Answer with Explanation:
These species have different structural composition therefore they possess bacterial resistance towards disinfectants. As Bacillus megaterium belongs to spore producing family therefor it is said that they are most difficult to kill followed by Mycrobacterium bovis having a thick wall with waxy coating. While Pseudomonas aeruginosa is a gram-negative cell and staphylococcus aureus is a gram-positive bacteria.
By following order, you rank their susceptibility to disinfection:
Least Susceptible > Most Susceptible
Bacillus megaterium>Mycrobacterium bovis>Pseudomonas aeruginosa>staphylococcus aureus
Bacteria and other microbes can be used to "clean up" an oil spill by breaking down oil into carbon dioxide and water. Two samples isolated from the Deepwater Horizon leak in the Gulf of Mexico were labeled A and B. The DNA of each was isolated and the percent thymine measured in each sample. Sample A contains 20.7 % thymine and sample B contains 30.9 % thymine. Assume the organisms contain normal double‑stranded DNA and predict the composition of the other bases.
Answer:
Sample A: Thymine= 20.7% ; Adenine= 20.7%; Guanine = 29.3% ; Cytosine= 29.3%
Sample B: Thymine= 30.9% ; Adenine= 30.9% ; Guanine= 19.1% ; Cytosine base= 19.1%
Explanation:
In a double-stranded DNA molecule, the percentage of adenine base is equal to that of the thymine base while the percentage of guanine base is equal to that of the cytosine base. It is called Chargaff's rule and is based on the complementary base pairing of purine and pyrimidine bases.
Sample A: Percentage of thymine= 20.7%
This means that the sample has= 20.7% adenine base
Total thymine and adenine bases in sample A= 20.7 + 20.7 = 41.4%
Therefore, proportion of guanine + cytosine bases in the sample= 100-41.4= 58.6%
Percentage of guanine base= 58.6/2 = 29.3%
Percentage of cytosine base= 29.3%
Sample B: Percentage of thymine= 30.9%
This means that the sample has= 30.9% adenine base
Total thymine and adenine bases in sample A= 30.9 + 30.9 = 61.8%
Therefore, proportion of guanine + cytosine bases in the sample= 100-61.8= 38.2%
Percentage of guanine base= 38.2/2 = 19.1%
Percentage of cytosine base= 19.1%
In this lesson we discussed the difference between strong and weak bases. Which of these bases would be classified as weak?
A. sodium fluoride
B. ammonia
C. potassium hydroxide
D. calcium hydroxide
E. none of the above
Answer:
E. none of the above i think
Explanation:
g Analysis of mitochondrial DNA can answer interesting and important questions. For which of the questions would analyzing mitochondrial DNA be futile? Are you my father? I have poor muscle function, is there something wrong with my ability to make lots of ATP? From biological samplings at a crime scene, can we keep this person as our primary suspect? Are you my long-lost brother? Which is more closely related to the great white shark, the tiger shark or the hammerhead shark?
Answer:
Are you my father?
Explanation:
Mitochondrial DNA is a good tool to identify and to evaluate maternal relatedness.
It can't be used in establishing paternity as barely a small portion of a person's father mitochondria can be inherited.
When Alex and Jason were confronted by a dangerous snake, Alex experienced a much greater level of autonomic nervous system arousal than Jason. Yet both report ed being equally scared by the experience. If true, this fact would be most consistent with the Oa. James-Lange theory. O b. Cannon-Bard theory. Oc. two-factor theory. Od. Yerkes-Dodson law
Answer: option A - James-Lange theory
Explanation:
The autonomic nervous system controls largely the involuntary body activities like sweating, heartbeat, breathing etc.
Alex experienced higher arousal induced by his autonomic nervous system despite being exposed to the equal external stimuli as Jason. This is perfectly described by James-Lange theory
Farmers use a variety of approaches in an effort to produce more successful yields. These may include hormones, antibiotics, herbicides, pesticides, and forms of biotechnology like plant breeding or genetic engineering. Sort each statement into the bin with the corresponding agricultural approach. Drag the appropriate items into their respective bins.
Answer:
Statements are missing. These are,
1-Consumers may reduce exposure to these by washing and peeling fruits and vegetables.
2-These may be used to increase weight gain and meat production in cattle.
3-These include organophosphates and sex pheromones.
4-Golden rice is yellow because it contains beta-carotene, which the plant can now manufacture because genes for certain enzymes were inserted in the rice genome.
5-Their use may be avoided in favor of integrated pest management (IPM).
6-Scientists inserted the Bt gene into corn to make it resistant to the corn borer.
7-These include recombinant bovine somatotropin (rbS
Hormones:
These include organophosphates and sex pheromonesThese include recombinant bovine somatotropin (rbS)These may be used to increase weight gain and meat production in cattle.Antibiotics:
NA.
