Answer:
The correct answer will be- There is some molecule which could have transformed the avirulent bacterial strains to become virulent and called it a transforming principle.
Explanation:
Griffith performed the experiment on two strains of Streptococcus pneumonia in which one strain is avirulent called R strain and another virulent strain called S strain.
When Griffith killed the S bacteria (virulent) by heat and added it in the R strain Petri dishes and then injected the bacteria to mice, the mice killed. This amazed him as to how the avirulent bacteria killed the mice ?
He concluded that there was some principle which transformed the avirulent bacteria to virulent bacteria and thus called the principle as transforming principle.
Brown eyes are dominant over blue eyes. If a woman with a dominant brown eyes ,(homozygous) reproduces with a man with recessive blue eyes , what is the chance they will have a child with brown eyes?
Answer:
The chances are 100%
Explanation:
A dominant allele is the allele that expresses itself in the offspring even if it is present in heterozygous condition(Bb). So when a woman with homozygous dominant(BB) condition of brown eyes reproduces with a man having recessive blue eyes(bb) then all the offsprings with heterozygous dominant condition for brown eyes will be produced.
B B
b Bb Bb
b Bb Bb
All the offsprings will have one dominant allele which is enough to produce brown eyes therefore all the offsprings will have brown eyes. So the chances are 100%.
Answer:
The chances will be 100%
Explanation:
List and describe four stages of food processing found in higher animals.
URGENT WILL NAME BRAINLIEST ANSWER NUMBER 28 EXTRA POINTS HELP PLEASE ASAP!
Answer:
So at point 1
Explanation:
at point 1 the rate of enzyme activity is increasing with increase in the concentration of sugar.
while at other points the enzyme activity remains constant with increase in concentration of sugar.
Answer:
Option 1
Explanation:
POINT 1 IS THE ACTIVATION POINT OF ENZYMATIC ACTIVITY THEREFORE AT THIS POINT ACTIVITY OF ENZYME WILL INCREASE WITH THE INCREASE OF SUGAR CONCENTRATION. WHILE ON POINT 2, 3, 4 THE ACTIVITY OF ENZYME REMAINS CONSTANT AS ENZYME ACTIVE SITE ARE ALREADY OCCUPIED BY THE SUGAR MOLECULES SO IF MORE SUGAR WILL BE ADDED IT DOSES NOT INCREASE ENZYME ACTIVITY.
Think of the number of different parasites (species of bacteria, viruses, fungi, flatworm, apicomplexan, etc.) that can attack humans. Is it reasonable to expect that every other organism faces a similar array of parasites?No--most species should have a single, generalist parasite. (Generalists can exploit more than 1 species.)No--most species should have a single, specialist parasite.Yes, humans are a relatively young species so should have a relatively low number of parasites.Yes--every species represents a resource that may be exploited by an array of parasites
Answer:
The correct answer will be option-Yes--every species represents a resource that may be exploited by an array of parasites
Explanation:
Parasites are the organism which for survival depends on the resources of the host organism and utilize them to the extent that it could lead to the death of the host.
The interaction between the parasite and host proves harmful to the host but beneficial to the parasite.
A number of parasites exist for human species which can directly harm humans. Similarly, a large number of hosts exist for different species which belongs to another kingdom also like even the bacteria has a parasite called bacteriophage which utilizes the resources of the host.
This indicates that every species has some resources which can prove beneficial to another organism in the response of which they become host to a large number of parasites.
Thus, the selected option is the correct answer.
You have added an irreversible inhibitor to a sample of enzyme and substrate. At this point, the reaction has stopped completely. What can you do to regain the activity of the enzyme? a. Removing the irreversible inhibitor should get the reaction working again.
b. Adding more substrate will increase the rate of reaction.
c. Adding more inhibitor should get the reaction up to speed again.
d. The enzyme is inactive at this point. New enzyme must be added to regain enzyme activity.
Answer:
The correct answer is d. The enzyme is inactive at this point. New enzyme must be added to regain enzyme activity.
Explanation:
This is because the inhibitor is irreversible, meaning that the enzyme is permanently modified, therefore, even if more substrate is added, the enzyme will no longer be able to work properly. If all of the inhibitor is consumed, adding more enzyme would regain the activity of the enzyme.
Irreversible inhibitor permanently inactivates enzymes. Removing the inhibitor, or adding more substrate or inhibitor will not recommence the enzyme’s activity. Only fresh, active enzyme will increase the rate of reaction.
Explanation:In biochemistry, an irreversible inhibitor permanently binds to an enzyme, rendering it incapable of catalyzing a reaction. As a consequence, the enzyme cannot regain its original function once it has encountered an irreversible inhibitor. Therefore, removing the irreversible inhibitor, adding more substrate or adding more inhibitor will not resurrect the enzyme’s activity.
