they are similar in their underlying structure. Therefore, an example of an evolutionary homology would be fins of a fish and flipper of a seal. correct option is d ) .
These structures are different in their function, but they have similar structures and the same embryonic origin.
The flipper of a seal and fins of a fish have the same basic structure. For example, they both have a humerus, radius, and ulna bones.
The structure is a result of their shared ancestry, and it is an example of an evolutionary homology.
Other examples of evolutionary homologies are wings of bats and wings of birds.
Even though they have different functions, they have similar structures. The bones, muscles, and other structures that make up their wings are similar because they are derived from a common ancestor.
Another example is the limbs of mammals.
Whether they are used for walking, swimming, or flying, they share a basic structure that is an example of evolutionary homology.
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You plan to test the hypothesis that students who don't work while attending biology 3 lab, perform better than students who do work. To do so, you recruit 100 students from the biology 3 labs at LAMC who will be taking the lab in the coming semester. Of these students, 50 will not work at all during the semester and 50 will work 40 hours per week. At the end of the semester, you will have access to the overall percent scores for each student in each course. A. What is the independent variable for this experiment? B. What is the dependent variable for this experiment? C. Indicate one standardized variable for this experiment? D. What is the control for this experiment? E. To graph the results for this experiment, which variable would be plotted on the horizontal (X) axis? F. Which variable would be plotted on the vertical (Y) axis? -How would you draw the graph below?
A. The independent variable for this experiment is the student's employment status, specifically whether they work or don't work during the semester while attending the Biology 3 lab.
B. The dependent variable for this experiment is the student's overall percent score in the Biology 3 lab course at the end of the semester.
C. One standardized variable for this experiment could be the grading criteria or rubric used to calculate the overall percent scores for each student. By using a standardized grading system, the scores can be compared accurately between the two groups.
D. The control for this experiment would be the group of students who don't work during the semester. This group serves as a baseline comparison to assess the performance of the group of students who work.
E. To graph the results for this experiment, the independent variable (employment status: work or no work) would be plotted on the horizontal (X) axis.
F. The dependent variable (overall percent score) would be plotted on the vertical (Y) axis.
Here are the explanations:
A. The independent variable for this experiment is whether or not the students work while attending Biology 3 lab. It has two levels: not working (group 1) and working 40 hours per week (group 2).
B. The dependent variable for this experiment is the overall percent scores of the students in the Biology 3 lab course. This variable will be measured and compared between the two groups.
C. One standardized variable for this experiment could be the grading criteria or rubric used to assign percent scores to the students. This ensures that the measurement of the dependent variable is consistent and standardized across all participants.
D. The control for this experiment would be the students' baseline performance or scores in the Biology 3 lab course before the experiment begins. This provides a comparison point to assess the impact of working or not working on their performance.
E. The variable plotted on the horizontal (X) axis would be the independent variable, which is whether the students work or not. The two levels would be represented as separate categories on the X-axis.
F. The variable plotted on the vertical (Y) axis would be the dependent variable, which is the overall percent scores of the students in the Biology 3 lab course. The scores would be plotted as continuous data points on the Y-axis.
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Positioning of the first Met tRNA on the mRNA is a critical step in translation inititiation. Selection of the first correct AUG is achieved by a complex of Met tRNA with a. initiation factors
b. initiation factors plus poly A binding protein
c. small subunit ribosome d.small subunit ribosome plus initiation factors
The positioning of the first Met tRNA on the mRNA during translation initiation is a crucial step. The correct selection of the first AUG codon is achieved by a complex of Met tRNA with the small subunit ribosome plus initiation factors. This complex ensures the accurate initiation of protein synthesis.
Translation initiation is the process by which protein synthesis begins in cells. It involves the assembly of the ribosome, mRNA, and initiator tRNA at the start codon of the mRNA molecule. The first Met tRNA, carrying the amino acid methionine, plays a crucial role in this process.
To ensure accurate initiation, a complex is formed between the small subunit ribosome, the initiator tRNA, and several initiation factors. These initiation factors help in the proper positioning of the components and facilitate the recognition of the start codon. Among the initiation factors, one important factor is the initiation factor 2 (IF2) that interacts with the initiator tRNA and the small subunit ribosome.
