This mutation is passed on to the offspring, leading to the evolution of limbless snakes. This mechanism explains how this leap could have occurred and why it occurs frequently because the mutation that causes limb loss is heritable and can be passed on to the offspring, leading to the evolution of limbless snakes in different lizard families.
1a. The statement, "most fossils of Mesozoic birds are from marine diving birds. Relatively few terrestrial species are known," means that there are more fossils of marine diving birds in the Mesozoic era than there are fossils of terrestrial birds. However, it doesn't necessarily imply that most Mesozoic birds were marine diving birds. There could have been more terrestrial bird species that are yet to be discovered, or maybe there were more terrestrial birds that didn't leave fossils behind. Hence, we cannot come to a conclusion based on incomplete evidence. 1b. The scientific method that could be used to determine whether the explosive radiation of Mesozoic birds is real or an artifact of the fossil record is "phylogenetic analysis." The process involves examining and comparing the DNA of different organisms to determine their evolutionary relationships. A phylogenetic analysis of Mesozoic bird fossils can help reveal their lineage and possible ancestors. If the sudden divergence of Mesozoic birds is real, we would see a rapid branching of their phylogenetic tree. On the other hand, if it's just an artifact of the fossil record, we wouldn't see such a rapid branching.2. The loss of limbs in snakes is an evolutionary leap that has happened in almost every lizard family at least once. The discovery of the developmental mechanism behind the loss of limbs in snakes explains how this leap could have occurred and why it occurs frequently. It's a result of a regulatory gene (Sonic Hedgehog) that determines the formation of limbs and other appendages in vertebrates. In snakes, there's a mutation in this gene that causes the limbs to stop developing. This mutation is passed on to the offspring, leading to the evolution of limbless snakes. This mechanism explains how this leap could have occurred and why it occurs frequently because the mutation that causes limb loss is heritable and can be passed on to the offspring, leading to the evolution of limbless snakes in different lizard families.
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A lot of attention has been dedicated to the so-called "cytokine storm" that can occur in patients with COVID-19. What are cytokines, and what is a cytokine storm? Why are they potentially life-threatening? What is one potential therapeutic that is being developed to combat the cytokine storm?
Cytokines are proteins produced by cells of the immune system that serve as signaling molecules to stimulate an immune response to fight off infections.
The cytokine storm is a severe immune reaction in which the body produces high levels of cytokines that can damage tissues and organs. This can cause fever, fatigue, and inflammation, which can lead to organ failure, respiratory distress, and potentially death.
Cytokine storm is a potentially life-threatening condition because it can cause severe damage to various tissues and organs in the body, leading to multiple organ failure and ultimately death. The cytokine storm is more likely to occur in individuals with weakened immune systems, and those with preexisting medical conditions such as diabetes, hypertension, and cardiovascular disease.
There is no cure for cytokine storm syndrome. Treatment typically involves supportive care to manage the symptoms and complications associated with the condition. However, researchers are currently working on developing a therapeutic called tocilizumab to combat the cytokine storm. Tocilizumab is a monoclonal antibody that targets a cytokine called interleukin-6, which is responsible for triggering the cytokine storm.
By blocking this cytokine, tocilizumab may help to reduce the severity of the cytokine storm and improve patient outcomes.
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Corepressors...
a. bind to the operator to prevent transcription
b. bind to the ribosome to prevent translation
c. bind to the repressor to activate it
d. none of the above
Repressors are proteins that inhibit the transcription of specific genes. Corepressors, or regulatory proteins that enhance repressor activity by binding to it, may also be used to control gene expression.
The answer is: a. bind to the operator to prevent transcription.Content-loaded Corepressors, which bind to the operator, prevent transcription. In molecular biology, gene expression regulation is the process by which a cell controls which genes are turned on (expressed) or off (silenced). Gene expression is managed by turning transcription on or off, which is the process of synthesizing RNA from DNA. Repressors are proteins that inhibit the transcription of specific genes. Corepressors, or regulatory proteins that enhance repressor activity by binding to it, may also be used to control gene expression.
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Phosphofructokinase is considered to be the enzyme responsible for controlling the rate-limiting step of the glycolytic pathway. Why would this step be considered rate-limiting step? Explain with reference to the feedback mechanisms that are occurring in cell respiration.
Phosphofructokinase is considered to be the enzyme responsible for controlling the rate-limiting step of the glycolytic pathway. This step is considered rate-limiting step because the activity of phosphofructokinase is subject to allosteric feedback control.
Phosphofructokinase is a regulatory enzyme of the glycolytic pathway. It is the enzyme that catalyzes the conversion of fructose 6-phosphate to fructose 1,6-bisphosphate. This reaction is an irreversible one and is a committed step in the glycolytic pathway. In order to maintain the appropriate levels of ATP in the cell, the activity of phosphofructokinase is subject to allosteric feedback control.
Feedback mechanisms in cell respiration refer to the regulatory mechanisms that exist in the cell that can regulate the rate of respiration. In cells, phosphofructokinase is inhibited by ATP, citrate, and high levels of NADH. These molecules are allosteric inhibitors that bind to the enzyme and change its conformation. This results in a decrease in the activity of the enzyme, which in turn slows down the rate of respiration. In contrast, ADP, AMP, and low levels of NADH are allosteric activators of phosphofructokinase. They bind to the enzyme and stimulate its activity, which in turn increases the rate of respiration
Phosphofructokinase is considered to be the enzyme responsible for controlling the rate-limiting step of the glycolytic pathway. This step is considered rate-limiting step because the activity of phosphofructokinase is subject to allosteric feedback control. In cells, phosphofructokinase is inhibited by ATP, citrate, and high levels of NADH. These molecules are allosteric inhibitors that bind to the enzyme and change its conformation. In contrast, ADP, AMP, and low levels of NADH are allosteric activators of phosphofructokinase. They bind to the enzyme and stimulate its activity, which in turn increases the rate of respiration.
