To calculate the number of dilution steps required, we can use the formula: Number of dilution steps = log10(target concentration / initial concentration) / log10(dilution factor)
In this case, the initial concentration is 10 bacteria per milliliter, and the target concentration is 100 bacteria per milliliter. The dilution factor at each step is 1:100.Let's calculate the number of dilution steps needed:
Number of dilution steps = log10(100 / 10) / log10(1/100) = log10(10) / log10(0.01) = 1 / (-2) = -1
Since we obtain a negative value for the number of dilution steps, we can convert it to a positive value by taking the absolute value:
Number of dilution steps = | -1 | = 1
Therefore, you would need 1 bottle of diluent to dilute the specimen to reach a concentration of 100 bacteria per milliliter.
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Deep nucleotide sequencing (NGS) is now cheap enough for general application. What can the technique mainly be useful for?
a. Quickly identifying new viruses
b. O speedy vaccine development
c. giving details on virus excretion in symptomless carriers
d. establishing the reproductive number of a virus
NGS (deep nucleotide sequencing) can mainly be useful for:
a. Quickly identifying new viruses.
c. Giving details on virus excretion in symptomless carriers.
A virus is an infectious submicroscopic creature that only reproduces inside of live cells. All living things, including plants, animals, and microbes like bacteria and archaea, are susceptible to virus infection. More than 11,000 of the millions of viral species have been characterised in detail since Dmitri Ivanovsky's 1892 publication revealing a non-bacterial disease infecting tobacco plants and Martinus Beijerinck's discovery of the tobacco mosaic virus in 1898. Viruses are the most common sort of living organism and may be found in practically all ecosystems on Earth. Virology is the study of viruses; it is a branch of microbiology.
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What are the two types of Speciation? 4.3
There are two pathways to speciation: PG: 137
1) Transformation: One species evolves into another species
2) Divergence: One or more species arise from a parent species
The two types of speciation are allopatric speciation and sympatric speciation.
Allopatric speciation: Allopatric speciation occurs when a population is geographically isolated, leading to reproductive isolation and the formation of new species. The physical barrier prevents gene flow between the separated populations, allowing for independent evolutionary changes to accumulate over time. The accumulation of genetic and phenotypic differences can eventually result in reproductive isolation, where individuals from the separated populations can no longer produce viable offspring if brought back into contact.
Sympatric speciation: Sympatric speciation occurs without geographic isolation, where a new species arises within the same geographic area as the parent population. Reproductive isolation is achieved through other mechanisms such as ecological, behavioral, or genetic factors. These mechanisms can lead to the development of reproductive barriers that prevent gene flow between different subgroups within the population. Over time, these subgroups accumulate genetic and phenotypic differences, eventually leading to the formation of new species.
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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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1. In both the male and female cath, identify structures that hure a urogcritil function. a. Male cat b. Fernale cat QUESTIONS 2-11: Match the cat teproductive structure in columin A with the apeopeia
In the male cat the urethra and the pe.nis have a urogenital function, while in the female cat, the va.gina has a urogenital function.
What organs have a urogenital structure?Organs have a urogenital structure if they are important for both the reproductive and the urinary systems:
The urethra, and pe.nis: These organs are important for the transportation of spermatozoids in the reproductive system but also for the transportation of urine.Va.gina: Similar to the urethra this structure has a reproductive purpose but also allows the urine to be transported out of the body.Note: Here is the complete question:
In both the male and female cath, identify structures that have a urogenital function.
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What is the cause of the evolution of senescence according to the two evolutionary hypotheses (Mutation Accumulation and Antagonistic Pleiotropy)? a)mutations accumulate in individuals as they grow older, leading to senescence
b)mutations that have pleiotropic fitness effects are removed by selection
c)mutations that cause deleterious fitness effects late in life are effectively neutral
d)mutations that cause deleterious fitness effects late in life experience a strong "force of natural selection"
The Antagonistic Pleiotropy theory postulates that senescence results from the presence of mutations that have advantageous early-life effects but unfavourable late-life consequences.