The initiation complex scans the mRNA molecule until it reaches the correct start codon, which is typically AUG. The start codon is recognized by the anticodon of the initiator tRNA, which is base-paired with the AUG codon. The interaction between the Met tRNA and the start codon is facilitated by the small subunit ribosome and the initiation factors. Once the correct start codon is recognized, the large subunit of the ribosome joins the complex, and protein synthesis begins. The initiator tRNA occupies the P-site of the ribosome, ready to receive the next amino acid and initiate the elongation phase of translation.
In conclusion, the positioning of the first Met tRNA on the mRNA during translation initiation is achieved by a complex consisting of the small subunit ribosome plus initiation factors. This complex ensures the accurate selection of the first AUG codon and facilitates the proper initiation of protein synthesis.
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The following are problems generated when a patient is given a false-positive diagnosis, except: Increased burden to the healthcare system. Generates anxiety and worry. Unnecessary treatments are administered. The opportunity for an effective intervention is missed. Generate unnecessary expenses, as those labeled positive often go through other diagnostic procedures.
A false-positive diagnosis does not directly cause an increased burden to the healthcare system or miss the opportunity for an effective intervention.
A false-positive diagnosis refers to a situation where a patient is incorrectly identified as having a particular condition or disease when they do not actually have it. This can have various negative consequences for the patient and the healthcare system. It generates anxiety and worry for the patient, as they may believe they have a serious health condition and may undergo unnecessary stress and psychological burden.
Furthermore, a false-positive diagnosis can lead to unnecessary treatments being administered to the patient. This can involve medications, therapies, or procedures that are not needed, potentially exposing the patient to side effects and risks without any benefit.
In addition, a false-positive diagnosis can result in unnecessary expenses. Patients who are labeled as positive often undergo further diagnostic procedures to confirm or rule out the condition. These additional tests and consultations can incur financial costs for the patient and the healthcare system.
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no explanation needed pls answer
quick.
Answer all short answer questions and the essay DQuestion 23 Crossing over occurs between..... Sister chromatids during prophase 1 of mitosis O Sater chromatics during prophase I of meiosis O Non-sist
Crossing over occurs between sister chromatids during prophase 1 of mitosis (option a).
Crossing over occurs during prophase 1 of meiosis. It is the exchange of genetic material between non-sister chromatids of homologous chromosomes. During meiosis, two rounds of cell division occur which produces haploid cells. During prophase 1 of meiosis, crossing over occurs between non-sister chromatids of homologous chromosomes. During this stage, the two homologous chromosomes exchange genetic material. This process leads to the creation of new combinations of genetic material and increases genetic diversity. The correct option is: Sister chromatids during prophase I of meiosis
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Quantity which refers to the number of reaction process that each active site of the enzyme catalyzes per unit time.
a. Turnover number
b. Catalytic efficiency
c. Enzyme activity
d. Specific enzyme activity
Quantity which refers to the number of reaction process that each active site of the enzyme catalyzes per unit time Turnover number. The correct option is a.
The turnover number refers to the number of reaction processes that each active site of an enzyme catalyzes per unit time. It is also known as kcat and is a measure of the catalytic activity of an enzyme. The turnover number provides information about how efficiently an enzyme can convert substrate molecules into product molecules. It is expressed as the number of substrate molecules converted per active site per second.
Option a, turnover number, accurately describes the quantity mentioned in the question. Options b, c, and d (catalytic efficiency, enzyme activity, and specific enzyme activity) are related concepts but do not specifically refer to the number of reaction processes per active site per unit time.
Therefore, the correct answer is a.
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Question 1
The difference between a nucleoside and a nucleotide is that
A. nucleotides contain a different sugar compared to nucleosides.
B. the bases in nucleotides are attached to sugars at different carbons compared to nucleosides.
C. nucleosides are used to synthesize DNA, whereas nucleotides are used to synthesize RNA.
D. nucleotides contain one or more phosphate groups, whereas nucleosides have none.
E. nucleosides contain purine bases, whereas nucleotides contain pyrimidine bases.
Question 3
Which statement is true regarding the relationship between replication and transcription of DNA?
A. Replication requires both a template and a primer, whereas transcription requires only a template.
B. The polymerases for both require a Mn2+ cofactor for activity.
C. Copies of both DNA strands are made during both processes.
D. Both have extensive processes to correct errors.
E. Both utilize the same nucleotides.
Question 5
In eukaryotes, nucleosomes are formed by binding of DNA and histone proteins. Which of the following is NOT true regarding histone proteins?
