Answer: Gram negative bacteria can appear gram positive if it is under decolorized I.e the alcohol is left for a very short time and the CV-I complex is not wash away, the bacteria can look like gram positive.
A gram positive bacteria can look like gram negative when it has been over decolorized, by allowing the decolorizing agent to stay too long on it or by using acetone.
Explanation:
Gram positive bacteria stained crystal violet when subjected to gram staining because it have cell wall composed of thick layers of peptidoglycan.
Gram negative bacteria have cell walls with thin layers of peptidoglycan and don't retain the crystal violet color when stained.
All of the following are post-translational modifications except A. Elongation B. Phosphorylation C. Proteolytic cleavage D. Glycosylation A. Elongation
Answer: Elongation
Explanation:
Elongation (in the context of DNA) is not a form of post translational modification. It is basically an increase in the length of a particular growing nucleotide chain during either replication, translation or transcription. Post translation modifications are basically changes that are most of the times chemical that can take place in a protein after translation and some of these occur to activate or to cleave some specific regions. Examples including phosphorylation, glycosylation (conjugation of carbohydrate by enzymes) etc
Exocytosis is a type of cellular transport that allows materials to move across the plasma membrane of a cell. Which of the statements describe properties of exocytosis? a. Exocytosis is the primary method of transporting large molecules out of the cell. b. Exocytosis is the primary method of transporting large molecules into the cell c. Exocytosis uses transport vesicles to export materials from the cell d. Exocytosis uses membrane channel proteins to import materials into the cell. e. Exocytosis uses transport vesicles to import materials into the cell.
Exocytosis is a cellular transport process that allows materials to move across the plasma membrane of a cell. It is the primary method of transporting large molecules out of the cell and uses transport vesicles to export materials from the cell. Option E
Explanation:Exocytosis is a cellular transport process that allows materials to move across the plasma membrane of a cell.
It is the primary method of transporting large molecules out of the cell and uses transport vesicles to export materials from the cell.
Exocytosis is used by cells to remove waste products and release chemical signals or substances, such as digestive enzymes and hormones, into the extracellular fluid.
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An experiment to assess the effect of watering on the life span of a certain type of root system incorporates three watering regimens. (a) How many populations involved in the study?
Answers: Two populations involved in the study. 1) Root system 2) Watering regimens
Explanation: The experiment to investigate the effect of watering on the life span of certain type of root system using three watering regimens are determined based on the life span of the root system. Therefore, two populations involved in the experiment. One is the population of watering regimens and the other is the root system.
The significance of checkpoints can be demonstrated by considering what happens when they are impaired. What would occur if there was a gain-of-function mutation in the promoter for the cyclin E gene such that cyclin E protein was always made at high levels even under conditions in which cyclin E would not normally be made?
The significance of checkpoints can be demonstrated by considering what happens when they are impaired. What would occur if there was a gain-of-function mutation in the promoter for the cyclin E gene such that cyclin E protein was always made at high levels even under conditions in which cyclin E would not normally be made?
a. Cells will pass through the G2/M checkpoint with damaged DNA.
b. Cells will skip the S phase and go directly to G2 phase and not complete DNA replication.
c. Cells will pass through the G1/S checkpoint even if conditions are not ideal for cell division.
d. Cells will pass the M checkpoint with chromosomes unattached to spindles.
Answer:
c. Cells will pass through the G1/S checkpoint even if conditions are not ideal for cell division.
Explanation:
Cyclins are the regulatory proteins that are formed during different stages of the cell cycle. A specific cyclin during each stage bind to corresponding cyclin-dependent kinases (CDKS). The activated CDKs phosphorylate the proteins required for the progression of a cell through a specific stage of the cell cycle.
For example, Cyclin E is synthesized at the peak near the G1 -S phase transition or G1/S checkpoint. They activate the CDK-2 and allow the cells to progress from G1 to the S phase. If the cell is not prepared for DNA replication, cyclin E is not formed or is inhibited by specific protein kinases which in turn does not allow the cell to enter the S phase.
Any gain of function mutation resulting in the constitutive synthesis of cyclin E would allow the cell to pass through the G1/S checkpoint irrespective of the conditions.