The two evolutionary explanations for senescence provide the following descriptions of the causes: According to the hypothesis of mutation accumulation, as people age, they accumulate harmful mutations that lead to senescence. These mutations can remain and cause a reduction in fitness and ageing because natural selection is less successful at getting rid of mutations that have late-life effects. The Antagonistic Pleiotropy theory postulates that senescence results from the presence of mutations that have advantageous early-life effects but unfavourable late-life consequences. Because the advantages of these mutations early in life outweigh the disadvantages of senescence later, they are kept in the population. Therefore, the appropriate response is: a) as people age, mutations increase, leading to senility
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which of the two if any does Digestion of food not occur? explain stomach or mouth?
The digestion of food occurs in the stomach and mouth. Digestion is the process of breaking down large molecules of food into smaller molecules that can be absorbed and used by the body.
Digestion begins in the mouth, where the food is physically broken down by chewing and mixed with saliva, which contains enzymes that begin the chemical breakdown of carbohydrates.The stomach is the next stop in the digestive process. It is a muscular sac that mixes the food with stomach acid and enzymes to further break down the food into a liquid called chyme. The stomach also releases the hormone gastrin, which triggers the release of more digestive juices in the small intestine, where the majority of digestion and absorption take place.In conclusion, both the mouth and stomach are involved in the digestion of food. The mouth is where the process begins, with the mechanical and chemical breakdown of food, while the stomach continues the process by mixing the food with digestive juices to break it down further.
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structures-structures with similar structure but differ in their function. The similar structure is due to having a common ancestor with that structure that was passed down. structures - structures with different structures/origin but the same function. These structures have a superficial resemblance due to convergent evolution, such as a bird and bat wing Convergent evolution is when two different species evolve for the same conditions (flying, swimming, etc.) parative Embryology The embryos of most animals develop in very similar S This is especially true for early stages of development cular Evidence (DNA & Proteins) Organisms share huge amounts of The is universal (used by all living things) DNA and proteins have been used to determine evolutionary Humans and chimpanzees share over. of their DNA W Most of the differences are in non-protein coding regions of the DNA
Structures with similar structure but differ in their function are known as homologous structures. The similar structure is due to having a common ancestor with that structure that was passed down. Homologous structures are used to find evolutionary relationships among organisms.
The study of comparative anatomy shows that the same basic structures of the body have been modified over time to serve various purposes. For example, the forelimbs of vertebrates are made up of the same bones, although they are used for different functions in different animals. This is because they have a common ancestor from which they evolved.Structures with different structures/origin but the same function are called analogous structures.
These structures have a superficial resemblance due to convergent evolution, such as a bird and bat wing. Convergent evolution is when two different species evolve for the same conditions (flying, swimming, etc.).Vestigial structures are structures that have no function but are remnants of structures that had a function in the ancestors of the organism. These structures may not have any function in the organism, but they may have had an important function in the organism's ancestors. DNA is a universal molecule that is used by all living organisms.
The genetic code is universal, and all organisms use the same code to build proteins. DNA and proteins have been used to determine evolutionary relationships among organisms. Organisms share huge amounts of DNA and proteins, and this similarity is used to determine their evolutionary relationships. Humans and chimpanzees share over 98% of their DNA. Most of the differences are in non-protein coding regions of the DNA.
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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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Examine this pedigree for a rare human disease and determine the most likely mode of inheritance. If person II-3 and person III-1 had a child, what would be the probability of that child having the disease? a) zero chance b) 1/4 c) 100 percent d) 1/2
Based on the pedigree, the most likely mode of inheritance for the rare human disease is autosomal recessive.