A. H1 functions as a monomer
B. Histone proteins have five major classes: H1, H2A, H2B, H3, and H4
C. Positively coiled DNA is wrapped around a histone core to form nucleosome
D. H1, H2A, H3 and H4 form the nucleosome histone core.
E. They are found in the nucleus.
Question 1:
Nucleosides are compounds composed of a nitrogenous base and a sugar, but without the phosphate group. Nucleotides, on the other hand, contain all three: nitrogenous base, sugar, and phosphate group. Hence, the difference between a nucleoside and a nucleotide is that nucleotides contain one or more phosphate groups, whereas nucleosides have none. The correct option is D.
Question 3:
Replication requires both a template and a primer, whereas transcription requires only a template. This statement is true regarding the relationship between replication and transcription of DNA.Question 5:
H1 functions as a monomer is the option that is NOT true regarding histone proteins. The histone proteins are proteins that help to package the DNA into the nucleus of the cell. They are found in the nucleus, and the DNA is wrapped around a histone core to form nucleosome. The histones are the major protein component of chromatin. Histone proteins have five major classes: H1, H2A, H2B, H3, and H4, and H1, H2A, H3 and H4 form the nucleosome histone core. The positively coiled DNA is wrapped around a histone core to form nucleosome.
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An embryo exposed to a new teratogen caused all the ectoderm cells to be stuck at the dorsal surface of the embryo. The most likely explanation for this would be
That teratogen activated migration
The teratogen prevented cleavage
The teratogen prevented cell division
The teratogen blocked epiboly
The most likely explanation for the ectoderm cells being stuck at the dorsal surface of the embryo after exposure to the teratogen would be that the teratogen blocked epiboly.
Epiboly is a process during embryonic development in which cells from the animal pole of the embryo migrate and spread over the surface of the yolk. This movement allows for the proper positioning and organization of the germ layers, including the ectoderm. If the teratogen interferes with the process of epiboly, it can disrupt the normal movement of cells and result in the ectoderm cells being unable to properly spread and differentiate to their correct locations. In this case, the teratogen's effect on blocking epiboly would explain why the ectoderm cells are stuck at the dorsal surface of the embryo. The teratogen is preventing the normal migration and spreading of cells, leading to this abnormal localization of the ectoderm cells.
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All the following about telomerase and telomeres are true except: there is no telomerase activity in somatic cells. the telomerase enzyme carries an RNA template. telomerase adds RNA nucleotides to the lagging strand. O telomere length and age-at-death appear to be correlated. human telomeres are repeats of 6 nucleotides. Question 10 1 pts Which of the following statements about DNA replication is correct? Does not require a primer to replicate the leading strand. The lagging strand is synthesised discontinuously in the 5' to 3' direction. Proceeds in one direction from the origin of replication. Topoisomerase unwinds and separates the parental DNA strands. O Uses DNA ligase to form the H-bonds holding the two complementary strands together.
Humans are capacle of suriving with duplicated and deleted portions of chromosomes. O True O False
true
duplication and deletion often produce offspring that survive but exhibit physical and mental abnormalities
3. Fill in each of the blanks below with the correct term:
a) The presence of fat and acid in chyme stimulates the
__________________ to release ______________ and _______________
into the bloodstream
Major Component of Food (macromolecule) at ingestion: End product of chemical digestion (ie, absorbed as): Transported away from digestive system by 2. On the back of the page or on a separate page, c
The presence of fat and acid in chyme stimulates the small intestine to release secretin and cholecystokinin into the bloodstream.Secretin and cholecystokinin are hormones released by the small intestine
. These hormones are stimulated by the presence of fat and acid in chyme. Secretin stimulates the pancreas to release bicarbonate ions into the small intestine. Bicarbonate ions neutralize the acidic chyme, which helps protect the small intestine from damage. Cholecystokinin stimulates the gallbladder to release bile into the small intestine.
Bile is important for the digestion and absorption of fat.Major Component of Food (macromolecule) at ingestion:FatEnd product of chemical digestion (i.e., absorbed as):Fatty Acids and GlycerolTransported away from the digestive system by:Lymphatic System.
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What are the infective stage and diagnostic stages for the following diseases?
1. Giardia Lamblia.
2. Leishmania.
3. Ascaris lumbricoides.
4. Toxoplasma Gondi.
5. african trypanosomiasis.