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Signal transduction- yeast genetics
in one sentence, what does alpha factor in the WT 'a' cell do?
(In terms of cell cycle/budding and FUS1 transcription)
In terms of cell cycle/budding and FUS1 transcription, the alpha factor in the WT 'a' cell induces the pheromone response pathway, leading to cell cycle arrest and activation of transcription factors that initiate FUS1 transcription.
In Saccharomyces cerevisiae, alpha factor is a peptide pheromone that activates a cell signaling pathway that controls mating and cell cycle progression. Alpha factor activates the G protein-coupled receptor, Ste2p, initiating a cascade of signal transduction events that result in the activation of the mitogen-activated protein kinase (MAPK) pathway. The pheromone response pathway results in cell cycle arrest and activation of transcription factors that initiate the transcription of mating-specific genes, including the FUS1 gene.
FUS1 encodes a protein involved in cell fusion and mating. The pheromone response pathway is a model system for studying signal transduction in yeast genetics, as many of the signaling proteins and pathways are conserved in higher eukaryotes.
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What is the function of Troponin C, Troponin I and Troponin T? How do they each cause muscle contraction? Include detail
Troponin C, Troponin I, and Troponin T are three subunits of the troponin complex found in muscle cells. They play crucial roles in regulating muscle contraction, specifically in skeletal and cardiac muscles.
Troponin C (TnC): Troponin C is a calcium-binding protein that is essential for muscle contraction. It binds to calcium ions (Ca2+) when the concentration of Ca2+ increases in the cytoplasm of muscle cells, triggering a series of events that lead to muscle contraction.
Troponin I (TnI): Troponin I is another subunit of the troponin complex that inhibits the interaction between actin and myosin, two key proteins involved in muscle contraction. Troponin I prevents muscle contraction in the absence of calcium ions. When calcium ions bind to troponin C, it causes a conformational change in troponin I, relieving its inhibitory effect on actin.
Troponin T (TnT): Troponin T is the third subunit of the troponin complex and plays a structural role in muscle contraction. Troponin T binds to tropomyosin, another protein that is associated with the actin filament. When troponin C binds to calcium ions, it induces a conformational change in troponin T, which in turn shifts the position of tropomyosin.
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When blood pressure increases, Multiple Choice O O O baroreceptors detect the change in the carotid arteries. the cardioregulatory center decreases parasympathetic stimulation heart rate and stroke vo
When blood pressure increases, baroreceptors detect the change in the carotid arteries, and the cardioregulatory center decreases parasympathetic stimulation, resulting in an increase in heart rate and stroke volume.
Baroreceptors are specialized sensory receptors located in the carotid arteries and aortic arch that detect changes in blood pressure. When blood pressure increases, these baroreceptors are activated and send signals to the cardioregulatory center in the brain.
The cardioregulatory center, which is part of the autonomic nervous system, responds to the increased blood pressure by decreasing parasympathetic stimulation and increasing sympathetic stimulation. This leads to a decrease in vagal tone (parasympathetic activity) and an increase in sympathetic activity.
The decrease in parasympathetic stimulation results in a decrease in the release of acetylcholine, which normally slows down the heart rate. As a result, the heart rate increases.
Additionally, the increase in sympathetic activity leads to the release of norepinephrine, which increases the force of contraction of the heart muscle, resulting in an increased stroke volume.
Overall, these responses work together to help normalize blood pressure by increasing cardiac output and maintaining adequate perfusion to the body's tissues.
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<The complete question is>
When blood pressure increases, Multiple Choice Option 1. baroreceptors detect the change in the carotid arteries. 2.the cardioregulatory center decreases 3. parasympathetic stimulation heart rate and stroke volume increase, 4.norepinephrine secretion increase
a) Compare and contrast the basal states of glucocorticoid and retinoid X receptors and their activation mechanisms by their cognate steroid hormones which lead to gene transcription. (20 marks)
Glucocorticoid Receptor (GR) and Retinoid X Receptor (RXR) are both nuclear receptors that function as transcription factors.
Here is a comparison and contrast of their basal states and activation mechanisms:
Basal State:
Glucocorticoid Receptor (GR): In the absence of its ligand (e.g., cortisol), the GR resides in the cytoplasm as part of a multiprotein complex.
Retinoid X Receptor (RXR): RXR can exist in both the cytoplasm and the nucleus.
Activation Mechanisms:
Glucocorticoid Receptor (GR): Upon binding of cortisol (the cognate hormone), the GR undergoes a conformational change, leading to dissociation from HSPs.
Retinoid X Receptor (RXR): RXR can be activated by its cognate ligand, 9-cis retinoic acid (9-cis RA), or through heterodimerization with other nuclear receptors.
Gene Transcription:
Glucocorticoid Receptor (GR): Activation of the GR by cortisol leads to the recruitment of coactivators to the GREs on target genes.
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How does the major difference between the heart of a frog and a
pig affect the blood?
The main difference between the heart of a frog and a pig is that a frog has a three-chambered heart while a pig has a four-chambered heart. This difference in heart structure affects how the blood flows through the body.