The best answer is option D: The enzyme is inactive at this point. New enzyme must be added to regain enzyme activity. To increase the rate of reaction, fresh, active enzyme must be introduced to the system.
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Please match the Key Word to the correct definition.
1. h
2. b
3. c
4. d
5. g
6. a
7. f
8. e
A woman with severe discoloration of her tooth enamel has four children with a man who has normal tooth enamel. Two of the children, a boy (B) and a girl (G), have discolored enamel. Each has a mate with normal tooth enamel and produces several children. G has six children, four boys and two girls. Two of her boys and one of her girls have discolored enamel. B has seven children, four girls and three boys. All four of his daughters have discolored enamel, but all his boys have normal enamel.Explain the inheritance of this condition
Answer:
Discolored tooth enamel appears to be an X-linked dominant trait in this family.
Explanation:
Which of the following might explain the lack of cloud formation? Choose one or more:
A. The air package rises too fast for condensation to occur.
B. It is too hot for adiabatic cooling to take place.
C. The air package never reaches its dew point.
D. There are too few aerosols present in the air package.
Answer:
The following might explain the lack of cloud formation
The air package never reaches its dew point. There are too few aerosols present in the air package.Explanation:
For the formation of the cloud the air in the higher position should have much colder than that of the surface. The vapours travels till it reaches the point where it has lower temperature as well as the pressure so that the water will condense into the droplets which will result in the formation of the cloud. When those vapours condenses to the liquid or to the solid it will release some of it energy to the air. For the formation of the cloud three things are mandatory and they are moisture, cooling air and the condensation nuclei. Also, for the formation of clouds, suspended particles like aerosols are required for the water to condense. As there are few aerosols, water doesn't get to condense.
Many long-distance swimmers "carb-load" before an endurance event like a 5 km swim. Carbohydrate loading involves eating carbohydrate-rich meals several days to one week before the event. In fact, many groups sponsor dinners the night before a race and serve high-carbohydrate foods like pasta. However, studies are mixed about the usefulness of this practice. One problem in practice is that it is difficult to consume such a high level of carbohydrates over several days. Event distance is a factor. Also, studies have failed to consistently show that performance - speed and/or endurance - improves, even when the diet is followed. Although some athletes report improved performance, it is possible that this is a placebo effect. Genetic and gender differences may also play a role. Regardless of its efficacy, what is the reasoning behind carbohydrate loading? Excess glucose enables the body to produce more ATP. which is stored in the body and can be used during the event. Decreasing the amount of fat and protein in the diet helps with weight loss, which can improve endurance and speed. Excess glucose is stored as muscle glycogen, reducing the amount of lactate generated in muscle during the event. Excess carbohydrates are stored as triacylglycerols, which can be oxidized during anaerobic metabolism. o Glucose is stored as glycogen, which can be broken down to supply energy during the event.
Answer:
Excess glucose is stored as muscle glycogen, reducing the amount of lactate generated in muscle during the event.
Explanation:
Carbohydrates are a main source of energy for the body and loading up on carbohydrates is believed to help store large amountsof energy and is used to help replenish and load muscle glycogen content. Athletes tend to load a few days prior to a long endurance athletic event to help provide adequate and extra energy to muscles.
Rhonda tells Jack that a solution is a type of mixture. Jack disagrees and responds that he does not think that a solution is a mixture.
Which explanation below explains who is correct?
A. Rhonda is correct because a solution is a type of mixture. It is a mixture because the substances are mixed together, and each substance loses its primary physical properties. The substances can each be easily seen.
B. Jack is correct because a solution is not a type of mixture. It is not a mixture because one of the substances settles at the bottom of the solution. Each substance keeps it primary physical properties.
C. Rhonda is correct because the substances keep their primary physical properties, but one dissolved into the other. Each substance is not easily seen.
D. Jack is correct because a solution is not a type of mixture because the substances stay mixed together and cannot be separated. Each substance loses its its primary physical properties.
Answer:
Jack is correct because a solution is not a type of mixture because the substances stay mixed together and cannot be separated. Each substance loses its primary physical properties.
Option: (D)
Explanation:
A mixture is composed of 'two or more different substances', physically combined and it is in the form of a solution, suspension or a colloid. A mixture may be homogeneous (salt and water) or heterogeneous (sand and water). A homogeneous mixture of two or more compounds is termed as solution. It has a solute and a solvent. The 'solute' is the substance dissolved in the solvent. Both lose their primary physical properties.When one allele is dominant to another, what is the most common biological mechanism for this dominance?