In an autosomal recessive inheritance pattern, individuals need to inherit two copies of the disease-causing allele (one from each parent) in order to express the disease. In this pedigree, individuals II-3 and II-4 are unaffected but both carry one copy of the disease-causing allele, making them carriers. Their child, III-1, expresses the disease, indicating that both II-3 and II-4 must have passed on their disease-causing alleles to III-1.If person II-3 and person III-1 were to have a child, the probability of that child having the disease would be 1/4 or 25 percent. This is because person II-3 is a carrier (heterozygous) and person III-1 is affected (homozygous recessive). When they have a child, there is a 25 percent chance that the child will inherit two copies of the disease-causing allele and therefore express the disease. The other possible outcomes include a 50 percent chance of the child being a carrier like II-3 or a 25 percent chance of the child being unaffected.
It's important to note that this probability assumes that both II-3 and III-1 are correctly identified as carriers and affected, respectively, based on their phenotypes and genetic testing.
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Which of the following is an incorrect statement about "calories"?
a. All one needs to know to accurately calculate one's daily calorie needs is knowledge of their sex and their weight. b. Two hundred calories from an avocado (which offers healthy fats and other nutrients) can be a better choice than eating 100 calories of deli meat. c. Fiber helps to slow the absorption of sugar. d. Healthy eating and drinking choices is about more than calories.
e. A zero-calorie pop/soda, for example, might also provide zero nutrients, and come packed with artificial sweetners. f. Consuming 100 calories in the form of an apple will provide a more "full" feeling than drinking 100 calories of pop/soda/Red Bull, etc.
An incorrect statement about "calories" is: a. All one needs to know to accurately calculate one's daily calorie needs is knowledge of their sex and their weight.
Option (a) is the incorrect statement. While sex and weight are important factors in estimating daily calorie needs, they are not the sole factors. Other variables, such as age, height, activity level, and metabolic rate, also play a significant role in determining individual calorie requirements. Therefore, relying solely on sex and weight to calculate daily calorie needs would not provide an accurate estimation.
Options (b), (c), (d), (e), and (f) are all correct statements about calories. Option (b) highlights that the nutrient content and overall health benefits of food should be considered along with calorie content. Option (c) emphasizes the role of fiber in slowing down sugar absorption, which is important for managing blood sugar levels. Option (d) acknowledges that healthy eating is not solely about calorie intake but also about nutrient quality and overall dietary choices. Option (e) points out that zero-calorie beverages may lack nutrients and contain artificial sweeteners. Option (f) highlights the difference in satiety and fullness between consuming calories from different sources, such as whole foods versus sugary beverages.
Therefore, option (a) is the incorrect statement among the given options.
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WHAT ARE THE LONG TERM EFFECT OF batrachotoxin TO THE
CIRCULATORY SYSTEM?
Batrachotoxin is a poisonous substance that is found in the skin of certain frogs and in some species of birds. This toxin acts on the sodium channels of the body. Batrachotoxin can have long-term effects on the circulatory system.
Batrachotoxin can lead to death, as it can cause respiratory and circulatory failure. Batrachotoxin causes sodium channels to remain open, allowing excessive amounts of sodium ions to enter the cells. As a result, the nerves and muscles of the heart are unable to function properly, leading to irregular heartbeat. Batrachotoxin can also lead to the accumulation of fluid in the lungs, making breathing difficult. The toxin can also cause swelling of the brain and seizures, leading to loss of consciousness.
the long-term effects of batrachotoxin to the circulatory system can be severe. Batrachotoxin is a poisonous substance that is found in the skin of certain frogs and in some species of birds. This toxin acts on the sodium channels of the body, causing nerves and muscles to be unable to function properly, leading to irregular heartbeat. Batrachotoxin can also cause the accumulation of fluid in the lungs, making breathing difficult, and swelling of the brain and seizures, leading to loss of consciousness. People who survive batrachotoxin poisoning may experience long-term effects, including heart disease, lung disease, and neurological problems. Therefore, the long-term effects of batrachotoxin on the circulatory system can be fatal and cause permanent damage.
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Tachycardia with marked rise of blood pressure occur after administration of : Select one O a. Isoprenaline b. Noradrenaline O c. Dobutamine d. Adrenaline d e. Atropine
Tachycardia with a marked rise in blood pressure is observed when adrenaline is administered. Adrenaline is a naturally occurring hormone, also known as epinephrine.
