The question is incomplete. The part of the question after this is: Assume that you can track the cellular locations of these two proteins from the time that translation is complete until the proteins reach their final destinations.
Answer:
PFK: cytoplasm
insulin: ER--> Golgi--> outside cell
Explanation:
The proteins which are made and have to function in the same cell like Phosphofructokinase (PFK) do not have to undergo the modification processes which are required fro transporting a protein. Such kind of proteins are translated in the free cytoplasmic ribosomes and released into the cytoplasm where they start to function.
The proteins like insulin need to be traveled to different cells where they have to function. Such kind of proteins are formed in the ribosomes which have rough Endoplasmic Reticulum attached to them. From here, they travel to the Golgi complex where they are modified and packaged. From the Golgi-complex, these proteins are moved out of the cell.
As described in Investigating Life 20.1, scientists found that the range of the alpine chipmunk has shifted about 500 meters up in elevation, leading to isolated subpopulations on individual mountaintops. According to the researchers, what has caused this shift?
A.harmful UV radiation due to the ozone holeB.habitat destruction due to acid precipitationC.introduction of nonnative speciesD.climate warming due to increased CO2 concentrations
Final answer:
The shift in the Alpine chipmunk's range is caused by climate warming due to increased CO₂ concentrations, pushing species to higher elevations.
Explanation:
According to researchers, the shift in the Alpine chipmunk's range upwards by about 500 meters, resulting in isolated subpopulations on individual mountaintops, is primarily due to climate warming due to increased CO₂ concentrations. This conclusion is based on the observation that, globally, many species are moving towards higher altitudes or latitudes in response to changing climate gradients. With the warming climate, species adapted to colder environments find their habitable zones moving uphill. As a result, these species, including the Alpine chipmunk, are forced to follow the shift in their preferred climate range, leading to higher elevations and, unfortunately, potential habitat isolation and range restriction.
If a small founder population of 12 people ends up on a small island, and one member of the population had a mutation that creates a new allele, what is the likelihood that the allele will be eliminated from the population? Enter your answer as a percentage (not decimal), omit the percent sign however.
Answer:
96
Explanation:
This is genetic drift question, we can tell because the variable here is the population size. The probability of an allele of getting fixed is determined by its initial frequency, which is actually the probability of the allele of being fixed. So we need to calculate the initial allele frequency of this mutation:
Mutation allele frequency = # mutation alleles / total # of alleles = 1/24 = 0.04%
That is the probability of this allele to get fixed, so the probability of this allele getting eliminated is:
Elimination probability = 1 - fixing probability = 1- 0.04 = 0.96%
Answer: The probability of this allele getting eliminated is 96
In general terms, normally a cell in your body crosses a checkpoint in the cell cycle when
A. It receives a "go-ahead" signal directly from your brain
B. Concentrations of proteins in the cytoplasm reach the appropriate level
C. It receives a "go-ahead" signal from your external environment
D. Concentrations of DNA in the nucleus reach the appropriate level
E. A mutation in the DNA occurs
Answer:
B. Concentrations of proteins in the cytoplasm reach the appropriate level
Explanation:
We can say that the cell cycle checkpoints is an important cell cycle control. In this regard, it is important to note that within a cell, the passage from one phase to another is controlled by regulatory factors that are characteristically proteinic and act at the so-called cell cycle checkpoints.
A cell passes the first checkpoint when protein concentrations in the cytoplasm reach the appropriate level.
Prolonged use of lithium to treat bipolar disorder can cause damage to the kidneys such that epithelial cells of the collecting duct cannot insert aquaporin channels in their membranes. Which one of the following symptoms would you least expect in a person with such kidney damage?A. polyuria (excessive urine production)B. polydipsia (frequent drinking)C. excessive thirstD. high levels of vasopressin secretionE. glucosuria (glucose in the urine)
Answer:
E. glucosuria (glucose in the urine)
Explanation:
Generally, glycosuria occurs in patients with kidney changes due to diseases such as Wilson's disease or cystinosis, can also be a hereditary problem, but is not expected in patients with kidney damage caused by prolonged lithium use.
Normally, the kidneys filter the blood, eliminating all substances that are not necessary for the body to function, while glucose is reabsorbed in the blood because of its importance in energy production, but people with renal glycosuria do not reabsorb glucose. , which causes it to be eliminated in the urine, occurring glucosuria.
"In the brain, vision originates in the rods and cones in the retina. Separate regions of the brain decode basic information, like color, shapes, intensity of light, and there are other regions that decode information like position in space, and awareness of patterns. As you use your visual system, all of these regions are working simultaneously. This simultaneous awareness of all regions working at the same time is due to which processing pattern listed below?"
a. oscillative processingb. reflexive processingc. parallel processingd. serial processing
Answer:
The correct answer will be option-Parallel processing
Explanation:
Parallel processing is the ability of the brain in which the brain processes the incoming stimuli of the different quality at the same time or simultaneously.