Frogs have a three-chambered heart that consists of two atria and one ventricle. The atria receive oxygen-poor blood from the body and oxygen-rich blood from the lungs, respectively. The ventricle then pumps the blood out to the rest of the body.
Because of the single ventricle, blood from both atria is mixed together before being pumped out. This means that oxygen-poor blood may mix with oxygen-rich blood, which lowers the overall oxygen content of the blood.
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Which of the following would decrease glomerular filtration rate? Vasodilation of the efferent arteriole Vasoconstriction of the afferent arteriole Atrial natriuretic peptide (ANP) All of the above
W
Vasoconstriction of the afferent arteriole would decrease the glomerular filtration rate.
Glomerular filtration rate (GFR) is the measure of the amount of blood filtered by the glomeruli of the kidneys per minute. The GFR helps in estimating the kidney's overall function. It is a key indicator of kidney function in both diagnosing and monitoring chronic kidney disease (CKD).
It is estimated by the rate of clearance of creatinine in a patient’s blood. Kidney function is severely impacted when the GFR falls below 15 mL/min.
There are three different factors that can affect glomerular filtration rate.
Efferent arteriole constriction
Afferent arteriole dilation
Decreased capillary blood pressure
All of the above-listed factors would increase the glomerular filtration rate.
Therefore, the only factor that would decrease the GFR is "Vasoconstriction of the afferent arteriole."
Thus, this is the correct option.
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The correct sequence of layers in the wall of the alimentary canal, from internal to external, is a.mucosa, muscularis, serosa, submucosa. b.submucosa, mucosa, serosa, muscularis. c.mucosa, submucosa, muscularis, serosa. d.serosa, muscularis, mucosa, submucosa.
The correct sequence of layers in the wall of the alimentary canal, from internal to external, is mucosa, submucosa, muscularis, serosa.
The correct option is C.
Mucosa, submucosa, muscularis, serosa.What is the alimentary canal?The alimentary canal is a muscular tube that begins at the mouth and extends through the pharynx, esophagus, stomach, small intestine, and large intestine to the anus. It is composed of four distinct layers of tissues that function together to perform digestion and absorption of nutrients from food.
These layers are referred to as mucosa, submucosa, muscularis, and serosa.The four layers of the alimentary canal are:Mucosa: The mucosa is the innermost layer of the alimentary canal. It is made up of three layers of tissues: the epithelium, the lamina propria, and the muscularis mucosae. It produces mucus, enzymes, and hormones that aid in digestion.Submucosa: The submucosa is the second layer of the alimentary canal. It is composed of connective tissues that contain blood vessels, nerves, and lymphatics. It also contains glands that produce mucus, enzymes, and hormones.
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prevent hemolytic anemia of the newborn, a Rh negative mother who has a Rh positive newborn is given this: _______________
To prevent hemolytic anemia of the newborn, a Rh negative mother who has an Rh positive newborn is given Rh immune globulin (RhIG).
Hemolytic disease of the newborn, also known as Rh disease, occurs when a Rh negative mother is sensitized to the Rh positive blood of her fetus during pregnancy or childbirth. This sensitization can lead to the production of antibodies that can cross the placenta and attack the red blood cells of subsequent Rh positive pregnancies, causing hemolytic anemia in the newborn. To prevent this condition, Rh negative mothers are typically given Rh immune globulin (RhIG), also known as Rho(D) immune globulin. RhIG is a blood product that contains antibodies against the Rh factor. When administered to a Rh negative mother, RhIG binds to any Rh positive fetal blood cells that may have entered her bloodstream during pregnancy or childbirth. This prevents her immune system from recognizing these cells as foreign and forming antibodies against them. As a result, the RhIG helps prevent sensitization and the subsequent development of hemolytic anemia in future pregnancies. By providing passive immunity against the Rh antigen, RhIG effectively reduces the risk of Rh disease in Rh negative mothers with Rh positive newborns, ensuring the health and well-being of the newborn.
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Which of the following statements is INCORRECT about mutualisms? Species do not usually engage in mutualisms for altruistic reasons, In some mutualisms, one or the other partner, under certain environmental conditions will withdraw the reward it usually provides There is an inherent conflict of interest between the partners in a mutualism, For an ecological interaction to be a mutualism, the net benchts must exceed the net costs for one partner but not the other.
The incorrect statement about mutualisms is that "There is an inherent conflict of interest between the partners in a mutualism".
Mutualism is a type of ecological interaction in which two species benefit from one another. There are different types of mutualisms that may vary in terms of the balance of benefits and costs between partners.
In mutualisms, there is generally not an inherent conflict of interest between the partners, as both species benefit from the interaction. In contrast, in some other types of ecological interactions such as predation or competition, there is often a conflict of interest between the interacting species. So, the third statement is the incorrect statement about mutualisms because it goes against the very essence of what mutualism stands for.
Therefore, the statement "There is an inherent conflict of interest between the partners in a mutualism" is incorrect about mutualisms.
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In contrast to Mitosis where the daughter cells are exact copies (genetically identical) of the parent cell, Meiosis results in genetically different cells, that will eventually also have the potential to create genetically unique offspring. But meiosis and mitosis are different in many other ways as well. Watch the videos and view the practical presentation. You will view stages of Meiosis in the Lily Anther EXERCISE 1: View the different stages of Meiosis occurring in the Lily Anther under the microscope. 1.1 Identify and draw Prophase I OR Prophase Il of Meiosis, as seen under the microscope. Label correctly (5) 1.2 What happens in Prophase I which does not occur Prophase II? (2) 1.3 Define: a. Homologous chromosome? (2) b. Synapsis (2) c. Crossing over (2) d. Chiasma (1) 1.4 Why is that siblings don't look identical to each other? (5)
Meiosis is the process in which genetically different cells are created, and they also have the potential to generate genetically unique offspring. The daughter cells produced in Mitosis are exact copies of the parent cell (genetically identical).