Pesticides/Herbicides:
Consumers may reduce exposure to these by washing and peeling fruits and vegetables. Their use may be avoided in favor of integrated pest management (IPM)Biotechnology/Genetic Engineering:
Golden rice is yellow because it contains beta-carotene, which the plant can now manufacture because genes for certain enzymes were inserted in the rice genome.Scientists inserted the Bt gene into corn to make it resistant to the corn borer.Why do all Krebs cycle reactions occur twice for each molecule of glucose that undergoes respiration?
Answer: This is because 2 pyruvate molecules were produced in Glycolysis,and each must be metabolize to extract one ATP molecule each contained.
In addition to produced required number of reduced Co-enzymes for Electron Transport Chain (ETC) reactions to take place for ATP production
It must also be twice/glucose molecule to generate required volume of oxygen for decaboxylation,because the oxygen in the reaction came from.from . redox and not from the atmosphere.
Therefore the both the link reaction and the kreb.cycle
must occur twice to produce mandatory 2 ATPs ,and generate the required amount of reduced Co Enzymes NADH and FADH2 for ETC(electron transport Chain ) to occur.
Explanation:
Answer:
The Krebs cycle reaction occurs twice because glucose produces two molecules of Pyruvate and then two acetyl CoA.
Explanation:
The citric acid cycle goes around twice for each molecule of glucose that enters cellular respiration because there are two pyruvates from the single glucose molecule and thus, two acetyl CoA is produced per glucose which is the starting molecule in the Krebs cycle.
Consider seven recessive genes, a, e, d, j, l, y and r. The seven loci are closely linked in the centromeric region of an autosome but their order is unknown. Six deletions are examined in an individual that is heterozygous for all seven genes. All six deletions resulted in pseudodominance of at least two genes, and two deletions spanned the region of the centromere. Deletion Centromere included in deletion Genes exhibiting pseudodominance del 1 No e and j del 2 Yes a, l and r del 3 No a and d del 4 Yes e, l and r del 5 No d and y del 6 No e and l Determine the gene order and the location of the centromere.
Answer:
Gene order: y,d,a,(r,l),e,j
Centromere: between l and r
Explanation:
If you see your data you can conclude that:
- deletions that include the centromere include r and l genes.
- At one side of the centromere is gene a and at the other side is gene e
- when you see mutation 3, the gene that is away from the centromere is d, then in mutation 5 the gene that is further is y
- On the other side, the mutation 6 tell us that e gene is by l gene side
- In mutation one you see that the gene further the centromere by that side is j
What can possibly kill sperm in the Great Sperm Race?
Answer:
Answer:
THE GREAT SPERM RACE:
A sperm's race to fertilize an egg is not so easy. Out of about 250 million sperm ejaculated into the human vagina during intercourse, not more than one in a hundred will survive the Great race to the end due to the hurdles it has to face in the hostile, ACIDIC CHAMBER to the cervix (it has hundreds of tiny branching tunnels that can trap, crush and slowly kill sperm).
If ovulation is not occurring soon the sperm will "drown in a thick flow of cervical mucus
"The correct answer is that a variety of factors can kill sperm in the Great Sperm Race, including an acidic pH, the presence of white blood cells, and the harsh conditions within the female reproductive tract.
In the Great Sperm Race, which is a metaphor for the journey sperm must undertake to fertilize an egg, sperm face numerous challenges that can impede their progress or lead to their demise. Here are some of the key factors that can kill sperm:
1. Acidic pH: The vagina is naturally acidic, which helps protect against infections. However, this acidic environment can be hostile to sperm. Sperm prefer a more alkaline pH, which is typically found in the cervical mucus during ovulation. If the pH is too acidic, it can damage or kill sperm.
2. White Blood Cells (WBCs): The presence of WBCs, or leukocytes, in the reproductive tract can be indicative of an infection or inflammation. WBCs can attack and kill sperm as part of their immune response to foreign bodies
3. Harsh Conditions: The female reproductive tract can be a challenging environment for sperm due to its complex structure and varying conditions. Sperm may encounter barriers such as thick cervical mucus, which can be difficult to penetrate, or they may be exposed to unfavorable temperatures or toxic substances.
4. Immunological Responses: Sometimes, the woman's immune system may mistakenly identify sperm as foreign invaders and mount an immune attack against them, leading to their destruction.
5. Sperm Defects: Not all sperm are capable of reaching the egg. Some may have defects such as abnormal morphology (shape) or poor motility (movement), which can prevent them from successfully navigating the reproductive tract.
6. Anti-Sperm Antibodies: Both men and women can produce antibodies that target sperm, leading to sperm agglutination (clumping) or immobilization, which can prevent sperm from reaching the egg.