That plays a vital role in the fight or flight response in the body.Tachycardia, an abnormally high heart rate, and hypertension, or high blood pressure, are among the possible side effects of the drug administration. Adrenaline is a medication that is often used in emergency situations.
The medication is administered by injection, and the dosage is determined by the patient's weight, the severity of their condition, and other factors. The medication stimulates beta receptors in the body, resulting in increased heart rate and constriction of blood vessels.
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1. Which biological science is the one that I told you is a complex subject ke immunology? & biochemistry A. microbiology B. biochemistry
D. biophysics C. neuroscience E. cell biology 2. Mucous membranes are a part of the_____________
A. adaptive immune system B. lymph node C. physical barrier D. bone marrow E. Ethymus
1. The biological science that is as complex as immunology is: B. Biochemistry
Immunology is the study of the immune system and how it works to fight off pathogens. Biochemistry is the study of chemical processes and substances in living organisms. Both fields can be quite complex, but biochemistry can be just as complex as immunology.
2. Mucous membranes are a part of the: C. physical barrier
Mucous membranes are a type of physical barrier in the body's defense against infection. They line various organs and body cavities, such as the nose, mouth, throat, lungs, and reproductive organs. The mucus produced by these membranes helps trap pathogens and prevent them from entering the body.
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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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An ORF is a continuous stretch of codons that begins with a start codon (usually AUG) and ends at a stop codon (usually UAA, UAG or UGA). The gene x has 920 codons. In a particular tissue, the base ‘C’ of 440th codon CAA (codes for glutamine) of gene x is edited to the base ‘U’. Answer the following question? Enter only a number in the provided space.
How many amino acids will be in the protein X from the un-edited mRNA Answer
How many amino acids will be in the protein Xedit from the edited mRNA? Answer
How many nucleotides will be in the open reading frame of gene Xedit from the edited mRNA?
An ORF is defined as a continuous sequence of codons that starts with a start codon and terminates at a stop codon. Gene X has 920 codons, and in a specific tissue, the C base of the 440th codon (CAA) of Gene X was replaced with a U base.
The number of amino acids in a protein is directly linked to the number of codons in the mRNA; since Gene X has 920 codons, Protein X will have 920/3 = 306 amino acids (since each codon codes for one amino acid, and there are three nucleotides in each codon).Therefore, the number of amino acids present in protein X from the unedited mRNA is 306 amino acids.
When the 440th codon (CAA) is edited by replacing the C base with a U base, the resulting codon becomes CUA, which codes for leucine rather than glutamine. The edited mRNA encodes a different protein, and the number of amino acids present in this protein is determined by the number of codons in the edited mRNA.
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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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If crossing over occurred in one cell and not another, how would the two cells compare?
A. Crossing over would have no effect on either cell.
B. Crossing over would increase the genetic diversity in one of the cells.
C. Crossing over would increase the chromosome number in one of the cells.
D. Crossing over would decrease the chromosome number in one of the cells.
E. Crossing over would cause one of the cells to stop dividing.
If crossing over occurred in one cell and not another, the two cells would have different genetic compositions.
Option B, "Crossing over would increase the genetic diversity in one of the cells," is the correct answer. Crossing over introduces new combinations of alleles by shuffling genetic material between the homologous chromosomes. This process promotes genetic diversity in offspring, as it creates novel combinations of genes that were not present in the parent cells. Options A, C, D, and E are not accurate in this context. Crossing over does have an effect on cell genetics, it does not affect the chromosome number, it does not decrease the chromosome number, and it does not cause one of the cells to stop dividing.