A black mamba snake has a length of 2.67 m and a top speed of 3.96 m/s . Suppose a mongoose and a black mamba find themselves nose to nose. In an effort to escape, the snake accelerates at 1.47 m/s 2 from rest. How much time does it take the snake to reach its top speed?
Answer:
time (t) = 2.69 sec
Explanation:
Given that;
The final top speed of 3.96 m/s ( final velocity V) = 3.96 m/s .
The initial speed is 0; &
The acceleration (a) = 1.47 m/s²
We can determine how much time does it take the snake to reach its top speed by using the equation of motion;
V = U +at
3.96 = 0 + (1.47×t)
3.96 = 1.47t
t = [tex]\frac{3.96}{1.47}[/tex]
time (t) = 2.69 sec
∴ It took the snake 2.69 sec to reach the its top speed.
Free energy in the cell creates disorder. Cells harness free energy to drive unfavorable reactions by using Activated Carrier Molecules. Which of the following act as carrier molecules in cells? A. AMP B. NADPH C. FAD D. ATP E. All of the F. Above A and C G. B and D
Answer:
G
Explanation:
NADPH (Nicotinamide Adenine Dinucleotide Phosphate) and ATP (Adenosine Triphosphate) are forms in which energy is stored in metabolic reactions
Final answer:
The question is about identifying carrier molecules among ATP, NADPH, FAD, and AMP in cells. ATP, NADPH, and FAD act as carrier molecules, facilitating energy transfer and electron shuttling in cellular processes. Therefore, the correct answer is that all of the mentioned molecules serve as carrier molecules, with a clarification on the unique roles of NADPH, FAD, and ATP.
Explanation:
The question asks which of the following act as carrier molecules in cells: AMP, NADPH, FAD, or ATP. In cellular metabolism, activated carrier molecules are crucial for transferring energy and electrons to drive endergonic reactions, which are reactions that require an input of energy. Notably, Adenosine triphosphate (ATP) is the primary energy currency of cells, acting as a direct source of energy for various cellular processes. NADPH and FAD are key electron carriers involved in redox reactions, including those in photosynthesis and cellular respiration, respectively. AMP, while related to ATP, does not serve in the same capacity as a direct energy carrier or electron shuttle in cells. Therefore, the correct answer is E. All of the Above, acknowledge the roles of NADPH, FAD, and ATP as activated carrier molecules in cellular processes.
The pKa for the side chain of histidine is 6.0. What is the ratio of the deprotonated imidazole side chain to the protonated side chain at pH 5.0 and at pH 7.5?
Answer:
When the pKa is 6.0, we can determine the fraction of protonated H is by:
pH = pKa + log [A]/[HA]
Where
A = Deprotonated imidazole side
HA = Protonated side
Given, pH = 5.0
5 = 6 + log [A]/[HA]
log [A]/[HA] = -1 (take antilog of both side)
[A]/[HA] = 0.1
The ratio of the deprotonated imidazole side chain to the protonated side chain at pH 5.0 = 0.1
Given, pH = 7.5
7.5 = 6 + log [A]/[HA]
log [A]/[HA] = 1.5 (take antilog of both sides)
[A]/[HA] = 31.62
The ratio of the deprotonated imidazole side chain to the protonated side chain at pH 5.0 = 31.62
Final answer:
The ratio of deprotonated to protonated histidine side chain at pH 5.0 is 0.1, and at pH 7.5 is 31.62, calculated using the Henderson-Hasselbalch equation.
Explanation:
The ratio of deprotonated to protonated histidine side chain can be calculated using the Henderson-Hasselbalch equation:
pH = pKₐ + log ([A⁻]/[HA])
where pKₐ is the acidity constant of the side chain, pH is the environment's pH, [A⁻] is the concentration of the deprotonated form, and [HA] is the concentration of the protonated form.
Calculation at pH 5.0:
pH = pKₐ + log ([A⁻]/[HA])
5.0 = 6.0 + log ([A⁻]/[HA])
log ([A⁻]/[HA]) = -1
[A⁻]/[HA] = 10⁻¹
[A⁻]/[HA] = 0.1
Calculation at pH 7.5:
pH = pKₐ + log ([A⁻]/[HA])
7.5 = 6.0 + log ([A⁻]/[HA])
log ([A⁻]/[HA]) = 1.5
[A⁻]/[HA] = 10^1.5
[A⁻]/[HA] = 31.62
At pH 5.0, the ratio of deprotonated to protonated histidine side chain is 0.1, meaning there is 10 times more protonated form present. At pH 7.5, the ratio is 31.62, indicating there is significantly more deprotonated form present.