7. Environmental Factors: External factors such as exposure to certain chemicals, radiation, or high temperatures can also damage or kill sperm.
Understanding these factors is crucial for diagnosing and treating infertility, as well as for developing effective contraceptive methods. Interventions may include adjusting the timing of intercourse to coincide with the more sperm-friendly environment during ovulation, treating infections, or using assisted reproductive technologies when necessary."
A famous experiment, Matthew Meselson and Franklin Stahl grew E. coli cells in a medium containing only the "heavy" isotope or nitrogen, ^15N. These cells were then transferred to a medium with the "light" isotope of nitrogen, ^14N. The results of the experiment supported the hypothesis of semi conservative replication, which was proposed by Watson and Crick. Heavy DNA (^15N DNA), hybrid DNA, and light DNA (^14N DNA) can be separated by centrifugation. If cells containing ^15N DNA are transferred to a medium with only ^14N NH_4CI as a nitrogen source, what percent of daughter molecules are composed of hybrid DNA after 3 generations? Predict what percent of daughter molecules would be composed of hybrid DNA if DNA exhibited conservative replication -that is, if a daughter DNA molecule were composed of newly synthesized DNA only.
Answer:
1. 25% percent of daughter molecules are composed of hybrid DNA after 3 generations
2. 0% of daughter molecules would be composed of hybrid DNA if DNA exhibited conservative replication
Explanation:
1. According to semi conservative model of replication, the two DNA strands separate during replication and each gives rise to a new DNA strand. So in a newly synthesised DNA molecule one strand is old and one stand is new.
Here, the medium contained 15N so all DNA molecules had 15N in beginning. Then they were transferred to 14N medium. After first round of replication all the molecules will be 15N14N hybrid since 14N will be used now for replication. After second round of replication 50% of molecules will be 14N14N and other 50% will be hybrid. After third round of replication, 75% molecules will be 14N14N and 25% will be hybrid.
2. According to conservative model of replication, parental DNA strands stay together after replication so a newly synthesised DNA molecule will have both new strands. In this case, a hybrid will never be formed since the 15N15N strand will always remain together and all the subsequent DNA molecules will have 14N14N composition.
Final answer:
After 3 generations in the Meselson-Stahl experiment, 25% of the daughter DNA molecules would be composed of hybrid DNA in semi-conservative replication. In contrast, conservative replication would result in no hybrid DNA molecules, as new DNA would be synthesized entirely from ¹⁴N without incorporating ¹⁵N.
Explanation:
The experiment conducted by Matthew Meselson and Franklin Stahl with E. coli grown in mediums containing heavy (¹⁵N) and light (¹⁴N) isotopes of nitrogen demonstrates the semi-conservative replication model of DNA. After transferring cells from a ¹⁵N medium to a ¹⁴N medium, the distribution of DNA after 3 generations reveals that a specific portion of the daughter molecules will be composed of hybrid DNA (containing both ¹⁵N and ¹⁴N).
After 1 generation, all of the daughter DNA molecules are hybrid, containing one strand of ¹⁵N DNA and one strand of ¹⁴N DNA (100% hybrid DNA).After 2 generations, 50% of the daughter molecules are hybrid DNA and 50% are light DNA (¹⁴N only).After 3 generations, 25% of the daughter molecules are hybrid DNA and 75% are light DNA.In contrast, if DNA replication were conservative, each generation would yield one original heavy (¹⁵N) DNA and one completely new light (¹⁴N) DNA molecule. After 3 generations in a conservative model, there would be no hybrid DNA molecules, as each generation would involve the creation of entirely new ¹⁴N DNA, without integration of ¹⁵N into new DNA strands.
Which naturalist was fascinated by the countryside around Concord, Massachusetts and wrote of his experiences living close with nature?
Select one:
a. Henry David Thoreau
b. Lewis Thomas
c. Rachel Carson
d. Ralph Waldo Emerson
Answer:A.
Henry David Thoreau.He was born in 1856 in Concord Massachusetts.And died in 1862.
He based his style of writing on close observation of nature and ehicla principles of life.
He advocated for conservation of natural resources on private lands of citizens,and preservation of wilderness as land area that belong to the public.
Throd detested consumption of animal food mateials..He regarded them as unclean.He concluded that animal food materials can not provide the major nutrients for his body needs;despite been costly than the desired plant mateials which are easy to get and well cleaned.
Explanation:
Which of the following statements about gene regulation concerning operons is INCORRECT? A) A negative repressible gene is controlled by a regulatory protein that inhibits transcription. B) For a gene under negative repressible control, a small molecule is required to prevent the gene's repressor from binding to DNA. C) For a gene under positive repressible control, the normal state is transcription of a gene, stimulated by a transcriptional activator. D) A regulator gene has its own promoter and is transcribed into an independent mRNA. E) Presence of operons where genes of related functions are clustered is common in bacteria but not in eukaryotes.