During the vision, the brain processes different stimuli at the same time like the shape, intensity, color and their spatial arrangement by different parts of the brain individually and then analyzing and comparing the stimuli to the stored memories. This parallel processing helps the brain to identify the vision.
Thus, parallel processing is the correct answer.
Which of the following is TRUE for BOTH mitosis and meiosis?
A. The parent and daughter cells have the same policy.
B. The daughter cells are genetically identical to each other.
C. Homologous chromosomes are separated during a division.
D. Crossing over occurs.
E. Sister chromatids are eventually separated.
Answer:
E. Sister chromatids are eventually separated.
Explanation:
Sister chromatids are separated and move towards the opposite poles of the cell during anaphase of mitosis and anaphase-II of meiosis. The splitting of centromere and shortening of microtubules assist the separation of sister chromatids and their segregation during mitosis and meiosis.
Crossing over and separation of homologous chromosomes occur during prophase-I and anaphase-I of meiosis respectively. Crossing over adds genetic variations in the daughter cells. Mitosis lack crossing over and produce genetically identical daughter cells.
The death cap mushroom, Amanita phalloides, contains several dangerous substances, including -amanitin. This toxin is a potent inhibitor of eukaryotic RNA polymerase II. Death, usually from liver dysfunction, occurs no earlier than several days after mushroom ingestion. Please speculate why
Answer:
Poisoning caused by the mushrooms like by eating Amanita phalloides can lead to the death of the organism after a few days. The chemical α- amanitin causes serious problems to humans as it shows its effect by inhibiting the RNA polymerase II enzyme which acts during the formation of mRNA.
If mRNA will not be formed, therefore the process of protein synthesis will not take place as a result, all the cellular processes will be halted and the organism will die.
The organisms do not die frequently after consumption of the mushroom instead dies few days only when the α- amanitin starts showing its effect on the liver and kidney and the liver and the kidney fails to perform their work.
Since the liver and kidney fails after a few days by the chemical α- amanitin, therefore, the organism dies after a few days.
The death cap mushroom contains α-amanitin, which inhibits RNA polymerase II, leading to delayed symptoms of poisoning such as liver dysfunction.
The death cap mushroom, Amanita phalloides, is notorious for its toxicity due to substances like α-amanitin. This particular toxin is a potent inhibitor of RNA polymerase II, which plays a crucial role in the synthesis of messenger RNA (mRNA) in eukaryotic cells. When α-amanitin inhibits RNA polymerase II, it disrupts the transcription process, eventually leading to cell death. However, the symptoms of poisoning, including liver dysfunction, do not manifest immediately because it takes time for a significant number of cells to be affected to the point of causing organ failure. Moreover, the body's compensatory mechanisms can temporarily mask the deterioration in function.
Two related misconceptions about photosynthesis and cellular respiration are that only plants perform photosynthesis and that cellular respiration is performed in animals as the equivalent of photosynthesis. Explain why these statements are inaccurate. Be sure to provide an accurate description of the relationship between photosynthesis and cellular respiration.
Answer:
Photosynthesis and cellular respiration is not restricted to just plants and just animals.
Explanation:
Photosynthesis is the opposite reaction of cellular respiration. The equation for photosynthesis is:
6CO2 + 6H2O + (energy) ---> C6H12O6 + 6O2
If you reverse the reaction:
C6H12O6 +6O2 ---> 6CO2 + 6H2O + (energy)
You get the chemical equation for cellular respiration.
The relationship between photosynthesis and cellular respiration is that they are opposite reactions of one another.
One example of these reactions not being limited to their respected plants/or/animals is that photosynthesis is also performed by protists, which include algae and other single-celled organisms, that are technically not considered plants OR animals.
Another example is that plants also participate in the process of cellular respiration. Plants use photosynthesis to grow and produce glucose, and they use cellular respiration to break down the glucose formed from photosynthesis to release energy.
Answer:
Exactly the statement is incorrect because both organism needs respration for energy production thus it's not restricted to one organism,
Addition**
Respiration takes place at the mitochondria and both they have
"An experiment was performed to determine the role that ATP plays in kinesin movement along microtubules. Kinesin and microtubules were incubated together in a test tube, but instead of ATP, a non-hydrolyzable analog of ATP was added to the tube. What impact on kinesin function do you expect to observe in the presence of this ATP analog?"
Answer:
There will be no movement of kinesin motor protein along the microtubules .
Explanation:
Microtubules is an important component of cytoskeleton which play an important role in maintaining cell shape,cell motility,cell division regulation etc.
Whereas kinesin belongs to the family of motor proteins present in the eukaryotic cell.Kinesin plays an important role axonal transport.