There are, however, several other distinctions between meiosis and mitosis. The stages of Meiosis in the Lily Anther are shown in the videos and the practical presentation.1.1 Prophase I of Meiosis, as seen under the microscope, is identified and sketched.
Correct labeling is done. 1.2 Unlike Prophase II, Prophase I involves synapsis and crossing over. 1.3 a. Homologous chromosomes are chromosomes that have similar genes, but they can carry distinct alleles. b. The pairing of homologous chromosomes is known as synapsis. c.
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SDS-PAGE can only efficiently separate proteins since:
- the pores of the polyacrylamide gel are smaller compared with
agarose gel
- DNA is more negative
- proteins are smaller compared with DNA
- SDS
SDS-PAGE can efficiently separate proteins because the pores of the polyacrylamide gel used in SDS-PAGE are smaller compared to an agarose gel, allowing for better resolution and separation of proteins based on their size and molecular weight.
SDS-PAGE (Sodium Dodecyl Sulfate-Polyacrylamide Gel Electrophoresis) is a widely used technique in molecular biology and biochemistry to separate proteins based on their molecular weight. It is a powerful tool due to several factors, one of which is the size of the pores in the gel matrix.
Polyacrylamide gels used in SDS-PAGE have smaller pore sizes compared to agarose gels, which are commonly used for separating nucleic acids like DNA. The smaller pore size of the polyacrylamide gel allows for more efficient separation of proteins. The proteins are forced to move through the gel matrix during electrophoresis, and their migration is impeded by the size of the pores. Smaller proteins can move more easily through the smaller pores, while larger proteins are hindered and migrate more slowly.
By applying an electric field, the proteins in the sample are separated based on their size and molecular weight. SDS (Sodium Dodecyl Sulfate) is a detergent used in SDS-PAGE that denatures the proteins and imparts a negative charge to them, making them move toward the positive electrode during electrophoresis. This further aids in the separation of proteins based on their molecular weight.
In summary, SDS-PAGE efficiently separates proteins due to the smaller pore size of the polyacrylamide gel, which allows for better resolution and separation based on size and molecular weight.
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Three genotypes in a very large population have, on average, the following values of survival and fecundity, regardless of their relative frequencies: Genotype A1A1 A1A2 A2A2 Survival to adulthood (viability) 0.80 0.90 0.50 Number of offspring 3.0 4.0 8.0 Absolute fitness 2.4 3.6 4.0 Which of the following best describes what will happen at this locus in the long run? There will be a stable polymorphism because the heterozygote has a higher survival rate than either homozygote. Nothing will happen because the differences among genotypes in survival and fecundity cancel each other out. Allele A2 will be fixed eventually. One allele will be fixed but we cannot predict which one. Allele Al will be fixed eventually.
The population under observation has three genotypes: A1A1, A1A2, and A2A2. These genotypes have survival rates of 0.80, 0.90, and 0.50, and fecundity rates of 3.0, 4.0, and 8.0, respectively.
The absolute fitness of these genotypes is 2.4, 3.6, and 4.0, respectively. Which of the following statements best describes what will happen to the locus in the long run? Allele A2 will eventually become fixed is the correct option. This is due to the fact that allele A2 has the highest fitness of the three alleles, with a fitness of 4.0, and will thus outcompete the other two alleles in the population over time. Eventually, A2 will become the only allele present in the population because it is more effective at reproducing and surviving than A1. Over time, A2 will increase in frequency while A1 will decrease, and ultimately, A2 will become fixed in the population because it will be the only allele remaining.
Therefore, allele A2 will be fixed eventually. The statement "There will be a stable polymorphism because the heterozygote has a higher survival rate than either homozygote" is incorrect.
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You are interested in developing CRISPR mutation alleles of human gene CCR5. You first look up the gene sequence on public database GenBank. Based on the sort of mutant alleles you want to create you decide to design 3 guide RNA target sites within the first 1000bp of the gene (shown below).
Each target site should be 20 bp long and it must have a protospacer adjacent motif (PAM), which has the form NGG, immediately downstream (3’) of the target site. N means any base. The DNA sequence below shows the coding strand only, in the 5’--> 3’ direction.