Answer:
The answer is B
Explanation:
For a gene under negative repressible control, a small molecule is required to prevent the gene's repressor from binding to DNA
The correct statement about gene regulation concerning operons is that D) a regulator gene has its own promoter and is transcribed into an independent mRNA.
Explanation:A regulator gene is a segment of DNA that controls the expression of one or more other genes, often by producing regulatory proteins or transcription factors. These proteins influence the rate and extent of gene transcription, affecting the synthesis of specific proteins within a cell.
A regulator gene has its own promoter and is transcribed into an independent mRNA. In prokaryotic cells, gene expression is regulated through the use of repressors and activators. Repressors bind to operators, blocking transcription, while activators bind to promoters, enhancing transcription. In both cases, the regulation happens at the level of transcription to control the expression of genes within operons.
Learn more about Gene regulation in operons here:https://brainly.com/question/31979270
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7. A bioengineer is trying to understand the biomechanics of a hole created in the skin for a transcutaneous implant. The engineer made a hole using a circular biopsy drill in the dorsal skin of a dog. The diameter of the drill is 5 mm. If the hole becomes an ellipse with a minor and major axis of 3 and 7 mm, answer the following questions. a. In which direction is the internal stress in the skin greater? b. In which direction are the collagen fibers more oriented? c. How can the bioengineer obtain a circular rather than an elliptical hole for the implant? d. Assuming the implant is non-deformable compared to the skin, what problems will arise between skin and implant when a load or force is applied to the skin or implant by handling accidentally?
The internal stress and collagen fiber orientation in the skin in response to a transcutaneous implant are greater and more oriented respectively along the major axis of the ellipse. To achieve a circular hole, drilling technique adjustments or material considerations may be necessary. Load or force application can cause complications due to deformability mismatch between the skin and a rigid implant.
A bioengineer analyzing the biomechanics of a transcutaneous implant in the skin of a dog is faced with a situation where a circular hole drilled by a biopsy drill has deformed into an elliptical shape with minor and major axes of 3 mm and 7 mm respectively. Such analysis is crucial in understanding the mechanical behavior of the skin in response to implants and the stress distribution around the implant site.
a. The internal stress in the skin is greater in the direction of the major axis, which is 7 mm. This is due to the fact that the skin stretches more along the major axis, causing increased tension and stress in that direction.
b. The collagen fibers within the skin are more likely to be oriented along the direction of least stress, which would be along the minor axis of the ellipse (3 mm).
c. To obtain a circular hole rather than an elliptical one, the bioengineer could consider altering the drilling technique, such as optimizing the speed and pressure, using different drill materials, or modifying the post-drilling processes to ensure the skin maintains its shape. Additionally, the mechanical properties of the skin and the drill's cutting edge sharpness should be considered to minimize deformation.
d. If a load or force is applied to the skin or implant, especially if the implant is non-deformable compared to the skin, there could be issues with alignment, pressure sores, and stress concentration at the interface. This mismatch in deformability can lead to inadequate load transfer, potential implant loosening or migration, and increased risk of skin breakdown or irritation.
What can be said about all the DNA molecules in a single Sanger sequencing reaction mix? Select All That Apply
DNA can be sequenced less expensively.
Fewer errors in the sequence are produced.
A whole-genome could be sequenced in a single day.
DNA can be sequenced much faster.
Answer:
The choices can be found elsewhere and as follows:
A) It adds ddNTP to the end of each DNA fragment.
B) It changes the length of the DNA fragments.
C) It separates DNA fragments based on their charge.
D) It separates DNA fragments generated during the sequencing reaction based on one-nucleotide differences in their size.
I think the correct answer is option D, It separates DNA fragments generated during the sequencing reaction based on one-nucleotide differences in their size. Hope this answers the question. Have a nice day.
Which of the following pairs of alleles would be most likely to cross over?
( ) A..........B
( ) A...B......
( ) A......B.......C
( ) AB.............
( ) ..........AB
Answer:
a) A..........B
Explanation:
Crossing over is the process that occurs during the pachytene stage of prophase-I of meiosis-I. During crossing over, two chromatids of a homologous pair of chromosomes exchange genetic material between themselves. It results in the production of new allele combinations on the chromatids. The process of crossing over mostly occurs between the alleles of two or more genes that are present far apart from each other on the same chromosome.
Among the given option, the arrangement of alleles in option A has the maximum distance between the alleles A and B. This increases the probability of crossing over between these two alleles. The alleles of genes present close together on a chromosome are least likely to undergo crossing over.