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Chain Reactions Linking Acorns to Gypsy Moth Outbreaks and Lyme Disease Risk Oak trees (Quercus spp.) produce large autumnal acorn crops every two to five years, and produce few or no acorns during intervening years. Acorns are a critical food for white-footed mice (Peromyscus leucopus). Mice are important predators of the pupal stage of the gypsy moth (Lymantria dispar). This introduced insect periodically undergoes outbreaks that defoliate millions of hectares of oak forests, decreasing tree growth, survival, and acorn crop production. An abundance of acoms provides food for white-tailed deer (Odocoileus virginianus). Mice and deer are the primary hosts of the black-legged tick (Ixodes scapularis), which carries Lyme disease.
The chain of events linking acorns to gypsy moth outbreaks and Lyme disease risk can be summarized as follows:
1. Acorn Production: Oak trees produce large acorn crops every two to five years, with intervening years having few or no acorns.
2. White-Footed Mice: Acorns are a critical food source for white-footed mice. During years with abundant acorn production, the mouse population increases due to the availability of food.
3. Gypsy Moth Predation: White-footed mice play a crucial role in controlling gypsy moth populations. They are important predators of the pupal stage of the gypsy moth.
4. Gypsy Moth Outbreaks: The gypsy moth, an introduced insect, periodically undergoes outbreaks. During these outbreaks, the gypsy moth larvae defoliate millions of hectares of oak forests. This defoliation can have detrimental effects on tree growth, survival, and acorn crop production.
5. Deer and Acorns: An abundance of acorns provides a food source for white-tailed deer. Deer play a role in dispersing acorns and can consume large quantities when they are available.
6. Tick Hosts: Both white-footed mice and white-tailed deer are primary hosts for the black-legged tick, also known as the deer tick (Ixodes scapularis). These ticks are responsible for transmitting Lyme disease.
7. Lyme Disease Risk: The black-legged tick carries the bacterium that causes Lyme disease. When ticks feed on infected hosts, such as white-footed mice and white-tailed deer, they can acquire the bacterium. Subsequently, these ticks can transmit the bacterium to humans when they bite.
The chain reaction begins with the production of acorns by oak trees, which supports an increase in the population of white-footed mice. These mice, in turn, prey on gypsy moth pupae, helping to control gypsy moth populations. However, when gypsy moth outbreaks occur, they can defoliate oak forests and impact tree health and acorn production. Abundant acorns also attract white-tailed deer, which are hosts for black-legged ticks carrying Lyme disease. Thus, the interconnectedness of these factors contributes to the risk of Lyme disease transmission to humans.
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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.
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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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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Research one genetic disorder, either something you have
experience with or something interesting, and discuss how it is
tied to meiosis. Include your understanding of how this disorder
occurs in the
One genetic disorder that is tied to meiosis is Down syndrome, also known as trisomy 21. It is caused by the presence of an extra copy of chromosome 21, which disrupts the normal chromosomal distribution during meiosis.
During meiosis, the process of cell division that produces gametes (sperm and eggs), chromosomes undergo recombination and segregation to create genetically diverse and haploid cells. However, in individuals with Down syndrome, there is an error in meiosis called nondisjunction, where chromosome 21 fails to separate properly. This results in one of the resulting gametes having two copies of chromosome 21 instead of one.
When a fertilized egg with an extra copy of chromosome 21 (trisomy) is formed, it leads to the development of Down syndrome. Individuals with Down syndrome typically exhibit physical characteristics such as distinct facial features, intellectual disabilities, and various health issues.
The occurrence of Down syndrome is directly linked to the abnormal distribution of chromosomes during meiosis, specifically the failure of proper separation of chromosome 21, resulting in an additional copy of this chromosome in the resulting offspring.
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28. In Chapter 12, we discussed different tests/assays we could use to identify the microbe(s) causing an infection in a host. Please describe one of these methods (or use your own example) and report (1 sentence/term):
a) The name of the method and how it works:
b) A type of control sample (either positive or negative control) you can run with your sample:
c) A false positive or false negative result that could occur when you run each assay:
The method described is Polymerase Chain Reaction (PCR), a molecular technique used to amplify specific DNA sequences. PCR involves a series of temperature cycles that allow DNA replication to occur in vitro.