6. Chagas disease.
7. Trichomoniasis Vagainalis.
8. Malaria.
9. Hookworms.
10. Enterobiasis.
11. Entermba Histolatika
1. Giardia Lamblia: Infective stage - Cyst; Diagnostic stage - Trophozoite.
2. Leishmania: Infective stage - Promastigote; Diagnostic stage - Amastigote.
Certainly! Here are the infective stages and diagnostic stages for the mentioned diseases:
1. Giardia Lamblia:
- Infective Stage: Cyst
- Diagnostic Stage: Trophozoite
2. Leishmania:
- Infective Stage: Promastigote
- Diagnostic Stage: Amastigote
3. Ascaris lumbricoides:
- Infective Stage: Eggs
- Diagnostic Stage: Adult worm
4. Toxoplasma gondii:
- Infective Stage: Oocyst
- Diagnostic Stage: Tissue cyst
5. African trypanosomiasis (Sleeping sickness):
- Infective Stage: Metacyclic trypomastigote
- Diagnostic Stage: Bloodstream trypomastigotes, intracellular amastigotes
6. Chagas disease:
- Infective Stage: Trypomastigote
- Diagnostic Stage: Amastigote
7. Trichomoniasis vaginalis:
- Infective Stage: Trophozoite
- Diagnostic Stage: Trophozoite (observed in vaginal secretions or urine)
8. Malaria:
- Infective Stage: Sporozoite
- Diagnostic Stage: Merozoite
9. Hookworms:
- Infective Stage: Infective larvae (L3)
- Diagnostic Stage: Adult worm, eggs in stool
10. Enterobiasis:
- Infective Stage: Infective eggs
- Diagnostic Stage: Adult pinworm or eggs on perianal swab or tape test
11. Entamoeba histolytica:
- Infective Stage: Cyst
- Diagnostic Stage: Trophozoite or cyst in stool sample.
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Imagine you are a health care professional and one of your patients just received a blood transfusion. When you go to check on her, you notice there is blood in her urine and is having breathing difficulties. You look at her chart and you notice that she is 0 - but received B+ blood! a. Explain your concern for the patient receiving B+ and why she is having these symptoms. Consider the respiratory, circulatory, and urinary systems and the role of antibodies in your answer. b. Erythropoietin may be used to correct this situation. Explain why.
The patient's symptoms of blood in urine and breathing difficulties after receiving B+ blood indicate a severe transfusion reaction due to an incompatible blood type. The presence of antibodies against the B antigen in the patient's blood, as a result of being blood type O, is causing the reaction. Erythropoietin can be used to help correct this situation by stimulating red blood cell production to compensate for the damage caused by the transfusion reaction.
a. The patient's symptoms of blood in urine and breathing difficulties suggest a severe transfusion reaction due to an incompatible blood type. The patient is blood type O but received B+ blood. Blood type is determined by the presence or absence of specific antigens on the surface of red blood cells. In this case, the patient's blood contains antibodies against the B antigen since blood type O individuals have naturally occurring antibodies against both A and B antigens.
When the patient received B+ blood, which contains the B antigen, the antibodies in the patient's blood recognized the foreign antigen and triggered an immune response. This immune response leads to the destruction of the transfused B+ red blood cells, causing the release of hemoglobin into the bloodstream. The presence of hemoglobin in the urine results in blood in the urine (hematuria).
The transfusion reaction can also lead to a systemic inflammatory response and damage to the respiratory and circulatory systems. The release of inflammatory mediators can cause fluid accumulation in the lungs, leading to breathing difficulties.
b. Erythropoietin is a hormone that stimulates the production of red blood cells in the bone marrow. In the given situation, erythropoietin may be used to correct the situation by stimulating red blood cell production. The transfusion reaction has likely caused significant damage to the patient's red blood cells, leading to a decreased number of functional red blood cells and subsequent anemia.
By administering erythropoietin, the production of new red blood cells can be increased, compensating for the damaged cells and improving oxygen-carrying capacity. This can help alleviate symptoms related to anemia and support the patient's overall recovery. However, it is important to address the underlying transfusion reaction and manage the patient's symptoms promptly and appropriately.