Answer:
Option D and E , AB............. and ..........AB
Explanation:
When the alleles are located far away from each other on the same chromosomes they remain unlinked and hence assort independently. On the other hand alleles which are lying close to each other on the same chromosomes are inherited as a single unit or are inherited together; therefore they are called as linked allele pair. Crossing occurs when the two chromosomes exchange their genetic material. When two genes lie very close to each other, there are very rare chances of them getting separated by crossing over.
Hence, option D and E are correct
A sample ( 885 mg ) of an oligomeric protein of M r 155,000 was treated with an excess of 1‑fluoro‑2,4‑dinitrobenzene (Sanger's reagent) under slightly alkaline conditions until the chemical reaction was complete. The peptide bonds of the protein were then completely hydrolyzed by heating it with concentrated HCl . The hydrolysate was found to contain 4.85 mg of DNP‑Val. 2,4‑Dinitrophenyl derivatives of the α‑amino groups of other amino acids could not be found. Calculate the number of polypeptide chains in this protein. Give the answer as a whole number.
Answer:
In 1945, Frederick Sanger described its use for determining the N-terminal amino acid in polypeptide chains, in particular insulin.[4] Sanger's initial results suggested that insulin was a smaller molecule than previously estimated (molecular weight 12,000), and that it consisted of four chains (two ending in glycine and two ending in phenylalanine), with the chains cross-linked by disulfide bonds. Sanger continued work on insulin, using dinitrofluorobenzene in combination with other techniques, eventually resulted in the complete sequence of insulin (consisting of only two chains, with a molecular weight of 6,000).[5]
Following Sanger's initial report of the reagent, the dinitrofluorobenzene method was widely adopted for studying proteins, until it was superseded by other reagents for terminal analysis (e.g., dansyl chloride and later aminopeptidases and carboxypeptidases) and other general methods for sequence determination (e.g., Edman degradation).[5]
Dinitrofluorobenzene reacts with the amine group in amino acids to produce dinitrophenyl-amino acids. These DNP-amino acids are moderately stable under acid hydrolysis conditions that break peptide bonds. The DNP-amino acids can then be recovered, and the identity of those amino acids can be discovered through chromatography. More recently, Sanger's reagent has also been used for the rather difficult analysis of distinguishing between the reduced and oxidized forms of glutathione and cysteine in biological systems in conjunction with HPLC. This method is so rugged that it can be performed in such complex matrices as blood or cell lysate.[6][7]
Explanation:
Example: A sample (525 mg) of an oligomeric protein of Mr 117,000 was treated COOH with an excess of1-fluoro-2,4-dinitrobenzene (Sanger's reagent) under slightly alkaline conditions until thechemical reaction was complete. The peptide bonds of the protein were then completelyhydrolyzed by heating it with concentrated HCI. The hydrolysate was found to contain 3.37 mgof DNP-Val (shown at the right), 2,4-Dinitrophenyl derivatives of the α- amino groups of otheramino acids could not be found H3C Calculate the number of polypeptide chains in this protein.Give the answer as a whole number Number A second oligomeric protein of M 230,000 wasshown by a similar endgroup analysis to consist of five polypeptide chains. SDS polyacrylamidegel electrophoresis in the presence of a reducing agent shows three bands: α (M, 30,000), β (M40,000) and γ(M-60,000), indicating three distinct polypeptides. SDS electrophoresis withoutreducing agent also yields three bands, with Mr of 30,000, 40,000, and 120,000 Which of the
Conservation biologists provide strong arguments about why we should all care about preserving biodiversity. When considering the resources we get from nature as well as the services the ecosystem perform, it is clear that life on Earth is threatened when biodiversity itself is threatened. Which of the following are services by natural, well-functioning ecosystems?
Nutrient cycling
Decomposition of dead orangisms
Dispersal of pollen and seeds
Increase in ambient global temperatures
Purification of air and water
Reduction in the severity of drought and floods
Pollination of crops and natural vegetation
An increase of erosion and siltation along waterways
Source of medicines
Recyling energy to be used again
Regulation of oxygen and carbon dioxide levels
Answer:Increase in ambient global temperatures.
Recyling energy to be used again
Regulation of oxygen and carbon dioxide levels
An increase of erosion and siltation along waterways
Explanation:
Conservational biologists think about the preservation of ecosystem by maintaining the environment in a human control way.
Increase in ambient global temperatures.: The humans must prevent the increase in global temperature worldwide by preventing the rise of greenhouse gases which can lead to global warming worldwide.
Recyling energy to be used again: The sources of energy like wood, waste water can be recycled again for reutilization.
Regulation of oxygen and carbon dioxide levels.: The oxygen and carbon dioxide levels must be regulated. As oxygen is the basic requirement for respiration. The increase in carbon dioxide levels due to human activities is likely to cause respiratory diseases and health hazards in living beings.
An increase of erosion and siltation along waterways.: The erosion and siltation will likely to deposit nutrients and debris which may either contaminate the waterway or may cause eutrophication.