Based on the following DNA sequence comparisons, which two unidentified microorganisms would you conclude are most closely related? Choose one: A. Both microbial genomes include a gene that encodes for streptokinase, an enzyme and virulence factor that activates human plasminogen. B. Both microbial genomes include genes that encode for protein components of the prokaryotic ribosome. C. Both microbial genomes include genes that encode for photosynthetic photoreceptor proteins. D. Both microbial genomes include a gene that encodes for nitrogenase, a nitrogen-fixing enzyme.
Answer:
A. Both microbial genomes include a gene that encodes for streptokinase, an enzyme and virulence factor that activates human plasminogen.
Explanation:
Streptokinase (Ska) is a single-chain 414 amino acid protein, produced by certain Streptococci species (group A, C, and G streptococci), which possess an activity closer to two host proteins (urokinase-type and tissue-type plasminogen activators), as it non-enzymatically process inactive plasminogen to proteolytically active plasmin.
Streptokinase is a species-specific virulence factor. If both microbial genomes produces streptokinase, then these microorganisms can both be grouped as Streptococci since they are closely related.
All other choices do not certainly ease the relationship beyond the domain level.
Answer:
A. Both microbial genomes include a gene that encodes for streptokinase, an enzyme and virulence factor that activates human plasminogen.
Explanation:
Streptokinase is an enzyme (also synthesize as a medication) produced specifically by streptococci spp. If the two unidentified microbes could produce this enzyme, then they should contain the gene that codes for streptokinase .Thus they share certain genome characteristics, and are related.
Other options related the similarities to the domain levels only
Effects of Cycles on Ecosystems
Pre-Test Active
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Trees obtain carbon from
a decaying organisms
b. sunlight
C the atmosphere
d. none of the above
Answer:
The atmosphere
Explanation:
The major reservoir of carbon in the abiotic environment is the atmosphere where it occurs as carbon dioxide. Atmospheric carbon dioxide is used by green plants for the synthesis of organic compounds through photosynthesis.
Decomposers provide mineral nutrients for: A. heterotrophs. B. autotrophs. C. the second trophic level. D. the third trophic level. E. omnivores
Answer:
The answer is option B. autotrophs
Explanation:
Decomposers release nutrients like nitrogen and carbon dioxide as well as water when they feed on dead plants and animals. Decomposers include fungi, bacteria and earthworms.
These nutrients are released into air, soil and water and serve as nutrients for autotrophs. Autotrophs produce their own food from sunlight energy, water and carbon dioxide. Example include green plants; which by the process of photosynthesis produce their own food using carbon dioxide and soil nutrients.
Measurements with small freshwater sponges such as Grantia have documented water flow rates around 3-4 ml/minute. Assuming that a sponge has a constant flow rate over the course of a day, how many ml of water could a single sponge such as Grantia filter in a 24-hour period?
A sponge with a flow rate of 3.5 ml/minute could potentially filter around 5040 ml of water in a 24-hour period.
Explanation:The process to find the total amount of water a single sponge such as Grantia filters in a 24-hour period involves simple multiplication. First, you need to know the flow rate of the sponge, which is given as 3-4 ml/minute. Let's take the average, which is 3.5 ml/minute. To find the total volume of water processed in a 24-hour period, we need to multiply the rate by the total number of minutes in a day, which is 1440 minutes. So, it's 3.5 ml/minute * 1440 minutes = 5040 ml/day. Therefore, this sponge could process around 5040 ml of water in a 24-hour period.
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What is a homogeneous structure and what are some examples
Answer:
A homologous structure can be described as structures present in similar and different species. These structures might not perform the same functions but are similar because they might have a common ancestor in the past.
An example of a homologous structure is the arms of a human and the wings of a bat. Although the arms of a human and the wings of a bat do not perform the same function yet they do have similarities because they have a common ancestor in the past.