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Dragonfly larvae are voracious predators that eat just about any living animal that can fit in their mouths - including small fish and other dragonflies. Liz wants to know whether dragonflies will eat each other if there are plenty of small fish around. She puts three dragonflies in a tank with no fish, and three dragonflies in a tank with lots of fish. She makes sure that the water temperature, the size of the dragonflies, and the size of the tanks are the same between the two tanks. After 24 hours, she counts how many dragonflies were eaten.
Liz's experiment demonstrated that when there are plenty of small fish around, dragonflies are more likely to engage in cannibalistic behavior due to heightened competition for resources.
Liz conducted an experiment to determine if dragonflies would eat each other when there were plenty of small fish available. She placed three dragonflies in a tank without fish and three dragonflies in a tank with lots of fish. After 24 hours, she observed that the dragonflies in the tank with fish exhibited cannibalistic behavior, while those in the tank without fish did not.
Dragonfly larvae are known for their predatory nature and their ability to consume various small animals, including other dragonflies. Liz set up two tanks with identical conditions, except for the presence or absence of small fish. In the tank without fish, the dragonflies did not resort to cannibalism, indicating that they may have sought alternative food sources or simply refrained from preying on each other in the absence of other options.
However, in the tank with an abundance of small fish, the dragonflies displayed cannibalistic behavior by consuming each other. This behavior could be attributed to increased competition for resources, where the availability of plentiful fish triggered predatory instincts and intensified the competition among the dragonflies for food. Consequently, the dragonflies turned to cannibalism as a means of securing sustenance.
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Uranus Which planet receives the least sunlight? Which planet has a moon with the thickest atmosphere? Saturn Mercury x Which planet has a moon with the largest fraction of the planet's mass? Neptune Which planet has a moon with liquid nitrogen geysers! Uranus Which planet has an axial tilt closest to 90 degrees? Venus Which planet rotates in an orientation that is closest to the opposite direction that it revolves around the Sun?
1. Neptune receives the least sunlight.
2. Saturn's moon, Titan has the thickest atmosphere.
3. Saturn's moon, Titan has the largest fraction of the planet's mass.
4. Neptune's moon, Triton has liquid nitrogen geysers.
5. Uranus has an axial tilt closest to 90 degrees.
6. Venus rotates in an orientation that is closest to the opposite direction that it revolves around the Sun.
1. Neptune receives the least amount of sunlight among all the planets in our solar system. Its distance from the Sun, which is about 30 astronomical units (AU), contributes to its low light levels. The large distance results in reduced solar radiation reaching Neptune's surface. Additionally, Neptune's thick atmosphere further diminishes the sunlight that penetrates through.
2. Titan, the largest moon of Saturn, possesses the thickest atmosphere of any moon in our solar system. Its atmosphere is predominantly composed of nitrogen, similar to Earth's atmosphere, but with additional components like methane and ethane. The presence of a dense atmosphere on Titan is a consequence of its relatively low temperature and high surface pressure. This atmosphere creates a dense smog-like haze, making it challenging to observe Titan's surface in visible light.
3. Titan, Saturn's largest moon, has the largest fraction of its parent planet's mass compared to any other moon in the solar system. Titan's mass is approximately 1.8% of Saturn's total mass. This substantial mass ratio is a result of Titan being one of the largest moons in our solar system and Saturn's massive size.
4. Triton, the largest moon of Neptune, is known for its geysers that emit liquid nitrogen. These geysers, discovered by the Voyager 2 spacecraft in 1989, are a consequence of Triton's unique geological activity. The moon's surface is covered in a layer of nitrogen ice, and beneath it, there is believed to be a subsurface ocean of liquid water. Heat from Triton's interior, possibly generated by tidal forces, causes the nitrogen to erupt through cryovolcanic geysers, releasing plumes of gas and liquid.
5. Uranus has the closest axial tilt to 90 degrees among all the planets in our solar system. It has a tilt of about 98 degrees, resulting in an extreme tilt that causes its rotational axis to be almost parallel to the plane of its orbit around the Sun. As a result, Uranus experiences extreme seasonal variations, with one pole facing the Sun directly while the other pole remains in constant darkness for long periods. This unique orientation of Uranus' axial tilt is believed to have been caused by a collision with a massive object during its early formation.