Answer:
Nutrient cycling
Decomposition of dead organisms
Dispersal of pollen and seeds
Purification of air and water
Reduction of the severity of droughts and floods
Pollination of crops and natural vegetation
Sources of medicines
Regulation of oxygen and carbon dioxide levels
The ________ extends through the hindbrain, midbrain, and forebrain.
a. Reticular formation
b. Medulla
c. Pons
d. Cerebellum
Answer: option C) Pons
Explanation:
Pons, or better still, the pons Varolii is a band of nerve fibers located within the brain stem. While, the brain stem is known to connect the spinal cord(Hind brain) to the fore brain
Thus, Pons extends through the hindbrain, midbrain, and forebrain.
Answer:
Reticular formation
It's correct
Which of the following represents a sensory input that is not part of both the somatic and autonomic systems?(A) Vision(B) taste(C) Baroreception(D) Proprioception
Answer:
the correct answer is (C) Baroreception
2. A likely explanation for an abnormal human phenotype associated with a trisomy is: A) the presence of multiple recessive mutant alleles. B) the extra chromosome has typically undergone significant rearrangements. C) the absence of genes necessary for certain cellular processes. D) altered gene dosage (also known as genetic balance). E) the random inactivation of a complete autosome.
Answer:
The correct answer is B the extra chromosome has typically undergone significant rearrangements.
Explanation:
Trisomy is a type of aneuploidy which is characterized by the presence or occurance of an extra copy of a particular chromosome.As a result the chromosomes in which the trisomy is occurring contain 3 arms instead of 2. Trisomy occur due to non disjunction of a particular chromosome during meiosis.
As a result a second copy of chromosome is present inside the cell containing that abnormal chromosome.If the same thing occur in the cell of gamet undergoing fertilization then there is a high chance that the embryo will carry that extra chromosome.
In case of humans Trisomy result in the rearrangement extra chromosome which ultimately give rise to
1 Trisomy 18 occur due to extra copy of chromosome no 18
2 Trisomy 13 occur due to extra copy of chromosome no 13
3 Trsomy 21 occur due to extra copy of chromosome no 21 which is also called down syndrome.
Final answer:
An abnormal human phenotype associated with a trisomy is likely due to altered gene dosage.
Explanation:
Trisomy is a state where humans have an extra autosome. The most common trisomy is of chromosome 21, which leads to Down Syndrome. Trisomic individuals suffer from an excess in gene dose, which disrupts the normal balance of gene dosage. This can lead to functional challenges and developmental delays. The most likely explanation for an abnormal human phenotype associated with a trisomy is altered gene dosage.
You are studying a disorder that is based on the genetic composition at three loci. Assume that a dominant allele at any locus adds 6 units of risk for the disorder and that a recessive allele at any locus adds 2 units of risk for the disorder. Individuals with 29 or more units of risk develop the disorder. The environment does not affect the presence or absence of this disorder. How many risk units will be present in an individual of genotype AABbCc?
Answer:
28 units
Explanation:
This disorder follows quantitative inheritance. It is controlled by three genes which do not show the usual dominant-recessive relationship . The six alleles individually contribute to the effect which add up to produce the cumulative phenotype. Dominant allele contributes 6 units of risk whereas recessive allele contributes 2 units of risk.
Individual with genotype AABbCc has four dominant alleles (AABC) and two recessive alleles (bc). So their total risk units =
(6*4) + (2*2) = 24 + 4
= 28 units
1. What information/structures were you able to glean from the Gram stain that you could not get from the methylene blue stain?
2. What information/structures were you able to glean from the methylene blue stain that you could not get from the Gram stain?
3. Is Saccharomyces cerevisiae Gram-positive or Gram-negative?
4. Research and describe the composition of yeast cell walls. How does the composition compare to the cell walls of Gram-positive or Gram-negative bacteria?
Answer 1: Gram staining can stain the outer structure of cell such as cell wall and cell membrane. This stain used to identify the gram positive and gram negative cell in a cluster of bacterial colony. These structure can not be stained by using methylene blue.
Answer 2: Methylene blue stain used to stain the internal structures of the cell including DNA in nucleus and RNA fractions in cytoplasm. These structures can not be stained and identified using gram staining.
Answer 3: A research paper (J. gen. Microbiol. (1963), 30, 223-235) stated that Saccharomyces cerevisiae is a gram positive bacteria.