The hydrolysis of ATP generates energy to power up the movement of kinesin along the microtubules.
According to the given question if kinesin and microtubules were incubated with an non hydrolyzable ATP analog instead of ATP,then the movement of kinesin along the microtubules will not occur due to lack of energy because the non-hydrolyzable analog of ATP cannot undergo hydrolysis to release energy.
After graduating from college, you decide to put your biology skills to work at a local company that does genetic counseling. Your first case is working with a couple that is trying to decide if it would be wise to conceive a child given the family’s genetic history with Huntington’s disease. This is a very damaging neurological disorder that usually strikes individuals later in life. The Huntington’s disease allele is a dominant allele (represented as "H"). Individuals who will not develop Huntington’s disease carry two copies of the recessive allele (represented as "h"). After testing both the wife and the husband, you determine that the wife’s genotype is hh and the husband’s genotype is Hh.
a)Which of them will eventually develop Huntington’s disease?
b)What are the possible genotypes for their children? What is the chance that their child will inherit Huntington’s disease?
Answer:
The correct answers are a) "the husband", b) The possible genotypes for their children are Hh and hh, while the child have a 50% chance of inherit Huntington’s disease.
Explanation:
Huntington’s disease is a very damaging neurological disorder that follows an autosomal dominant inheritance. This means that a healthy person has two copies of the recessive allele "h", while a person with Huntington’s disease has at least a copy of the dominant "H" allele. Since the husband genotype is Hh, he is going to develop the Huntington’s disease eventually. Also, it should be informed to the couple that their children could either have the Hh or the hh genotype with a 50% chance of having either of the genotypes, which means that their child has a 50% chance of developing the Huntington’s disease.
Final answer:
The husband with genotype Hh will develop Huntington's disease. Their children could inherit genotype Hh or hh, with a 50% chance of inheriting the Hh genotype and thus the disease.
Explanation:
In the scenario where a wife's genotype is hh and a husband's genotype is Hh, we can deduce the following:
a) Which of them will eventually develop Huntington’s disease?:
The husband will eventually develop Huntington's disease as he has one copy of the dominant allele H. Since Huntington's disease is an autosomal dominant disorder, only one copy of the dominant allele is required for the disease to be expressed.
b) What are the possible genotypes for their children? What is the chance that their child will inherit Huntington’s disease?:
The possible genotypes for their children are Hh and hh. There is a 50% chance that each child will inherit Huntington's disease -- that is, the child inherits the Hh genotype. This can be determined through a simple Punnett square, where one axis represents the husband's alleles (H and h) and the other axis represents the wife's alleles (h and h), leading to two possible offspring genotypes: Hh and hh.
The DNA in a cell's nucleus encodes proteins that are eventually targeted to every membrane and compartment in the cell, as well as proteins that are targeted for secretion from the cell.
For example, consider these two proteins:
1. Phosphofructokinase (PFK) is an enzyme that functions in the cytoplasm during
glycolysis.
2. Insulin, a protein that regulates blood sugar levels, is secreted from specialized
pancreatic cells.
Assume that you can track the cellular locations of these two proteins from the time that translation is complete until the proteins reach their final destinations.
For each protein, identify its targeting pathway: the sequence of cellular locations in which the protein is found from when translation is complete until it reaches its final (functional) destination. (Note that if an organelle is listed in a pathway, the location implied is inside the organelle, not in the membrane that surrounds the organelle.)
Options:
Cytoplasm only, ER --> cytoplasm, ER --> Golgi --> outside cell, cytoplasm --> ER --> outside cell, Golgi --> ER --> outside cell, cytoplasm --> Golgi --> outside cell, nucleus --> cytoplasm, ER --> Golgi --> cytoplasm
Protein Targeting Pathway
PFK _______________
Insulin ____________
Answer:
PFK (protein): cytoplasm only
Insulin (protein): ER->Golgi->outside cell
Explanation:
The protein Phophofructokinase (PFK) has to function inside the cytoplasm of the cell where it is made. Hence, it does not have to go through the modification process needed for transport. Such kind of proteins which are made and function in the same cell are translated on free cytoplasmic ribosomes.
The proteins which have to be transported to other areas like insulin are translated in the rough endoplasmic reticulum. From there, they travel to the Golgi complex where they are modified and packaged so that they can travel outside the cell.
How does the enzyme telomerase meet the challenge of replicating the ends of linear chromosomes?
A) It adds a single cap structure that resists degradation by nucleases.
B) It causes specific double-strand DNA breaks that result in blunt ends on both strands.
C) It catalyzes the lengthening of telomeres, compensating for the shortening that could occur during replication without telomerase activity.
D) It adds numerous GC pairs, which resist hydrolysis and maintain chromosome integrity
Answer:
The correct answer is C) It catalyzes the lengthening of telomeres, compensating for the shortening that could occur during replication without telomerase activity.