1 cttcagatag attatatctg gagtgaagaa tcctgccacc tatgtatctg gcatagtgtg 61 agtcctcata aatgcttact ggtttgaagg gcaacaaaat agtgaacaga gtgaaaatcc 121 ccactaagat cctgggtcca gaaaaagatg ggaaacctgt ttagctcacc cgtgagccca 181 tagttaaaac tctttagaca acaggttgtt tccgtttaca gagaacaata atattgggtg 241 gtgagcatct gtgtgggggt tggggtggga taggggatac ggggagagtg gagaaaaagg 301 ggacacaggg ttaatgtgaa gtccaggatc cccctctaca tttaaagttg gtttaagttg 361 gctttaatta atagcaactc ttaagataat cagaattttc ttaacctttt agccttactg 421 ttgaaaagcc ctgtgatctt gtacaaatca tttgcttctt ggatagtaat ttcttttact 481 aaaatgtggg cttttgacta gatgaatgta aatgttcttc tagctctgat atcctttatt 541 ctttatattt tctaacagat tctgtgtagt gggatgagca gagaacaaaa acaaaataat 601 ccagtgagaa aagcccgtaa ataaaccttc agaccagaga tctattctct agcttatttt 661 aagctcaact taaaaagaag aactgttctc tgattctttt cgccttcaat acacttaatg 721 atttaactcc accctccttc aaaagaaaca gcatttccta cttttatact gtctatatga 781 ttgatttgca cagctcatct ggccagaaga gctgagacat ccgttcccct acaagaaact 841 ctccccggta agtaacctct cagctgcttg gcctgttagt tagcttctga gatgagtaaa 901 agactttaca ggaaacccat agaagacatt tggcaaacac caagtgctca tacaattatc 961 ttaaaatata atctttaaga taaggaaagg gtcacagttt ggaatgagtt tcagacggtt 1021 ataacatcaa agatacaaaa catgattgtg agtgaaagac tttaaaggga gcaatagtat
Come up with 3 guide RNA target sites
Three guide RNA target sites within the first 1000 base pairs of the CCR5 gene, each 20 bp long with a PAM (NGG) immediately downstream: Target Site 1: 61-80 bp (AGTCCTCATAAATGCTTACT), Target Site 2: 101-120 bp (CCACCTAAGATCCTGGGTCC), Target Site 3: 181-200 bp (TAGTTAAAACTCTTTAGACA).
What are three guide RNA target sites within the first 1000 base pairs of the CCR5 gene, each 20 bp long with a protospacer adjacent motif (PAM) in the form of NGG immediately downstream?Based on the given DNA sequence, we need to design three guide RNA target sites within the first 1000 base pairs (bp) of the CCR5 gene. Each target site should be 20 bp long and have a protospacer adjacent motif (PAM) in the form of NGG immediately downstream of the target site.
Here are three possible guide RNA target sites:
Target Site 1: 61-80 bp
Target sequence: AGTCCTCATAAATGCTTACT
PAM sequence: GGT
Target Site 2: 101-120 bp
Target sequence: CCACCTAAGATCCTGGGTCC
PAM sequence: AGA
Target Site 3: 181-200 bp
Target sequence: TAGTTAAAACTCTTTAGACA
PAM sequence: AAA
For Target Site 1, we selected the sequence starting from position 61 and ending at position 80. The target sequence is AGTCCTCATAAATGCTTACT, and the PAM sequence is GGT.
For Target Site 2, we chose the sequence starting from position 101 and ending at position 120. The target sequence is CCACCTAAGATCCTGGGTCC, and the PAM sequence is AGA.
For Target Site 3, we selected the sequence starting from position 181 and ending at position 200. The target sequence is TAGTTAAAACTCTTTAGACA, and the PAM sequence is AAA.
These guide RNA target sites can be used for CRISPR-Cas9 gene editing experiments to introduce specific mutations in the CCR5 gene.
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Fatty acid breakdown generates a large amount of acetyl CoA. What will be the effect of fatty acid breakdown on the activity of the Pyruvate Dehydrogenase (PDH) Complex?
a. The activity of the PDH complex would remain the same b. The activity of the PDH complex would decrease c. The activity of the PDH complex would increase
Pyruvate dehydrogenase (PDH) complex is a cluster of multienzyme that facilitates the conversion of pyruvate into acetyl-CoA.
Acetyl-CoA is a critical energy-generating molecule that helps provide energy to the human body. The PDH complex is regulated via negative feedback inhibition, which helps to control the rate of metabolism of pyruvate. Negative feedback inhibition happens when high energy levels in the body act as an inhibitor to metabolic pathways, leading to a reduction in enzyme activity.
Acetyl-CoA is a compound that is produced by a range of metabolic pathways, including fatty acid breakdown. When there is an increase in acetyl-CoA, the body will increase the activity of the Pyruvate Dehydrogenase (PDH) Complex. It's because Acetyl-CoA also serves as a key regulator of PDH activity.Acetyl-CoA regulates the activity of the PDH complex by inhibiting its activity. When there is an increase in acetyl-CoA, the PDH complex will be inhibited, which will help to control the rate of metabolism of pyruvate. Thus, we can say that the activity of the PDH complex would increase when there is an increase in acetyl-CoA.
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rect Question 7 0/0.57 pts Which is NOT true of attachment or adherance? may be due to fimbriae may be due to capsules (glycocalyx) may be due to biofilms only normal microbiota can attach necessary i
Adherence can be due to different factors such as fimbriae, capsules, and biofilms. However, normal microbiota is not the only microbe that can attach itself to a surface.
Attachment or adherence is the ability of bacteria to attach themselves to a surface. Adherence can be due to different factors such as fimbriae, capsules, and biofilms. However, normal microbiota is not the only microbe that can attach itself to a surface. Below are the detailed explanations of each factor that can cause adherence: Fimbriae: Bacteria have hair-like structures known as fimbriae that are used to attach themselves to surfaces. The fimbriae help the bacteria to stick to cells and surfaces within the host. Capsules (Glycocalyx): Capsules, a thick layer of polysaccharides, are used by bacteria to form a physical barrier around themselves to prevent phagocytosis. The capsules aid in the adhesion process by binding to host cells.