Protein function is lost or reduced when a protein is denatured. Which of the environmental factors listed can cause protein denaturation? excessive heat extreme pH exposure to water protein‑digesting enzymes
Answer:
Option A, Excessive heat
Explanation:
A denatured protein is the one whose structure changes with the loss of an activity. Generally, a denatured protein unfolds itself when the 2 D network of water around the protein molecule breaks by the breaking of hydrogen bonds. This primarily happens when the temperature is high. Both hydrogen bonds and hydrophobic interactions are disturbed by the heat thereby breaking hydrogen bonds.
Answer:
excessive heat
extreme pH
Explanation:
Protein denaturation can occur when they are exposed to abruptly increased temperature conditions and extremes of pH. Heat denature proteins by affecting the weak interactions that stabilize their secondary or tertiary structures. It makes the proteins less functional or completely non-functional. Similarly, extremes of pH alter the net charge on the protein. This causes electrostatic repulsion between the amino acids and disrupts some hydrogen bonds. Loss of these weak interactions denatures proteins.
Plants are set up in a greenhouse. 1/3 of potted plants are grown under red light, 1/3 under green light, and 1/3 under natural light. Plant biomass is measured at the beginning of the experiment and after a 28 day period.
In this study, is the amount of plant growth the independent or dependent variable?
dependent
independen
Answer:
Plants are set up in a greenhouse. 1/3 of potted plants are grown under red light, 1/3 under green light, and 1/3 under natural light. Plant biomass is measured at the beginning of the experiment and after a 28 day period.
In this study, is the amount of plant growth the independent or dependent variable?
Dependent variable
Explanation:
Plant is the independent variable as the light varies in color which makes the plant to depend on them for growth
What term describes an individual possessing two of the same alleles at a gene locus? monohybrid dihybrid homozygous wild type heterozygous
Question was't arranged i have arranged it in ask for detail section.
Answer:
Option e. homozygous is the correct answer.
Explanation:
A gene which has two identical alleles on homologous chromosomes is called homozygous. It is denoted by XX (capital letters) for dominant character (alleles) and xx (lowercase letters) for recessive character (alleles).
An individual possessing two of the same alleles at a gene locus is described as "homozygous."
Homozygous:
Homozygous refers to a genetic condition where an individual carries two identical alleles at a particular gene locus on a pair of homologous chromosomes.
These alleles can be the same for a particular trait, whether dominant or recessive.
For instance, if a person has two identical alleles for brown eyes (let's denote this as "BB"), they are considered homozygous for the eye color gene at that specific locus.
Understanding Alleles and Genetic Loci:
Genetic loci are specific positions on a chromosome where a particular gene is located. Each gene at a given locus can have multiple forms, known as alleles.
Alleles can be either identical (homozygous) or different (heterozygous) for a specific gene.
Contrast with Heterozygous:
In contrast, heterozygous individuals have two different alleles at a specific gene locus.
For example, if an individual has one allele for brown eyes and one allele for blue eyes (denoted as "Bb"), they are considered heterozygous for the eye color gene.
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1. Although generally not considered to be alive, a _______ is studied alongside other microbes such as bacteria2. The protein coat that surrounds the nucleic acid of a virus3. A viral life cycle that results in bursting of the host cell4. A viral life cycle in which the virus inserts its genome into the genome of its host, where it may remain dormant for long periods________.5. Viral genome that has inserted itself into the genome of its host________
Answer:
The correct terms are as - 1. virus, 2. capsid, 3. lytic cycle, 4. lysogenic cycle, 5. prophage.
Explanation:
A virus is considered as nonliving by various scientists as it required a living host to divide, however, it is studied with the microbes such as the bacteria and others. The virus is made up of nucleic acid surrounded by the protein coat called a capsid.
Viruses normally show two types of life cycles inside the host cell that are the lysogenic cycle and the lytic cycle.
The lytic cycle is characterized as the life cycle that ends with the destruction or bursting of the host cell. The lysogenic cycle includes being dormant inside the host genome until the favorable condition reestablished. Prophage is a virus that inserts itself into the host genome on its own.
Thus, the correct answer is - 1. virus, 2. capsid, 3. lytic cycle, 4. lysogenic cycle, 5. prophage.
The questions are about different aspects of virology including viral structure, and the lytic and lysogenic cycles. A virus, although not considered truly alive, is studied as a microbe. Its structure includes a protein coat (capsid) and its genetic content or nucleic acid. The lytic and lysogenic cycles describe how a virus interacts with its host.