Explanation:
While replicating, the ends of the chromosomes always lose a part of the ends, therefore, the telomerase catalyzes this growing process, somehow compensating the information that will be lost.
Final answer:
Telomerase catalyzes the lengthening of telomeres, compensating for the shortening that could occur during replication without telomerase activity.
Explanation:
The telomerase enzyme meets the challenge of replicating the ends of linear chromosomes by catalyzing the lengthening of telomeres, compensating for the shortening that could occur during replication without telomerase activity. Telomerase contains a catalytic part and an RNA template. It attaches to the end of the chromosome and adds complementary bases to the RNA template on the 3' end of the DNA strand. Once the lagging strand template is sufficiently elongated, DNA polymerase can add nucleotides complementary to the ends of the chromosomes, ensuring the replication of chromosome ends.
A drug that prevents disassembly of the mitotic spindle would stall the cell cycle
A. in the middle of G1
B. just before DNA replication
C. just before chromosome condensation
D. just before or just after the separation of sister chromatids
E. at the G2 checkpoint
Answer:
D. just before or just after the separation of sister chromatids
Explanation:
The mitotic spindle is a structure of the eukaryotic cell cytoskeleton involved in mitosis. Their function is to separate the sister chromatids during cell division so that they are included in the daughter cells. It consists of longitudinally aligned microtubule bundles that together have an ellipsoid-like shape.
For this reason, we can state that if a drug that prevents disassembly of the mitotic spindle paralyzes the cell cycle immediately before or shortly after separation of the sister chromatids.
An electron loses potential energy when it:
A. Shifts to a less electronegative atom
B. Shifts to a more electronegative atom
C. Increases its kinetic energy
D. Increases its activity as an oxidizing agent
E. Moves further away from the nucleus of the atom
Answer:
B that is electron Shifts to a more electronegative atom
Explanation:
Electron loses potential energy when Shifts to a more electronegative atom. In an reduction-oxidation reaction, when an electron comes in proximity of high electronegative atom, it tends to lose some energy in form of potential energy. Electrons have more energy potential when associated with less electronegative atoms (such as C or H) and less energy potential when associated with a more electronegative atom (such as O).
hence the correct answer is B that is electron Shifts to a more electronegative atom
An electron loses potential energy when it shifts to a more electronegative atom, as the increased attraction to the atom's nucleus lowers potential energy and stabilizes the system.
Explanation:An electron loses potential energy when it shifts to a more electronegative atom. Electronegativity refers to the ability of an atom to attract electrons. When an electron moves from a less to a more electronegative atom, the electron is drawn closer to the atom's nucleus, resulting in a decrease in potential energy. This is because a more electronegative atom exerts a stronger pull on the electron, stabilizing the system by lowering the electron's potential energy.
Moreover, within an isolated atom, an electron loses potential energy when it moves closer to the nucleus (n decreases), making the system more stable and the atom ionized to a lower energy state if it captures an electron. This entire process enables the cell to incrementally use energy through the movement of high-energy electrons, which is crucial for metabolic pathways and the extraction of energy from food.
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Which of the following statements about the central dogma is correct? Drag "True" or "False" to the end of each statement. ResetHelp The central dogma predicts that mRNAs are transcribed into DNA. False Transcription is the process of using a single strand of DNA as a template to produce a complementary sequence of RNA. True The central dogma predicts that a change in a DNA sequence will result in a change in an RNA sequence. False The central dogma summarizes how information is transferred from DNA to RNA to protein in cells. True The arrows connecting DNA, RNA, and protein in the central dogma model indicate a conversion of one type of molecule to another. True
Answer:
The central dogma predicts that mRNAs are transcribed into DNA. FalseTranscription is the process of using a single strand of DNA as a template to produce a complementary sequence of RNA. TrueThe central dogma predicts that a change in a DNA sequence will result in a change in an RNA sequence. TrueThe central dogma summarizes how information is transferred from DNA to RNA to protein in cells. TrueThe arrows connecting DNA, RNA, and protein in the central dogma model indicate a conversion of one type of molecule to another. TrueExplanation:
The Central Dogma of Molecular Biology was postulated by Francis Crick in 1958. It explains how the flow of information from the genetic code occurs. This model mainly shows that a sequence of a nucleic acid can form a protein, but the opposite is not possible. According to this dogma, the flow of genetic information goes along the following lines: DNA → RNA → PROTEINS.
Since one molecule originates from the configuration of the other, if an error in the DNA sequence occurs, a change in the RNA sequence occurs, which in turn results in a change in protein.
According to the central dogma in DNA, where genetic information is contained, which can be transcribed into RNA molecules. In the transcription process, a DNA molecule serves as a template for the creation of an RNA molecule. It is in this RNA molecule that the code used to organize the amino acid sequence and to form proteins in the translation process is found. This process consists in the union of amino acids, obeying the order of codons presented in a messenger RNA.