Biofilms: Biofilms are bacterial communities that form on surfaces. Biofilm formation is one of the most important virulence factors of bacteria. The biofilm allows the bacteria to form a sticky matrix that attaches to a surface and helps in the accumulation of nutrients. Normal microbiota: The bacteria that are commonly found in the human body or other animals are called normal microbiota. Normal microbiota, such as Lactobacillus in the vagina, produce an acidic environment that discourages the growth of pathogenic bacteria. However, it is not the only microbe that can attach itself to a surface. In summary, the statement that only normal microbiota can attach is not true.
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1. Select the ncRNA that facilitates the binding of telomerase
to the telomere and acts as a template for DNA replication.
Select one:
a. TERC
b. snRNA
c. SRP RNA
d. Xist RNA
The ncRNA that facilitates the binding of telomerase to the telomere and acts as a template for DNA replication is TERC.
ncRNA stands for non-coding RNA which does not have protein-coding instructions but perform various important cellular functions including RNA splicing, regulation of gene expression, RNA processing, and stability.The TERC RNA (telomerase RNA component) is an RNA molecule that acts as a template for the DNA replication.
It serves as a functional and structural subunit of telomerase, a ribonucleoprotein that adds a specific DNA sequence repeat to the 3′ end of DNA strands of chromosomes.The binding of telomerase to telomeres is facilitated by TERC RNA. In addition to TERC RNA, telomerase comprises a protein catalytic subunit (TERT) and associated proteins. TERC RNA provides the template for the synthesis of new DNA strands that add repeats of telomeric DNA to the ends of the chromosome.
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there are 4 rows of DNA sample the first row is my professor, the
second is mines and the last 2 are my classmates
the
first row is the sample that was at the crime sence(joe sample) and
t
Crime Scene lab homework assignment for Unit Name: Complete your homework assignment directly on this page, tear it from the lab book and hand it into the Instructor at the beginning of the next lab p
The given information mentions the presence of 4 rows of DNA samples with each row associated with a different person. The first row refers to the DNA sample that was found at the crime scene (Joe Sample). The second row is yours, the third row belongs to one of your classmates, and the fourth row belongs to another classmate. The task seems to involve analyzing the DNA samples to identify the perpetrator of the crime.
As per the given information, Joe Sample's DNA sample was found at the crime scene. Therefore, the DNA samples from the remaining three rows need to be compared with Joe Sample's DNA to identify the perpetrator. If a match is found, then the person whose DNA matches with Joe Sample's DNA is the perpetrator of the crime. If there is no match, then the perpetrator is none of the three people whose DNA samples were analyzed.
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After cloning an insert into a plasmid, determining its orientation is best accomplished with ... O Two restriction endonucleases that cut in the insert. O Two restriction endonuclease, one that cuts once within the insert and the other that cuts once in the plasmid backbone. A single restriction endonuclease that cuts twice to release the insert. A single endonuclease that cuts twice in the plasmid backbone.
The answer is that when a foreign DNA fragment is inserted into a cloning vector, the orientation of the insert is crucial.
After cloning an insert into a plasmid, determining its orientation is best accomplished with two restriction endonucleases, one that cuts once within the insert and the other that cuts once in the plasmid backbone.
The correct orientation of the insert guarantees that the promoter and terminator sequences in the plasmid will be effective. The incorrect orientation of the insert will result in the inactivation of the promoter and terminator sequences in the plasmid. Therefore, to ensure the correct orientation of the insert, it is necessary to perform a diagnostic restriction enzyme digestion. The two enzymes selected should have recognition sites that cut the plasmid in one site and the insert in another site. The end result is to get two bands on a gel, which confirms the orientation of the insert. One band should correspond to the uncut plasmid, while the other should correspond to the plasmid cut by the restriction enzyme. The band's size will differ depending on the position of the restriction enzyme site in the insert. Determining the orientation of the insert in the vector is crucial because if the insert's orientation is reversed, the inserted gene's reading frame may be disrupted, leading to a complete loss of function. A gene inserted in reverse orientation with respect to the promoter and terminator is in the opposite orientation, making it impossible to transcribe and translate the protein properly. Diagnostic restriction enzyme digestion is one of the techniques used to determine the orientation of the insert in the plasmid. Two different restriction enzymes are used to digest the plasmid DNA. One of the restriction enzymes must cleave the insert DNA, while the other must cleave the plasmid DNA. As a result, two fragments are generated, one of which is the original, unaltered plasmid, while the other is a plasmid containing the inserted DNA. The length of the fragment with the insert and the distance between the restriction enzyme cleavage site in the insert and the site in the plasmid will determine the insert's orientation in the plasmid. In conclusion, determining the insert's orientation in the plasmid is critical for efficient expression of the inserted gene. Therefore, it is best accomplished using two restriction enzymes, one that cuts once within the insert and the other that cuts once in the plasmid backbone.
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Question 21 Fertilization makes a ... cell, which is called .... O haploid-zygote Ohaploid- ovum O diploid-ovum O diploid-zygote 2.5 pts
Fertilization makes a diploid cell, which is called a zygote.Fertilization is the fusion of the male and female gametes to form a zygote. It takes place in the oviduct. During fertilization, a haploid sperm nucleus fuses with a haploid egg nucleus to create a diploid zygote. It is one of the most important reproductive processes.
When the egg and sperm combine, it creates a single cell that contains all of the genetic material needed to create a human being.A zygote is the initial diploid cell that results from the fusion of two haploid gametes during fertilization. It is a single cell that contains all of the genetic information required to produce a human being. The zygote is the first stage of embryonic development.