Answer 4: A research paper (The Chemical Composition and Structure of the Yeast Cell Wall) provided the actual and true composition of yeast cell wall. According to this paper, yeast cell wall is composed of protein ( 13%), lipids (8.5%), and Polymaccharides (78%). While on the other hand cell wall of gram positive bacteria is composed of peptidogylcane (50-60%), lipids (1-4%), teichoic acid and lipoteichoic acid. Further, cell wall of gram negative bacteria composed of lipopolysacharides, lipoproteins, peptidoglycane and lipids (11-12%).
two of the four principle Classes of organic compounds are proteins and nucleic acid’s. What is the relationship between proteins and nucleic acid
Answer:
Nucleic acids combine to synthesize proteins, from DNA which becomes RNA and when entering the ribosome generates such synthesis.
Explanation:
Protein synthesis is a complex process that begins in the cell nucleus and begins when the protein gene is encoded and expressed by the transcription process. The transcript is the one that transmits the information from the DNA to the RNA (the two nucleic acids).
There is also an tRNA, which carries the amino acids for each messenger RNA where they bind to the proper position, within the protein synthesis
Answer:
As a biomolecules, proteins and nucleic acids are not closely similar in structure. The major relationship between the two has to do with protein production -- DNA deoxyribose nucleic acids contains the information that a cell uses, with the help of RNA, to make protein
Explanation:
Which of the following describes a situation of scarcity?
A. Someone offers free advice about getting into college.
B. A person lets the kids in the neighborhood use his pool.
C. Someone distributes free bottles of water at the beach.
D. A child charges friends for a ride on his new bike.
Answer:
A child charges friends for a ride on his new bike
Explanation:
Answer:
D.
Explanation:
A child charges friends for a ride on his new bike.
The hydrolysis of pyrophosphate to orthophosphate is important in driving forward biosynthetic reactions, such as the synthesis of DNA. This hydrolytic reaction is catalyzed in E. coli by a pyrophosphate that has a mass of 120 kd and consists of six identical subunits. For this enzyme, a unit of activity is defined as the amount of enzyme that hydrolyzes 10 umol of pyrophosphate in 15 min at 37 degrees C under standard assay conditions. The purified enzyme has a Vmax of 2800 units per milligram of enzyme.
a) How many moles of the substrate is hydrolyzed per second per milligram of enzyme when the substrate concentration is much greater than Km? (Draw a velocity versus substrate
concentration graph to show your point)
b) How many moles of active sites is there in 1 mg of enzyme? Assume that each subunit has
one active site.
c) What is the turnover number of the enzyme? (Don't bother comparing it to other enzymes;
just show all your own, which equations you use, what numbers you use, where you get them
from, etc.)
Answers:
a) How many moles of the substrate is hydrolyzed per second per milligram of enzyme when the substrate concentration is much greater than Km?
Answer:
31.1 µmol PPi s-1 mg -1
b) How many moles of active site are there in 1 mg of enzyme?
Answer:
5.0 x 10-8 mol
c) What is the turnover number of the enzyme?
Answer:
3732 s-1 per mol enzyme but there are 6 active sites per enzyme therefore turnover is 622 s-1 (per active site)
Explanation:
i have attached two pictures that explains the methodology as i can not write formulas here that's why i solved it and created a picture for attachment.
The turnover number (kcat) of the enzyme can be calculated using the Vmax value and the enzyme concentration, the number of active sites can be determined using the subunit information, and the moles of substrate hydrolyzed per second per milligram of enzyme at Vmax can be calculated.
Explanation:a) When the substrate concentration is much greater than Km, the enzyme is operating at Vmax. The turnover number (kcat) represents the number of substrate molecules converted to product by a single enzyme active site per unit time at Vmax. In this case, Vmax is given as 2800 units per milligram of enzyme. To calculate the moles of substrate hydrolyzed per second per milligram of enzyme, we need to convert the units:
Turnover number (kcat) = Vmax / (enzyme concentration in mg)
Assuming a 1 mg enzyme concentration:
Turnover number (kcat) = 2800 units per milligram of enzyme / 1 mg = 2800 s-1 per milligram of enzyme
b) If each subunit of the enzyme has one active site, then there would be six active sites in the enzyme with a mass of 120 kd. Therefore, the number of moles of active sites in 1 mg of enzyme would be:
Moles of active sites = Number of active sites / Avogadro's number
Number of active sites = 6
Moles of active sites = 6 / 6.022 x 1023 = 9.96 x 10-24 mol
c) The turnover number (kcat) represents the efficiency of an enzyme. It can be calculated using the equation:
Turnover number (kcat) = Vmax / (enzyme concentration in M * moles of active sites)
Using the given Vmax of 2800 units per milligram of enzyme, the enzyme concentration of 1 mg, and the moles of active sites as calculated in part b), we can calculate the turnover number:
Turnover number (kcat) = 2800 units per milligram of enzyme / (1 mg * 9.96 x 10-24 mol) = 2.81 x 1027 s-1 per mol
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How are the Earth's crust and humans the most similar according to these graphs? a The largest percentage of both is oxygen. b Silicon is rarely found in nature. c Chemical reactions occur with similar frequency in humans and the Earth's crust. d The smallest percentage of both is potassium.