A. Dominant alleles often encode functional products, whereas recessive alleles often do not encode functional products
B. Dominant alleles subdue or repress recessive alleles
C. Cellular machinery recognizes which allele is dominant and only reads the
instructions from that one
D. A dominant allele on one chromosome causes the other, homologous
chromosome (containing the recessive allele) to be completely inactivated
Answer:
A. Dominant alleles often encode functional products, whereas recessive alleles often do not encode functional products
Explanation:
One gene can have it's variant forms called allele but in one individual only two allele form of a gene can be present. These two allele form can be the same(homozygous) or different(heterozygous). A dominant allele often encode a function protein of product while recessive allele often code for non- functional product.
Therefore when the dominant allele is present in the chromosome then function products are formed which can be seen in the phenotype and even a single dominant allele produce the functional product like in heterozygous condition so phenotype in heterozygous and homozygous dominant condition appears same. So the correct answer is A.
The basic hereditary unit of information in biology is the
A. cell
B. gene
C. chromosome
D. genome
E. gamete
Answer: Option B
Explanation: A gene is the specific portion or sequence on the DNA (Deoxyribonucleic acid) that is able code for a functional protein. Thus, the entire DNA sequence in each cell consists of several genes and some non-coding portions that all makes up the complete hereditary information usually called genome.
Observe this linear chain
Cell -> Chromosomes -> DNA -> Gene.
The above illustrates that a gene is the basic carrier of hereditary information.
This will definitely help.
The cell cycle must be regulated in a multicellular organism in order to
A. maintain the proper cell number in tissues
B. activate cell division when repair of a tissue is needed
C. halt cell division when repair of a tissue is complete
D. prevent cancer
E. all of the above
Answer:
The correct answer is E. all of the above
Explanation:
The cell cycle has several checkpoints that helps in the regulation of the cell cycle. Cell cycle regulation is important because if the cell cycle is not regulated then the cell will constantly divide which will be a waste of energy and an abnormal increase in the number of cells can do damage to the tissue. So regulation is important to maintain a proper cell number in the tissue.
Cell division helps in repairing the damaged tissue, therefore, cell division must be stopped after tissue is repaired and regulation is also important to prevent cancer because cell division stops when the mutation occurs in the cell's genetic material. Therefore the correct answer is E.
It has been observed that cancer cells take in much more glucose than normal cells. In fact, labeled glucose can be used to detect cancerous tumors through PET scans. Explain the changes in the metabolism of cancer cells that cause them to use so much glucose. Include the following terms in your answer: flux, regulation, ATP, and oxygen.
Answer:
Explanation:
Cancer cells are characterized by higher proliferation and replication of cells than normal cells. This causes a high demand for energy and nutrients. Cancer cells also thrive under hypoxic or low oxygen environment and they depend highly on glucose consumption for ATP production, that it, glycolysis is the main pathway for generating energy. Glucose transporters (GLUT's) are integral membrane proteins that mediate the flux of glucose in cells and have been implicated in cancer. For example GLUT2 has been found to be up-regulated in human carcinoma cells.
Which statement regarding ATP is false? Compared with other phosphate‑containing compounds in biochemistry, ATP has a higher free energy than nearly all of the others. Most enzymes that use ATP bind it in a form complexed with Mg2+. ATP concentrations in a cell are usually much higher than the equilibrium concentration for ATP hydrolysis. ATP can act as a group donor in adenylylation reactions.
Answer:
The correct answer will be option- ATP concentrations in a cell are usually much higher than the equilibrium concentration for ATP hydrolysis
Explanation:
Adenosine Triphosphate or ATP is known as the energy currency of the cell as it provides energy during the metabolic reactions.
The study on the structure of ATP revealed that it contains ribose sugar attached with three phosphates group bonded to each other via two phosphoanhydride bonds.
The metabolic reaction usually uses Magnesium bonded ATP as Magnesium covers the negative charge of the triphosphate which allows the ATP to bind hydrophobic clefts of the active site.
The ATP also transfers the adenyl group to the nucleophile which helps generate the pyrophosphate.
Which of the following locations is LEAST likely to harbor commensal microbes?A. Blood B. Mouth C. Skin D. Vagina
Answer: Option A
The blood is the least that can harbour commensal microbes.
Explanation:
Commensal microbes are microbes that form a relationship with an organism in which one benefits and neither is hurt. These bacteria are necessary for a healthy and developed immune system. They are part of the microplora in the mouth. Commensal bacteria are least in the blood because the blood is a sterile environment and is not conducive for their habitation. The immune response of the blood to the bacteria can cause septic shook which can lead to death.