It starts with denaturation, where the DNA strands are separated by heating. Then, primers specific to the target DNA sequence are annealed to the separated strands. Next, DNA polymerase extends the primers, synthesizing new DNA strands. This process of denaturation, annealing, and extension is repeated multiple times, resulting in the exponential amplification of the target DNA sequence if present.
A positive control sample that can be run alongside the test sample is a known sample containing the target DNA sequence of the microbe being tested. This positive control should yield a positive result, confirming that the PCR assay is working correctly and capable of detecting the target DNA sequence.
False positive results in PCR can occur if there is contamination in the laboratory. Contaminating DNA, such as stray DNA from previous experiments or reagents, can be amplified, leading to a positive signal even in the absence of the target microbe. False negatives, on the other hand, can occur if the primers used in the PCR assay do not match the DNA sequence of the microbe causing the infection. If the primers fail to bind to the target DNA, amplification will not occur, resulting in a negative result despite the presence of the microbe.
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2. What is the advantage of using Sabouraud agar?
The advantage of using Sabouraud agar is it relatively easy to prepare, has higher yield rate than other fungal growth media, will change colour as the pH changes, and can differentiate between fungal species.
Sabouraud agar is a selective and differential medium used for fungal growth, it has a distinct composition with lowered pH that inhibits bacterial growth and improves fungal growth. One of the advantages of using Sabouraud agar is that it's relatively easy to prepare and relatively inexpensive. Secondly, it has a higher yield rate than other fungal growth media, which is why it is commonly used for clinical specimen cultures. Sabouraud agar can also be used for the isolation and identification of fungi from food, water, soil and other materials, this is due to the fact that it has a high degree of sensitivity and specificity.
Another advantage is that the pH indicator of Sabouraud agar will change colour as the pH changes, which makes it easier for the detection of fungal colonies. Lastly, it can differentiate between fungal species by the color and texture of their growth, which helps identify the type of fungal infection a patient has. So therefore the advantage of using Sabouraud agar is it relatively easy to prepare has higher yield rate than other fungal growth media, will change colour as the pH changes, and can differentiate between fungal species.
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Targeting an Antibiotic Resistance Gene using CRISPR-Cas9
The rise and spread of antibiotic resistance in bacteria are alarming because of the impact on the cost, complications, and outcomes of treatment. Of particular concern are resistant bacteria that cause hospital-acquired infections (HAIs). Enterococcus faecalis, a member of the intestinal normal microbiota, is now a leading cause these infections. This organism is an opportunist, meaning that if the normal microbiota population is disturbed (for example by antibiotic treatment), it proliferates and becomes pathogenic. The pathogenic strains usually exhibit larger than normal genomes, having acquired mobile genetic elements such as plasmids, transposons, or phages (viruses that infect bacteria). Some of these elements contain antibiotic resistance genes.
Now a collaborative research team from the University of Texas at Dallas and the University of Colorado is investigating the use of CRISPR-Cas9 for overcoming antibiotic resistance in E. faecalis. Recall that bacteria use the CRISPR-Cas system as a defense mechanism, protecting them against the foreign DNA of mobile gene elements. CRISPR-Cas9 consists of an endonuclease (Cas9) that uses a guide RNA (gRNA) to locate and cleave foreign double stranded DNA at a specific site. For example, if a phage injects its DNA into a bacterial cell, that cell uses its CRISPR-Cas9 system to identify and destroy that phage DNA. The system also creates "memory" so that the bacterial cell is protected against future encounters with that same type of phage. Scientists can manipulate the CRISPR-Cas9 system by inserting specific gRNAs to target the Cas9 endonuclease to exactly where they want it to go in a genome, a potential tool for gene silencing or editing.