Answer: Option (a) is the correct option that is largest percentage of oxygen is present in both earth's crust and human body.
Explanation: Oxygen is present in large amounts both in earth's crust and human body. The percentage of oxygen in earth's crust is 46.6% and the percentage of oxygen in the body is about 66.6%. So it is concluded that both human body and earth's crust have high percentage of oxygen.
In the metabolism of fats, the fatty acids enter the reactions of the _________.
Answer: cytosol
Explanation:
Fatty acid metabolism is also known as fat oxidation. The fatty acid to be metabolized e.g Stearic acid enters the reactions of the CYTOSOL of the cell to become ACTIVATED, then it forms Stearyl-CoA.
The Stearyl-CoA then is transported to the cell mitochondria where it is reduced sequentially in form of units of acetyl-CoA until oxidation is completed
Many studies have indicated that there is some genetic influence on body weight and composition. Researchers have established a variety of theories that attempt to explain why some individuals gain weight more easily than others and why weight can be so difficult to change long-term. Choose the correct statement about theories regarding genetic influence on weight status.
a. The thrifty gene theory suggests that some individuals have a "thrifty metabolism," which could help a person survive when food is scarce.
b. According to recent research, the FTO gene may increase feelings of satiety.
c. The set point theory indicates that bodies are meant to stay within a narrow weight range, which cannot be changed long-term.
d. Drastically reducing calorie intake causes BMR to increase, which helps individuals deviate from their "set point" for weight
Answer:
A
Explanation:
The thrifty gene hypothesis postulates that due to dietary scarcity during human evolution, people are prone to obesity by storing energy as fat, an ability to take advantage of rare periods of abundance and this is advantageous during times of food availability. individuals with greater adipose reserves would more likely survive famine. This tendency to store fat could lead to obesity.
FTO is an enzyme that is encoded in humans by FTO gene located on chromosome 16. Increase in this enzyme is associated with the regulation of energy intake but not feeding reward. Also, the set point theory indicates that bodies are meant to stay within a narrow weight range but it is susceptible to chnage overtime.
What structure is used to seal the DNA into an opening created by the restriction enzyme during recombinant DNA technology
Answer:
DNA Ligase
Explanation:
DNA ligase is a specific type of enzyme, that facilitates the joining of DNA strands together by catalazing the formation of phophodiester bond.
Is used both DNA repair and DNA replicaation. It has extended use in molecular biology for recombinant experiments.
Answer:
DNA ligase
Explanation:
DNA ligase is an enzyme responsible for the joining of DNA strand ends. If two pieces of DNA have matching ends, ligase can link them to form a single, unbroken molecule of DNA.
A fictional bacterial operon is responsible for the production of the biosynthetic enzymes needed to make the theoretical amino acid Tisophane (Tis). In the wild? type condition, when Tis is present, no enzymes are made. In the absence of Tis, the enzymes are made. The operon is regulated by a separate gene, R, deletion of which causes loss of enzyme production, regardless of whether Tis is present. A second class of regulatory mutations has been identified and these map to a region (the RB site) near the promoter of the Tis operon; these so?called RB? mutations also block production of the enzymes, regardless of the availability of Tis.
a. Is the operon under positive or negative control? Justify your answer.
b. Propose a model for regulation of the operon which explains the lack of enzyme production in the presence of Tis in wildtype cells, the role of the R protein and the effects of the RB? mutations.
Answer:
a. The operon is under positive control
a.) Loss of enzyme synthesis results from loss of regulatory R gene product which is regulated in the presence of Tis ophane for transcription. Absence of (Tis), transcription is positively controlled by regulatory Rprotein
b.) The proposed model reveals Tis / B complex prevents Tis enzymes from undergoing transcription through the RB operator by binding to the B region
the role of the R protein is
-decrease affinity of Tis / B complex to B region of RB
-bind to the R region on RB
- change the conformation of B region
- yield decreased affinity of Tis / B complex when R gene is lacking preventing transcription in the process
Effects of the RB mutations is that R protein will not bind to Tis / B complex to reduce transcription
The greatest resistance to blood flow and therefore the greatest drop in pressure occurs as blood passes through the ______.
Answer:
i tried i do not know
Explanation:
The greatest resistance to blood flow and therefore the greatest drop in pressure occurs as blood passes through the arterioles.
They are the major location of total peripheral resistance, arterioles play a crucial role in maintaining mean arterial pressure and tissue perfusion.
A force that opposes a fluid's flow is called resistance. The majority of blood vessel resistance is caused by the size of the vessel. Blood flow reduces as vessel diameter shrinks due to an increase in resistance.
The largest reduction in blood pressure is caused by arterioles, which also have the highest resistance to rise.
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