The central dogma of biology states that genetic information is transferred from DNA to RNA to proteins, with the process of transcription playing a key role.
Explanation:The central dogma of biology describes how the information in genes is transferred from DNA to RNA, and then from RNA to proteins. It does not predict that mRNAs are transcribed into DNA. Instead, transcription involves a single strand of DNA serving as a template to produce a complementary sequence of RNA. A change in a DNA sequence may consequently result in a change in an RNA sequence, and thus, potentially a change in a protein sequence as well. The arrows linking DNA, RNA, and protein in the central dogma model symbolize this conversion process, or flow of information, from one molecule type to another.
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An individual with untreated type II diabetes has 400mg/100ml plasma glucose. Which of the following statements about this individual is INCORRECT?
-This individual will exhibit glucosoria
-This person's rate of glucose filtration will be greater that the rate of glucose reabsorption
-This person's SGLT carriers have reached saturation
-This person's rate of glucose excretion will be greater than the rate of glucose filtration
Answer:
The INCORRECT statement is This person's rate of glucose excretion will be greater than the rate of glucose filtration
Explanation:
The INCORRECT statement is This person's rate of glucose excretion will be greater than the rate of glucose filtration
It is incorrect because it is the opposite.
The incorrect statement is that the individual's rate of glucose excretion will be greater than the rate of glucose filtration.
An individual with untreated type II diabetes and a plasma glucose level of 400mg/100ml will indeed exhibit glucosuria, which occurs when the blood glucose level exceeds the renal threshold for glucose at 180mg/100ml.
At this point, the kidney's capacity to reabsorb glucose is surpassed, leading to the excretion of glucose in the urine. The statement that this person's rate of glucose excretion will be greater than the rate of glucose filtration is incorrect.
Part C - Breast-Feeding For most women, the American Academy of Pediatrics recommends exclusive breast-feeding for the first 6 months of the baby’s life. What are the benefits of breast-feeding? Please choose the correct statements about breast-feeding.
Please choose the correct statements about breast-feeding.Select all that apply.Select all that apply.1- Infants receive colostrum via breast milk during the first 6 months of breast-feeding.2- Breast-feeding is typically less expensive than formula feeding.3- Breast-feeding can reduce an infant’s risk of infection, allergies, and certain chronic diseases.4- All mothers should consume 500 kcal extra daily while breast-feeding until weaning of the infant.5- Women with AIDS or active tuberculosis should feed formula rather than breast-feed.
Final answer:
Breast-feeding provides critical benefits including nourishment through colostrum and cost savings over formula. It also reduces risks of various infant health issues, though extra maternal caloric intake may vary. Mothers with certain infectious diseases should opt for formula.
Explanation:
The American Academy of Pediatrics advocates for exclusive breast-feeding during the first 6 months of an infant's life due to its numerous benefits. Colostrum, which is produced in the first 48-72 hours postpartum, is an essential part of breast milk, providing vital immunoglobulins to the infant.
Breast-feeding is typically less expensive than formula feeding.
It can reduce an infant's risk of infection, allergies, and certain chronic diseases.
Mothers are recommended to consume additional calories while breastfeeding, but the exact amount may vary based on the individual's metabolism and the infant's needs.
Women with AIDS or active tuberculosis should not breastfeed due to the risk of transmitting the infection to the infant, and should instead use formula.
On the planet Trogus, the population of Trogites have an autosomal gene, HDZ, with two alleles, T and t. T is dominant to t. Individuals with the T allele in their genotype show a propensity for sorbet, while the remainder prefer ice cream. In multiple population studies, 60% of individuals with the TT genotype prefer sorbet, 25% of Tt individuals prefer sorbet, and 5% of tt individuals prefer sorbet.
In an F1 cross of Tt x Tt genotyped individuals, in the F2population of individuals who prefer sorbet, what is the probability that a random individual has the Tt genotype?
A. 0.991.
B. 0.141.
C. 0.435.
D. 0.044.
E. 0.522.
Answer:
The correct option is C. 0.435
Explanation:
To get the results, let's make a punnet square and check the results. The parents in the cross will be heterozygous prefering sorbet. We can say that the alleles for sorbet are dominant over the alleles which prefer ice cream.
T t
T TT Tt
t Tt tt
The results show that there is 25% (0.25) probability that the individual will be homozygous preferring sorbet (TT). 25% (0.25) will have the probability of preferring ice- cream (tt).
50% (0.5) which is nearest to 0.435 will have the probability for having heterozygous alleles (Tt) and preferring sorbet,
Stanozolol is an anabolic, androgenic (testosterone-like) steroid that acts as an agonist at androgen receptors. Unlike testosterone, it is metabolized to inactive form very slowly. What happens to a man who takes large doses of stanozolol to bulk up? (Please select all correct answers.)