6. Venus rotates in an opposite direction to the majority of other planets in our solar system. While most planets, including Earth, have a counterclockwise rotation when viewed from above the Sun's north pole, Venus rotates clockwise. This retrograde rotation of Venus is also referred to as "retrograde motion" or "retrograde spin." The exact reason for Venus' retrograde rotation is not yet fully understood, but it is believed to be the result of a complex series of interactions and collisions that occurred during the planet's formation.
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In order to do tissue culture, you need to understand your media. Address the following questions about your medias.
Most tissue culture cells are incubated in the presence of 5% CO2 and 95% air. Although CO2 is required for optimal cell growth, high concentrations of dissolved CO2 in the media will lower the pH, which will inhibit cell growth.
What causes the drop in pH and what substance is added to media to maintain the pH at 7.4? Please show chemical equations in your answer.
Why is Fetal Bovine Serum is heat-inactivated and what is the procedure for heat-inactivation and subsequent storage?
The heat-inactivation procedure kills the complement without damaging the other components in the FBS, thus ensuring that the cells are not destroyed during culture.
Tissue culture is a technique that is used to grow and maintain cells in vitro under controlled laboratory conditions. In order to do tissue culture, it is necessary to understand the media being used. The following questions address various aspects of the media that are used in tissue culture.
Most tissue culture cells are incubated in the presence of 5% CO2 and 95% air. While CO2 is necessary for optimal cell growth, high concentrations of dissolved CO2 in the media will lower the pH, which will inhibit cell growth. The drop in pH is caused by the accumulation of carbon dioxide (CO2) in the media. Carbon dioxide reacts with water in the media to produce carbonic acid (H2CO3), which ionizes to release hydrogen ions (H+) and bicarbonate ions (HCO3-):CO2 + H2O ⇌ H2CO3 ⇌ H+ + HCO3-Bicarbonate is added to media to maintain the pH at 7.4.
It acts as a buffer by binding to excess hydrogen ions and removing them from the media, thereby preventing the pH from dropping too low. The following equation shows how bicarbonate acts as a buffer:HCO3- + H+ ⇌ H2CO3Fetal bovine serum (FBS) is used in tissue culture as a supplement to provide nutrients, growth factors, and other critical components that are required for cell growth. FBS is heat-inactivated to inactivate any complement, which is a group of proteins that can destroy cells.
This is done by heating the FBS at 56°C for 30 minutes. Following heat-inactivation, the FBS is stored at -20°C until it is ready to be used.
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Yersinia pestis the causative agent of the bubonic plague, killed 50% of Europe's population in the middle ages. It is considered the most successful pathogen infecting humans. When engulfed by a macrophage the bacterium prevents acidification of the phagosome and is transported to the lymph nodes unharmed. Here it multiplies and creates the buboes (extremely swollen lymph nodes) that characterize this disease.
Though Y. pestis can grow in a macrophage, it is destroyed inside dendritic cells that process the antigen and initiate an adaptive immune response. If a person is to survive this infection without antibiotics, what type of cells would need to be activated by the dendritic cell to combat this pathogen as it replicates inside a macrophage.
plasma cells
Cytotoxic T cells
T-helper 1 cells
neutrophils
mast cells
As a result, activating Th1 cells by dendritic cells to combat this pathogen as it replicates inside a macrophage is crucial for the survival of an infected person without antibiotics.
When engulfed by a macrophage, the bacterium prevents the acidification of the phagosome and is transported to the lymph nodes unharmed, where it multiplies and creates the buboes (extremely swollen lymph nodes) that characterize this disease.
Though Y. pestis can grow in a macrophage, it is destroyed inside dendritic cells that process the antigen and initiate an adaptive immune response.
If a person is to survive this infection without antibiotics, T-helper 1 cells would need to be activated by the dendritic cell to combat this pathogen as it replicates inside a macrophage.
The T-helper 1 (Th1) cells are a subpopulation of T-helper cells, which are involved in the immune system's response to infectious agents, including viruses, intracellular bacteria, and parasites.
They stimulate macrophages to kill phagocytosed microbes, and the cytokines they produce, such as IFN-γ and IL-2, play a critical role in the response to intracellular pathogens.
The Th1 response is crucial for the host defense against Y. pestis, which is known to survive and replicate inside macrophages.
T-helper 1 cells activate macrophages through a series of signaling events, resulting in increased phagocytosis, enhanced intracellular killing, and the release of microbicidal molecules such as reactive oxygen species (ROS) and nitric oxide (NO).
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