Explanation:1. Although generally not considered to be alive, a virus is studied alongside other microbes such as bacteria. 2. The protein coat that surrounds the nucleic acid of a virus is known as a capsid. 3. A viral life cycle that results in bursting of the host cell is referred to as the lytic cycle. 4. A viral life cycle in which the virus inserts its genome into the genome of its host, where it may remain dormant for long periods is known as the lysogenic cycle. 5. A viral genome that has inserted itself into the genome of its host is referred to as a provirus or prophage.
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"The total concentration of receptors in a sample is 10 mM, and the concentration of free ligand is 2.5 mM. Calculate the percentage of receptors that are bound to ligand. Enter your answer as a number only (no percent symbol)."
To calculate the percentage of receptors that are bound to the ligand, subtract the concentration of free ligand from the total concentration of receptors (this gives the concentration of receptors that are bound), divide by the total concentration of receptors and multiply by 100. In this case, 25% of the receptors are bound to the ligands.
Explanation:To calculate the percentage of receptors that are bound to a ligand, you first need to determine how many receptors are not bound to the ligand. This value is the total concentration of receptors minus the concentration of free ligand. In this case, it would be 10 mM - 2.5 mM = 7.5 mM.
To find the percentage of receptors that are bound, we need to divide the receptors that are bound (free receptors) by the total concentration of receptors and then multiply by 100.
Thus, the calculation is as follows: (2.5 mM / 10 mM) x 100 = 25%.
The total concentration of receptors is 10 mM, and the free ligand concentration is 2.5 mM, therefore, 25% of receptors are bound to the ligand.
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The energy content and biomass of ________ is lowest in a terrestrial food web. Group of answer choices a. detritivores and decomposers b. top carnivores producers c. small carnivores such as spiders and lizards
Answer:
Option b. top carnivores is correct.
Explanation:
organism that found at the top of the PYRAMID OF NUMBERS that feed (preys) on other organisms but which is not itself preyed on are called top carnivore. They actually regulate terrestrial ecosystems. For example, tiger. (See attached Image)
In a terrestrial food web, producers have the lowest energy content and biomass.
Explanation:In a terrestrial food web, the energy content and biomass is lowest in producers. Producers, such as plants, convert solar energy into chemical energy through photosynthesis. They are the primary source of energy in a food web, and as energy flows through the system, some energy is lost at each trophic level, resulting in a decrease in energy content and biomass.
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Most naturally occurring selective pressures do not eliminate reproduction by the affected individuals. Instead, their reproductive capacity is reduced by a small proportion. How would your results differ if there was only 20% negative selection pressure rather then 100%?
Answer:
The result would differ in the sense that the reproductive success would not decrease at the same rate at 20% as it would at 100%. At 20% it would decrease faster.
Final answer:
A 20% negative selection pressure results in a slower rate of allele elimination from the population, leading to a more gradual evolutionary change compared to 100% negative selection pressure. This allows for greater genetic diversity and adaptability in facing future environmental changes.
Explanation:
Naturally occurring selective pressures, such as predators, diseases, or environmental conditions, tend to reduce an organism's reproductive capacity rather than completely negate it. If there was only 20% negative selection pressure, as opposed to 100%, this would mean that the affected individuals still retain a significant proportion (80%) of their reproductive capability. Consequently, the alleles or genetic traits subject to this pressure would be eliminated from the population at a slower rate compared to a scenario with 100% negative selection pressure. Over time, this could lead to a more gradual change in the allele frequencies within the population, affecting the pace at which evolution occurs under these selective pressures.
The evolutionary impact of differing levels of selective pressure illustrates how natural selection can shape populations over time. A 20% selective pressure allows for a greater diversity of genetic material to remain within the population, potentially enabling it to adapt to future changes in the environment or new selective pressures. Thus, understanding the nuances of selective pressure helps us grasp the complex dynamics of evolution and the survival of species in changing environments.
Why did agriculture become the preferred way of supplying basic human needs?