It begins to divide rapidly and undergoes numerous rounds of cell division, which leads to the formation of an embryo.Zygotes are diploid cells, which means they contain two complete sets of chromosomes. One set is inherited from the mother, and the other is inherited from the father. The zygote divides into two cells during the first stage of embryonic development. These two cells divide into four, and so on, as the embryo continues to grow and develop.
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Which of the following will most likely disrupt the Hardy-Weinberg equilibrium that xists for a population of small rodents ving in a habitat with ample resources? a. The rodents reproduce frequently and have large litters, so the population size is increasing. b. Mate selection is completely random within the population of rodents. c. The population continues to remain isolated from other populations of the rodent. d. The coding region of a gene is altered in sperm produced by a particular male that mates with several of the female rodents, which produce many progeny as a result.
The option that is most likely to disrupt the Hardy-Weinberg equilibrium in a population of small rodents living in a habitat with ample resources is: The coding region of a gene is altered in sperm produced by a particular male that mates with several of the female rodents, which produce many progeny as a result. So, option D is accurate.
The Hardy-Weinberg equilibrium describes the genetic equilibrium that occurs in an ideal, non-evolving population. It is based on several assumptions, including random mating, no genetic drift, no gene flow, no mutation, and no selection.
In this scenario, if the coding region of a gene is altered in the sperm produced by a male and is passed on to a large number of progeny, it introduces a genetic change into the population. This alteration can disrupt the equilibrium by changing the allele frequencies. As the altered gene spreads through the population, it can result in a departure from the expected genotype frequencies predicted by the Hardy-Weinberg equilibrium.
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4 The hypothalamus * O acts as a link between the nervous and endocrine systems. releases hormones that travel to the pituitary gland. is actually part of the brain. all of the above Which statement about steroid hormones is correct? * They are very soluble in blood. They are derived from cholesterol. They are hydrophilic. They are composed of amino acids. . The endocrine system releases * electrical messages that travel through neurons. hormones that travel through the bloodstream. proteins that alter gene regulation. all of the above.
The hypothalamus is a part of the brain that acts as a link between the nervous and endocrine systems, releases hormones that travel to the pituitary gland, and is actually part of the brain.
Steroid hormones are derived from cholesterol. The endocrine system releases hormones that travel through the bloodstream.An explanation is needed to understand these answers and why they are correct. So, let's get started:The hypothalamus * O acts as a link between the nervous and endocrine systems. releases hormones that travel to the pituitary gland. is actually part of the brain.
The hypothalamus is actually a part of the brain that functions as a link between the nervous and endocrine systems. It regulates homeostasis, hunger, thirst, body temperature, circadian rhythms, sleep, emotional behavior, and other autonomic activities, as well as the release of hormones. It produces hormones such as oxytocin and vasopressin, which are released into the bloodstream by the pituitary gland. Steroid hormones are derived from cholesterol.
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Which of the following cells is haploid? O Daughter spermatogonium O Primary spermatocyte O Secondary spermatocyte O Mother spermatogonium
Option c is correct. The haploid cell in question is the secondary spermatocyte because they contain half the number of chromosomes compared to the original diploid cells.
In the process of spermatogenesis, which occurs in the testes, diploid cells called spermatogonia undergo mitotic divisions to produce primary spermatocytes. These primary spermatocytes then undergo the first meiotic division, resulting in the formation of haploid cells known as secondary spermatocytes. The secondary spermatocytes are haploid because they contain half the number of chromosomes compared to the original diploid cells. These haploid cells further undergo the second meiotic division to generate spermatids, which eventually mature into sperm cells.
It is important to note that the daughter spermatogonium and mother spermatogonium are diploid cells, as they have the same number of chromosomes as the original spermatogonium. The primary spermatocyte is also diploid because it has not undergone meiosis yet. Only after the first meiotic division does the cell become haploid, resulting in the formation of secondary spermatocytes.
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What are the implications for exercise training with aging,
mitochondrial myopathies, diabetes, and obesity?
As an individual ages, mitochondrial function naturally declines, which has implications for exercise training. Additionally, mitochondrial myopathies, diabetes, and obesity all impact mitochondrial function and can affect exercise training differently.
Implications for exercise training with agingAs people age, their mitochondrial function decreases, leading to reduced aerobic capacity, a reduction in muscle mass, and a decrease in overall exercise performance. However, regular exercise can help preserve mitochondrial function, increase muscle mass, and improve overall health.
Implications for exercise training with mitochondrial myopathiesMitochondrial myopathies are a group of diseases caused by a malfunction in the mitochondria. Because the mitochondria produce the energy necessary for exercise, individuals with mitochondrial myopathies may experience fatigue, muscle weakness, and difficulty exercising.
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Q10 How does transferring the mating mixtures from YED to CSM-LEU-TRP plates allow us to select for diploids (i.e. why can only diploids survive on this media)? ( 2 )
Q11 What does the colour and growth of colonies on these plates suggest to you about the gde genotype and mating type of the strains X and Y ? Explain your answer. (6) Q12 Suggest two advantages that diploidy has over haploidy (for the organism concerned) Q13 Why do you think the ability of yeast to exist as haploid cells is an advantage to geneticists? ( 2 )
Transferring the mating mixtures from YED (yeast extract dextrose) plates to CSM-LEU-TRP (complete synthetic medium lacking leucine and tryptophan) plates allows us to select for diploids because the CSM-LEU-TRP plates lack these two essential amino acids, The color and growth of colonies on the CSM-LEU-TRP plates can provide information about the gde genotype and mating type of the strains X and Y.