The research team previously showed that drug resistant E. faecalis does not have an intact CRISPR-Cas system; it lacks the Cas9 component and is thus susceptible to the uptake of foreign DNA. Now the team has developed a novel way of getting a functional CRISPR-Cas9 into those organisms in an effort to rid them of their antibiotic resistance genes. They engineered a plasmid, inserting genes for CRISPR-Cas9 along with gRNA sequences that are homologous to a resistance gene for the antibiotic erythromycin. The engineered plasmid was then introduced into a donor strain of E. faecalis that has conjugation ability. The presence of the CRISPR-Cas9 in the donor strain makes it immune to acquiring foreign DNA. When the donor strain conjugated with the drug resistant E. faecalis strain, the resistant strain gained a copy of the engineered plasmid containing the modified CRISPR-Cas9 system. The CRISPR-Cas9 in that cell then targeted its erthromycin resistance gene.
The team was able to show that the introduced plasmid significantly reduced the resistance of the resistant E. faecalis to erythromycin, making it sensitive to this drug. The work indicates that it may be possible in the future to use conjugation delivery of CRISPR-Cas9 antimicrobials.
Rodrigues, M. et. al. 2019. Conjugative Delivery of CRISPR-Cas9 for the Selective Depletion of Antibiotic-Resistant Enterococci. Antimicrob Agents Chemother. 63(11). pii: e01454-19.
Why is the genome of pathogenic Entercoccus aerogenes slightly larger than that of their nonpathogenic counterparts?
a. Pathogenic strains of Enterococcus have a CRISPR-Cas9 cassette and this makes them larger.
b. Pathogenic Entercoccus strains make the enzyme Dicer, so have an additional gene for this enzyme.
c. Pathogenic strains have acquired extra DNA in the form of a mobile genetic element (MGE).
d. Pathogenic strains do not have a larger genome; they have a double copy of their single chromosome.
e. Pathogenic strains of any bacterium are larger than nonpathogenic strains, and have larger genomes.
Pathogenic strains have acquired extra DNA in the form of a mobile genetic element (MGE).
The correct option to the given question is option c.
The genome of pathogenic Enterococcus aerogenes is slightly larger than that of their nonpathogenic counterparts because pathogenic strains have acquired extra DNA in the form of a mobile genetic element (MGE).
In particular, some of these elements contain antibiotic resistance genes. The pathogenic strains usually exhibit larger than normal genomes, having acquired mobile genetic elements such as plasmids, transposons, or phages (viruses that infect bacteria).
The acquisition of extra DNA in pathogenic strains makes it possible for them to proliferate and become pathogenic, especially if the normal microbiota population is disturbed, for example, by antibiotic treatment.Therefore,Pathogenic strains have acquired extra DNA in the form of a mobile genetic element (MGE).
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Which stores more potential energy: one molecule of glucose or two
molecules of pyruvate? explain.
One molecule of glucose stores more potential energy than two molecules of pyruvate. Glucose is a six-carbon molecule that is broken down through a series of chemical reactions into two molecules of pyruvate. This process is called glycolysis, which takes place in the cytoplasm of the cell. During glycolysis, glucose is oxidized and converted into two molecules of pyruvate.
This process generates a small amount of energy in the form of ATP and NADH. However, the majority of the energy is still stored in the chemical bonds of the two molecules of pyruvate. After glycolysis, the two molecules of pyruvate are transported into the mitochondria, where they are further oxidized through a process called the citric acid cycle. During this process, more ATP and NADH are generated, and the energy stored in the bonds of the pyruvate molecules is gradually released.
In conclusion, while two molecules of pyruvate do store some potential energy, they do not store as much as one molecule of glucose. This is because glucose has more carbon atoms and more chemical bonds than pyruvate, and therefore has a higher potential energy content.
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Which variable rises after capillary beds?
a. Blood pressure
b. Blood vessel area
c. blood velocity
d. blood volume
A blood velocity After capillary beds, the blood velocity of the blood rises because of the following reasons. After capillary beds, blood velocity increases. Explanation:Capillaries are the smallest and the thinnest vessels in the human body. These vessels play a vital role in the body as they help in the exchange of various substances such as nutrients, waste, and oxygen. It connects the arterial and venous systems.