(Blank) organize the elements to form glucose
You are a virologist interested in studying the evolution of viral genomes. You are studying two newly isolated viral strains and have sequenced their genomes. You find that the genome of strain 1 contains 25% A, 55% G, 20% C, and 10% T. You report that you have isolated a virus with a single-stranded DNA genome. Based on what evidence can you make this conclusion? PLEASE EXPLAIN YOUR ANSWER!A. single-stranded genomes always have a large percentage of purinesB. Double-stranded genomes have equal amounts of A and TC. Single-stranded genomes have a higher rate of mutation
Answer:
B
Explanation:
The evidence that prompt this conclusion is the reported amount of nucleotide present which is varied because double stranded DNA genome which is usually heavier, more stable usually contains Adenine to thymine ratio of 1 to 1 and guanine to cytosine ratio of 1 to 1 and an over all of 1 to 1 ratio of purines to pyrimidines ( Double-stranded genomes have equal amounts of A to C )compared to what is reported which has a varied ratio ( A to T is 1 : 3 and G to C is about 0.77 for ss stranded DNA).
Describe the experiments that revealed the structure of the genetic material. If a sample of DNA contains 8% adenine (A), then what are the percentages of thymine (T), cytosine (C), and guanine (G)
Answer:
Explanation:
One of the earlier experiments that confirmed DNA and not proteins as genetic material was conducted by Alfred Hershey and Martha Chase
Hershey and Chase experiment showed the genetic material of virus is DNA not protein. Their experiments demonstrated that Phage are composed of DNA and protein, when phage infect bacteria, their DNA enters the host bacterial cell but not protein before infection
Hershey and Chase experiment prove that DNA is the hereditary material.
From chargaff base pairing rule, the ratio of purine to pyrimidine is a cell is 1:1. Since adenine pair with thymine and cytosine pair with quanine, this means the percentage of adenine will equal the percentage of thymine.
%A = %T and %G = %C.
%A + %T + %G + %C =100
If a DNA contains 8% adenine, therefore the percentage of thymine is 8%
8% + 8% + %G +%C = 100%
16% + %G +%C = 100%
%G +%C = 100%-16%
%G +%C = 84% = 84%/2= 42%
Since %G = %C
Then %G = 42%, %C = 42
By Chargaff's law, the percentage of a base is always equal to the base from which it pairs.
So the percentage of adenine is 8%, thymine is 8%, cytosine is 42%, and guanine is 42%.
Chargaff's lawAccording to the law :
If the percentage of adenine is 8%, the thymine concentration will also be 8% because adenine pair with thymine. The combine percentage of both pairs is 16%.The remaining percentage is 84%, so the percentage of cytosine and guanine will be 42% each. The 16% of adenine and thymine and 84% of guanine and cytosine makes 100% of concentration of four bases together.Thus, the percentage of adenine is 8%, thymine is 8%, cytosine is 42%, and guanine is 42%.
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Soon after the island of Hawaii rose above the sea surface (somewhat less than one million years ago), the evolution of life on this new island should have been most strongly influenced by _____.
A) genetic bottleneck.
B) sexual selection.
C) habitat differentiation.
D) founder effect.
Answer:
Soon after the island of Hawaii rose above the sea surface (somewhat less than one million years ago), the evolution of life on this new island should have been most strongly influenced by habitat differentiation
Explanation:
Habitat differentiation has to do with when new environment is carved out of old one to form a new one, just as the description that occurs in island of Hawai Rose.
Suppose that rat liver expresses a protein called Yorfavase. This enzyme is composed of 192 amino acid residues, and thus the coding region of the yfg gene consists of 576 bp. However, the rat genome database indicates that the yfg gene consists of 1440 bp. Select which type of DNA does not contribute to the additional 864 bp found in the yfg gene.
O O O 5-end untranslated regulatory region centromeric DNA
DNA coding for a signal sequence noncoding intron O promoter sequence
Answer:
Centromeric DNA
Explanation:
Genes are responsible for formation of proteins. A codon produces a single amino acid and it comprises of three base pairs. The whole DNA of organism does not contribute to encode protein. There are sequences in DNA that does not contribute in coding known as non-coding DNA and perform different functions. These may include the O O O 5-end, DNA coding for a signal sequence, noncoding intron and O promoter sequence .
The centromere is the part of DNA that links the sister chromatids. The centromere is not converted to mRNA or proteins. It is also not involved in regulation of genes. So it did not contribute to 864 bp found in the yfg gene.