Q10: Only diploid cells that have undergone mating and successfully fused their nuclei will have the ability to grow on CSM-LEU-TRP plates since they can complement each other's auxotrophic (deficient) mutations.
The diploid cells contain two copies of each gene, so if one copy carries a mutation causing an auxotrophy for leucine and the other copy carries a mutation causing an auxotrophy for tryptophan, the diploid cell will be able to grow on the CSM-LEU-TRP plates.
Q11: If the colonies on the plates appear white and exhibit good growth, it suggests that both strains carry functional copies of the GDE genes and are mating type "a" (or "α"). If the colonies appear pink or have reduced growth, it suggests that one or both of the strains have a mutation in the GDE genes or may have a different mating type.
Q12: Two advantages of diploidy over haploidy for the organism concerned (likely referring to yeast) are:
Genetic Redundancy: Diploid organisms have two copies of each gene, providing redundancy in case one copy contains a harmful mutation. This redundancy helps ensure that at least one functional copy of each gene is present in the organism, reducing the impact of deleterious mutations on survival and reproduction.Genetic Variation and Adaptability: Diploidy allows for the shuffling and recombination of genetic material through sexual reproduction. This increases genetic diversity within the population, enabling the organism to adapt and respond better to changing environmental conditions. The presence of two copies of each gene also allows for the exploration of different combinations of alleles, potentially leading to advantageous traits.Q13: The ability of yeast to exist as haploid cells is advantageous to geneticists because it simplifies genetic analysis and manipulation. Haploid cells have a single copy of each gene, making it easier to study the effects of specific mutations or to introduce targeted genetic modifications.
Haploidy allows for straightforward genetic crosses and the isolation of pure genetic strains. Additionally, the presence of a single allele simplifies the interpretation of phenotypic traits, as the observed trait can be directly linked to a specific mutation or genetic change.
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Please make a prediction about how the following species could evolve in the future, based on current pressures:
- medium ground finch
- snake
However, based on current pressures, medium ground finch might adapt further to changes in food availability and habitat, while snakes could potentially evolve in response to changes in prey distribution or climate.
Pressures can have both positive and negative impacts on individuals. They can motivate and drive people to achieve their goals, pushing them to perform at their best. However, excessive or constant pressures can lead to stress, anxiety, and burnout. The pressure to succeed academically, professionally, or socially can create a significant burden on individuals, affecting their mental and physical well-being. It is important to find a balance and manage pressures effectively to maintain a healthy and fulfilling life. Seeking support, setting realistic expectations, and practicing self-care can help alleviate the negative effects of pressures.
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Which of the following statements on selection bias is correct? (Multiple answers allowed.)
A. If cases are selected from a single hospital, the identified risk factors may be unique to that hospital.
B. If the cases are drawn from a tertiary care facility, the risk factors identified may be only in persons with severe forms of the disease.
IC. t is generally preferable to use incident cases of the disease in case-control studies of disease etiology.
D.A mother who has had a child with a birth defect often tries to identify some unusual event that occurred during her pregnancy with that child.
The correct statements on selection bias are: A. If cases are selected from a single hospital, the identified risk factors may be unique to that hospital. B. If the cases are drawn from a tertiary care facility, the risk factors identified may be only in persons with severe forms of the disease. The correct answer is options (A) and (B).
A. When cases are selected from a single hospital, the identified risk factors may be specific to that particular hospital. This is because the patient population and characteristics of that hospital may differ from other hospitals, leading to unique risk factors associated with the disease. B. Selecting cases from a tertiary care facility can introduce selection bias, as the risk factors identified may be applicable only to individuals with severe forms of the disease. Tertiary care facilities often deal with complex and severe cases, which may have different risk factors compared to milder cases seen in primary or secondary care settings.
C. The statement regarding incident cases in case-control studies is not correct. Case-control studies compare cases (individuals with the disease) to controls (individuals without the disease) and are retrospective in nature. Therefore, using incident cases (newly diagnosed cases) is not a requirement for case-control studies.Regarding the additional statement about a mother trying to identify unusual events during her pregnancy, it describes a situation where recall bias may occur. Recall bias refers to the tendency for individuals, in this case, a mother, to selectively remember and report specific events or exposures that they believe might be linked to an outcome, such as a birth defect.
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Assuming brown or blue eye color is determined by different alleles of a single gene. A woman with brown eye marries a man who also has brown eye color. Their daughter has blue eye color. The daughter then married to a man with blue eye color vision. What is the probability of the daughter's first child to have brown eye color?
50%
0%
100%
25%
The probability of the daughter's first child having brown eye color can be determined by considering the inheritance patterns of eye color alleles. The correct answer is option b.
If brown eye color is determined by a dominant allele and blue eye color is determined by a recessive allele, and both the daughter and her husband have blue eyes, it suggests that they both carry two copies of the recessive blue allele. In this case, the probability of their child inheriting the dominant brown allele from either parent would be zero, as neither parent possesses the brown allele.
Therefore, the probability of the daughter's first child having brown eye color would be 0%. However, it is important to note that eye color inheritance can be more complex and involve multiple genes, so this simplified explanation assumes a single gene model for eye color determination.
The correct answer is option b.
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Complete Question
Assuming brown or blue eye color is determined by different alleles of a single gene. A woman with brown eye marries a man who also has brown eye color. Their daughter has blue eye color. The daughter then married to a man with blue eye color vision. What is the probability of the daughter's first child to have brown eye color?
a. 50%
b. 0%
c. 100%
d. 25%