Answer:
It will result /lead to side effect that could be fatal and life threatening since its inactivation time is longer thus the effect of the large dose will be more pronounced and long lasting.
Explanation:
An hormone agonist is a compound/molecule that can bind to the hormone specific receptor leading to the activation of the endocrine gland, the biosynthesis of secreted hormone and the eliciting the physiological effects of such hormone. Androgenic anabolic efficacy of steroids is usually goverened by their receptor binding andspecificity and their biotransformation from active to inactive form. Possible side effect include elevated blood pressure, voice deepening, decease libido, early myocardial infarction and many more
Jerry is a professional football player who injured his knee while running during a game. Jerry has had surgery and his knee is fully healed, but he is afraid to put pressure on it because he doesn’t want to re-injure it. The trainer implements a plan in which he first differentially reinforces standing, then walking, then jogging, and finally running at full speed. This is an example of using shaping to:
Answer:
reinstate a previous behavior
Explanation:
The rehabilitation to stop the pain and inflammation, for the recovery of the mobility and the muscular tone and the realization of a program of sports rehabilitation are essential to achieve a good recovery.
The recovery process must be carried out in a precise and multidisciplinary manner, the load that the knee must support and tolerate at all times, as well as rest and recovery times.
Anthocyanin is a pigment that gives flowers and leaves purple colors. The M gene codes for a transcription factor (Myb) that promotes expression of an enzyme that produces anthocyanin. The W gene codes for a different enzyme (Chs) that allows anthocyanin to be deposited in plant leaves and flowers. The dominant phenotype is the production of functional Myb and Chs. Use this information to answer the following question. Plants that have the mm genotype do not show any purple color. What is the best explanation for why this is?
Answer:
Anthocyanin is not produced in the plant cells
Explanation:
Anthocyanin is not produced in plant cells with the genotype mm.
As you can see from the question above, anthocyanin is responsible for the purple color of the flowers. Anthocyanin is encoded by the M gene, which is a dominant gene. Because it is a dominant gene, we know that it will be expressed in plants with the Mm and MM genotype, but will not be encoded by plants with the mm genotype. With this we can conclude that plants that have the mm genotype do not have purple color, because anthocyanin is not produced in the plant cells of these plants, since they do not have the M gene.
A cross is made between homozygous wild-type female Drosophila (a^+ a^+ b^+ b^+ c^+ c^+) and triple-mutant males (aa bb cc) (the order here is arbitrary). The F1 females are test crossed back to the triple-mutant males and the F2 phenotypic ratios are as follows: a^+ b c = 18 a b^+ c = 112 a b c = 308 a^+ b^+ c = 66 a b c^+ = 59 a^+ b^+ c^+ = 320 a^+ b c^+ = 102 a b^+ c^+ = 15 total = 1000 Map these gene to a chromosome in a correct order and determine the map distance between them. Show all your work.
Answer:
a is the middle gene.
Distance [b-a]= 24.7 mu
Distance [a-c]= 15.8 mu
Distance [b-a} = 40.5 mu
Explanation:
A homozygous wild-type female drosophila (a⁺b⁺c⁺/a⁺b⁺c⁺) is crossed with a homozygous recessive male (abc/abc). The order of the genes here is arbitrary.
The F1 is heterozygous for the three genes (a⁺b⁺c⁺/abc). The F1 females were test crossed (crossed with abc/abc males).
The F2 shows the following phenotypic ratios:
320 a⁺b⁺c⁺308 a b c 102 a⁺ b c⁺112 a b⁺ c66 a⁺ b⁺ c59 a b c⁺18 a⁺ b c 15 a b⁺ c⁺Total = 1000
The male parent is homozygous recessive for the 3 genes, so the observed phenotypes of the offspring correspond to the gametes received from the mother.
Recombination during meiosis is a rare event, so the most abundant gametes are always the parentals: a⁺b⁺c⁺ and abc.
The least abundant gametes, following the same logic, are the double crossovers (DCO): a⁺bc and ab⁺c⁺.
1st. Determine the gene orderCompare the parental and the DCO gametes. The allele that is switched corresponds to the middle gene. In this case, gene a is in the middle of the other two.
2nd Determine the single crossover gametes
The F1 mother that generated all 8 types of gametes had the genotype b⁺a⁺c⁺/bac (correct order of genes).