A father with myotonic dystrophy has three daughters who are all carriers of the mutant allele and two sons who are unaffected noncarriers. the three daughters have six sons, of which four are affected and two are not, and four daughters, of which two are carriers and two are not. From this description, what type of mutation is probably responsible for myotonic dystrophy?
Answer:
Inherited Autosomal Dominant Mutations
Explanation:
The mutations in DMPK gene or ZNF9 gene contribute to the onset of Muscular Dystrophy. This is a type of inherited disorder that can run in families and due to it's autosomal transfer nature, it affects both sexes equally.
Multicellular organisms use cell division, mitosis, for growth and the maintenance and repair of cells and tissues. There are few cells in the body that do not undergo mitosis: most somatic cells divide regularly, some more than others. Single-celled organisms may use cell division as their method of reproduction. Regardless of the reason for mitosis, the process ensures genetic continuity. Consider the model of the cell cycle. Which detail(s) from the model best support the argument that cell division promotes genetic continuity?
The model of the cell cycle provides details about how cell division promotes genetic continuity, including DNA duplication, accurate segregation of chromosomes, and cytokinesis.
Explanation:The model of the cell cycle provides several details that support the argument that cell division promotes genetic continuity:
During interphase, the cell grows and the nuclear DNA is duplicated. This ensures that each daughter cell receives an exact copy of the genetic material, maintaining genetic continuity. In the mitotic phase, the duplicated chromosomes are segregated and distributed into daughter nuclei. This process ensures that the genetic information is accurately passed on to the next generation of cells. Following mitosis, the cytoplasm is divided through cytokinesis, resulting in two genetically identical daughter cells. This further ensures that each cell receives a complete set of genetic information. Learn more about Genetic continuity here:https://brainly.com/question/33255930
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In addition to providing yogurt with its unique flavor and texture, lactic acid-producing bacteria also provide which additional benefit during food production?a. Providing xenobiotics b. Lowering the pH to kill pathogenic bacteria c. Pasteurizing milk products d. Breaking down lactose for lactose-intolerant individuals
Answer:
Option b. Lowering the pH to kill pathogenic bacteria is correct answer.
Explanation:
bacterial motors are sensitive to pH. By decreasing the pH bacterial motors stops working. This was identified in a new research. But, with the weak acids and a lower internal pH they slow and ultimately stop moving (became dead).
Reference: Powell, K. Acid stops bacteria swimming. Nature (2003).
Which of the following statements is/are correct? 1. The secondary oocyte (ovum) contains most of the cytoplasm and organelles from the oogonium. 2. The secondary oocyte contains the diploid chromosome complement. 3. The polar body forms from discarded DNA during oogenesis.
In oogenesis, the secondary oocyte indeed inherits most cytoplasm and organelles from the oogonium, making statement 1 correct.
The question revolves around oogenesis, the process of ovum (egg) formation, and includes three statements for validation. Let's address each:
The secondary oocyte contains most of the cytoplasm and organelles from the oogonium. This statement is correct. During oogenesis, the secondary oocyte indeed inherits most of the cytoplasm, nutrients, and organelles to ensure the zygote has sufficient resources following fertilization.
The secondary oocyte contains the diploid chromosome complement. This statement is incorrect. The secondary oocyte does not contain a diploid set of chromosomes, but rather a haploid set. After the first meiotic division, the chromosome number is halved.
The polar body forms from discarded DNA during oogenesis. This statement is correct. Polar bodies are formed from uneven cell division, where they receive a minimal amount of cytoplasm and are mostly composed of unneeded DNA, eventually degenerating.
In summary, statements 1 and 3 are correct, while statement 2 is incorrect within the context of oogenesis.
If a steam that is septic (anaerobic) receives further human waste containing carbon, nitrogen and sulfur containing materials, what does the carbon become, the nitrogen, and finally the sulfur?
Answer:
Carbon convert into methane, Nitrogen converts into Ammonia and Sulfur converts into hydrogen sulfide gas
Explanation:
As the flow time in a wastewater treatment plant increase the color of waste water converts from grey to black as the condition becomes more anaerobic and flow becomes more septic. In such scenario if more waste water is added the sulfur and nitrogen extracts from the waste are converted into ammonia and hydrogen sulfide. Carbon is present in the form of organic waste and in the absence of oxygen it gets converted into methane gas. However , in presence of oxygen this carbon is released as carbon dioxide.