The narrowing of the capillaries from tiny vessels increases the resistance of the flow of blood. This then leads to a drop in blood pressure and an increase in blood velocity.The sum of the cross-sectional area of the capillaries is bigger than the cross-sectional area of the arteries and veins; this means that the blood that flows in the capillaries moves at a much slower pace than the blood that flows in the arteries and veins.
This causes a drop in the speed of blood flow, thus leading to a rise in blood velocity.Along with blood velocity, the blood pressure and blood volume may also change after capillary beds. Blood pressure decreases after capillary beds while the blood vessel area and blood volume vary depending on the situation or condition.
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pitenesin 6. In this lab, we reviewed numerous fossil species and their defining characteristics. To help you make compari- sons across these species and understand larger trends in our evolutionary history, complete the Australopith and Early Homo Chart on pp. 446-447. AUSTRALOPITH AND EARLY HOMO CHART Fossil Species Dates and Geographic Region Cranial and Dental Traits Postcranial Traits Suggested Behavior Australopithecus anamensis Australopithecus afarensis LAB 15 | The Australopiths and Early Members of the Australopithecus africanus Australopithecus garhi Australopithecus sediba Australopithecus (Paranthropus) aethiopicus AUSTRALOPITH AND EARLY HOMO CHART (continued) Fossil Species Dates and Geographic Region Cranial and Dental Traits Postcranial Traits Suggested Behavior Australopithecus (Paranthropus) boisei Australopithecus (Paranthropus) robustus Australopithecus deyiremeda Homo habilis (including H. rudolfensis)
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In this lab, we have examined many fossil species and their defining characteristics. To help you make comparisons across these species and understand larger trends in our evolutionary history.
let us complete the Australopith and Early Homo Chart. The Australo pith and Early Homo Chart is a tabular presentation of some Australopith and Early Homo fossils. This chart allows you to make comparisons across these fossils, to identify some of their similarities and differences.
Understand some of the significant trends in the evolution of these hominins.The following is a sample of the Australopith and Early Homo Chart that we have completed in this lab: Fossil Species Dates and Geographic Region Cranial and Dental Traits Postcranial Traits Suggested Behavior Australopithecus anamensis .
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Explain the common cold and flu of viral origin.
Explain the disease caused by the varicella-zoster virus
Explain the disease measles, mumps and rubella
Common cold and flu are respiratory illnesses caused by viral infections, with the common cold usually caused by rhinoviruses and the flu by influenza viruses. Varicella-zoster virus causes chickenpox (varicella) and later reactivates as shingles (herpes zoster).
Measles, mumps, and rubella are viral infections with distinct symptoms, with measles causing fever and a characteristic rash, mumps affecting the salivary glands, and rubella causing a rash and posing risks to pregnant women.
Common Cold and Flu:
The common cold and flu are both respiratory illnesses caused by viral infections. The common cold is usually caused by rhinoviruses, while the flu is caused by influenza viruses. These viruses are highly contagious and primarily spread through respiratory droplets when an infected person coughs, sneezes, or talks.
Varicella-Zoster Virus (VZV) Disease:
The varicella-zoster virus causes two distinct diseases. The primary infection results in chickenpox, also known as varicella. Chickenpox is highly contagious and spreads through direct contact or respiratory droplets. It is characterized by a blister-like rash, itching, fever, and general malaise.
Measles, Mumps, and Rubella:
Measles, mumps, and rubella are all viral infections that can cause distinct diseases. Measles, caused by the measles virus, is highly contagious and spreads through respiratory droplets.
Mumps, caused by the mumps virus, is also highly contagious and spreads through respiratory droplets or direct contact with infected saliva. It affects the salivary glands, leading to swelling and pain in the cheeks and jaw.
Rubella, caused by the rubella virus, is generally a mild infection but can have severe consequences if contracted by pregnant women. It spreads through respiratory droplets and causes a rash, low-grade fever, and swollen lymph nodes.
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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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