The single crossover (SCO) gametes resulting from recombination between genes b and a are b⁺ac and ba⁺c⁺.The single crossover (SCO) gametes resulting from recombination between genes a and c are b⁺a⁺c and bac⁺.3) Calculate the recombination frequencies between genesRecombination frequency (RF) = #Recombinants/Total progeny
RF [b-a]= (102+112+18+15)/1000= 0.247RF [f-br]= (66+59+18+15)/1000= 0.1584) Calculate the distance in map unitsDistance (mu) = RF x 100
Distance [b-a]= 0.247 × 100 = 24.7 mu
Distance [a-c]= 0.158 × 100 = 15.8 mu
Distance [b-a} = 24.7 mu + 15.8 mu = 40.5 mu
The gene map therefore looks like:b------------24.7 mu--------------------------a---------15.8 mu-----------c
The Punnett square above shows a cross between two sweet pea plants in Mendel's greenhouse. Both parents have blue flowers (Bb).
Which statement describes the offspring expected from this cross?
Answer:
answer is 75% blue-flower and 25% white- flower pea plants
Explanation:
As Blue is dominant trait,
both parents are heterozygous dominant
Bb (father ) bb( mother )
So crossing between
Bb x bb
B b
B BB Bb
b Bb bb
Offspring will be BB, Bb, Bb, bb
so 75% blue-flower and 25% white- flower pea plants
Scientific evidence suggests that the entire human population descended from just several thousand African migrants about 50,000 years ago." What sort of evidence would you expect to find if this claim were true? List the types of evidence you would expect to find if all humans living today came from a small population in Africa 50,000 years ago.
Final answer:
If the claim that all humans descended from a small population in Africa 50,000 years ago is true, we would expect to find genetic evidence, genetic variance, and archeological evidence supporting this claim.
Explanation:
Scientific evidence suggests that the entire human population descended from just several thousand African migrants about 50,000 years ago. If this claim were true, we would expect to find several types of evidence:
Genetic Evidence: Researchers have found that all human genomes tested outside of Africa have close ties to the genomes of people in Africa. This indicates that there is a genetic link between humans from different parts of the world, supporting the idea of a common African ancestry.
Genetic Variance: Genetic studies have also shown that there is a wider genetic variance in Africa compared to the rest of the world. This suggests that the ancestral population in Africa was larger and more genetically diverse, supporting the idea that all humans trace their ancestry back to Africa.
Archeological Evidence: Archeological findings such as fossils and artifacts can provide clues about human migration patterns. If all humans originated from Africa, we would expect to find evidence of early human settlements and migration routes out of Africa to different parts of the world.
Vitamin K, one of the fat-soluble vitamins, plays a role in both bone health and blood clotting. There are two primary forms of the vitamin: phylloquinones and menaquinones. Choose the correct statement about vitamin K.a. Menaquinones are the primary form of vitamin K in the diet.b. Vitamin K deficiency inhibits the blood’s ability to clot.c. Excessive use of broad-spectrum antibiotics may cause vitamin K toxicity.d. The major source of vitamin K in our diets is from fortified grains.
Answer:
b. Vitamin K deficiency inhibits the blood’s ability to clot.
Explanation:
The main form is vitamin K1 (phylloquinone); followed by vitamin K2 (menaquinone), formed from the bacterial action on the large intestine tract and a third compound, vitamin K3 (menadione), a synthetic fat-soluble molecule. These vitamins can be found in many foods: in green vegetables (lettuce, cauliflower and spinach), in tomatoes, Brazil nuts, cashews, potatoes, contained in soybean oil, egg yolk, milk and to a lesser extent in wheat and oats.
Vitamin K participates in blood clotting, its deficiency can make it difficult to stop bleeding. Also causing damage to the body when in excessive concentration, for example: dyspnea (shortness of breath), and chest pain in adults with high vitamin K1 disorder and hyperbilirubinemia in newborns whose mothers in gestation underwent treatment with based on vitamin K3.
A differential agar medium: Select one: a. only selects for one specific kind of organism b. none of the above c. contains a single substrate that shows differences between groups of microbes on the basis of different chemical reactions producing different appearances d. displays specific characteristics of different species of bacteria
Answer:
C
Explanation:
Differential media contains single substrate that shows differences between groups of microbes on the basis of different chemical reactions producing different appearances. Example include MacConkey agar, Blood agar.
In blood agar microbes are differentiated on the types of hemolysis produce during the break down of red blood cells. They are alpha hemolysis characterize by partial destruction of red blood cells with greenish to brownish discoloration of the medium, beta hemolysis characterize by clear zone of destruction of red blood cells and gamma or no zone of clearing.
In normal mitochondria the rate of electron transfer is tightly coupled to the demand for ATP. When the rate of use of ATP is relatively low, the rate of electron transport is low. When demand for ATP increases, electron-transport rate increases. Under these conditions of tight coupling, the number of ATP molecules produced per gram of oxygen consumed when NADH is the electron donor-the P/O ratio is about 2.5-3.0.
Ingestion of uncouplers causes profuse sweating and increase in body temperature. Explain this phenomenon in molecular terms. What happens to the P/O ratio in the presence of uncouplers.
To study the molecular mechanism of DNA replication, you incubated soluble E. coli extracts with a mixture of dATP, dTTP, dGTP, and dCTP, all of which are labeled with the -phosphate group. After a while, the incubation mixture was treated with trichloroacetic acid, which precipitate the DNA but not the nucleotide or very short oligonucleotides. The precipitate was collected and the extent of the radioactively label nucleotide incorporation into the DNA was determined.
a.) If any one of the four nucleotides is omitted from the incubation mixture, would you detect radioactivity in the precipitate? Explainb.) Would 32P be incorporated into the DNA if only dTTP is labeled? Explainc.) Would radioactivity be found in the precipitate if 32P labeled the β and γ rather than the α phosphate of the deoxyribonucleotides? Explain.
Answer:
Explanation:
A) Since, mixture of dATP, dTTP, dGTP, and dCTP all were labeled; radioactivity can be detected even if one of the four nucleotides is omitted from the incubation mixture. This is because all other three nucleotides contain labeled phosphate and hence, radioactivity can be detected.
B) Yes. If 32P is labeled only dTTP, it can be detected in DNA. Since, dTTP can pair with 'A' nucleotide in DNA.
C) Mostly radioactivity can be detected in DNA when the alpha or terminal phosphates are labeled. It is difficult to detect radioactivity when phosphate groups other alpha like beta or gamma phosphates are labeled. Even if the radioactivity is detected it can be detected in rare cases like 5`end labeling of DNA on gamma phosphate.
To understand DNA replication, omitting any nucleotide from the mix or changing the radiolabel's location alters the assimilation of radioactivity in precipitated DNA. Presence of all nucleotides is crucial for complete DNA synthesis, and only the α-phosphate group incorporation leads to radioactive DNA.
Explanation:To study the molecular mechanism of DNA replication in E. coli extracts using nucleotides labeled at the α-phosphate, several scenarios illustrate how the radioactivity incorporation would be affected by experimental modifications.
If any one of the four nucleotides (dATP, dTTP, dGTP, dCTP) is omitted from the incubation mixture, radioactivity in the precipitate would likely be reduced or absent because DNA synthesis requires the presence of all four nucleotides. Each is necessary for building the complementary strand, thus, omitting one would halt DNA polymerase activity, and no or significantly less labeled DNA would be precipitated.If only dTTP is labeled with 32P, 32P would be incorporated into the DNA where thymine bases are incorporated. As DNA synthesis proceeds, dTTP is incorporated opposite adenine (A) in the template strand, resulting in radioactive labeling of the DNA.Should 32P label the β and γ rather than the α phosphate of the nucleotides, radioactivity would not be found in the precipitate. This is because, during DNA synthesis, the α-phosphate group is incorporated into the DNA backbone, while the β and γ phosphates are cleaved off during the formation of the phosphodiester bond.You treat bacteria with an extract made from peanuts that had a fungus growing on it. You then plate 10s of these bacteria onto a plate that contains penicillin. As a control you leave out peanut extract. You obtain 2 colonies from the control and 3 colonies after treatment with the corn extract. You conclude:
A. The peanut extract is not mutagenic.
B. The peanut extract is mutagenic.
C. Penicillin is mutagenic.
D. None of the answers listed above (ABC) are correct.
Answer:
C. Penicillin is mutagenic.
Explanation:
Mutagens are agents that change genetic material. They multiply the frequency of mutations of an organism over their standard level. The peanut extract that was left as control shows this natural or standard level, but when tenths of these bacteria were put into a plate that contained penicillin, the bacteria colonies increased the frequency of mutations of these bacteria after a determined time.
Considering the nitrogen cycle, how do all organisms depend on bacteria to produce and maintain adequate nitrogen in the environment?
I NEED HELP ! Choose the correct answer..
A. Bacteria eat the roots of plants that emit nitrogen.
B. Animals eat plants that contain bacteria, which contain nitrogen.
C. Since plants cannot use atmospheric nitrogen directly, they rely on bacteria in the soil to incorporate nitrogen into nitrates.
D.Plants and animals work together to create the appropriate amount of decay in the soil for nitrogen to form.
Answer:
Considering the nitrogen cycle, all organisms depend on bacteria to produce and maintain adequate nitrogen in the environment by eating the plants that contain bacteria which contain nitrogen.
Option: (B)
Explanation:
Nitrogen is a main element in the nucleic acid of both RNA and DNA which is most important for all living creatures and biological molecules. When plants doesn’t get enough nitrogen it doesn’t produce amino acid, without amino acid plant cannot make special proteins. Amino acid is building block for DNA which tends to the generation of organism. Nitrogen fixation is a process where bacteria convert atmospheric nitrogen to usage form for plant and animals get nitrogen by eating those plants.