Match the appropriate labels to their respective targets.
a. Hormonal stimulus
b. Neural stimulus
c. Humoral stimulus
d. Humoral and/or hormonal stimulus
1. Aldosterone production
2. Parathyroid hormone production
3. Testosterone production
4. Epinephrine production
Answer:
Hormonal stimulus [tex]\rightarrow[/tex] testosterone production
Neural stimulus [tex]\rightarrow[/tex] epinephrine production
Hormonal stimulus [tex]\rightarrow[/tex] aldosterone production
Humoral stimulus [tex]\rightarrow[/tex] parathyroid hormone production
Explanation:
The correct matching for the given unmatched is given below
Hormonal stimulus [tex]\rightarrow[/tex] testosterone production
Neural stimulus [tex]\rightarrow[/tex] epinephrine production
Hormonal stimulus [tex]\rightarrow[/tex] aldosterone production
Humoral stimulus [tex]\rightarrow[/tex] parathyroid hormone production
Final answer:
Hormone release is regulated by humoral, hormonal, and neural stimuli. Aldosterone production is a result of a humoral stimulus, parathyroid hormone is regulated by hormonal stimulus, testosterone by both humoral and hormonal stimuli, and epinephrine by a neural stimulus.
Explanation:
The regulation of hormone release can be triggered by a variety of stimuli. Here, we match the appropriate labels to their respective targets representing the hormonal release mechanism:
Aldosterone production - c. Humoral stimulusParathyroid hormone production - a. Hormonal stimulusTestosterone production - d. Humoral and/or hormonal stimulusEpinephrine production - b. Neural stimulusEach of these targets are regulated by one of the three mechanisms:
Humoral stimuli are changes in ion or nutrient levels in the blood.Hormonal stimuli refer to the release of hormones in response to other hormones.Neural stimuli occur when a nerve impulse prompts the secretion of a hormone.All the cells shown are part of the same organism. Which can you infer to be
true?
O
All of the cells came from the same cell.
O
All of the cells are able to produce food molecules.
O
All of the cells have the exact same function.
O
All of the cells have the exact same structures.
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Answer:
hi there!
Explanation:
• Variation light and variation in temperatura
For me these are the problem, because they had to move some plant to the outdoors, because they don´t have enough space inside. So this changed of place could affect the plants.
Through genetic engineering, you replace a specific second messenger in a cell’s signaling system with a new second messenger whose molecule has a greatly increased lifetime. How will this change most likely affect the ability of the cell to respond to signaling events that depend on this pathway?
Answer:
The correct answer is - less efficient signaling.
Explanation:
Second messengers are the molecules that help in transferring the information in the form of intracellular signal that has come from external stimulus.
Second messenger molecules should be short-lived as the concentration of these molecules should be decreased fastly so the cell can be available for other stimuli mediated by the same second messenger.
With increased lifetime, signaling will be less efficient as another stimulus cannot be detected.
Thus, the correct answer is - less efficient signaling.
A longer lifetime of a second messenger in a cell's signaling system will likely prolong the cell's response to signaling events that depend on this pathway, potentially affecting cellular processes such as metabolic changes, gene activation, and protein synthesis. However, it could also disrupt the balance of signaling events, potentially leading to alterations in cell function.
Explanation:Genetic engineering and replacement of a second messenger in a cell's signaling system with a new one that has a greatly increased lifetime would have significant effects on cell signaling. Second messengers, such as cyclic adenosine monophosphate (CAMP) and inositol triphosphate (IP3), play a crucial role in activating metabolic changes within the cell once produced by the effector protein. These changes can involve the activation of other enzymes, modification of ion channels, and changes such as the activation of genes in the nucleus which leads to an increased synthesis of proteins.
With an increased lifetime of the second messenger, the modifications and activations made by this messenger can be prolonged, leading to the prolongation of the cell's response to signaling events that depend on this pathway. This can result in longer lasting cellular responses like protein synthesis, cell metabolism changes, and ion channel modifications. Interestingly, these kinds of changes in neurons often form the basis of stronger connections between cells at the synapse and serve as the basis of learning and memory. However, it's also important to note that an increase in the lifetime of the second messenger could potentially disrupt the careful balance of signaling events. Cell signaling needs to be carefully regulated to ensure appropriate response intensity and timing. Hence, any drastic changes in the components of a signaling pathway can lead to complications or alterations in cell function.
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Koch's postulates are used to establish the infectious agent that causes a particular infectious disease. In 1937, Thomas Rivers modified Koch's postulates so that they could be applied to viral pathogens.
Which of Koch's postulates was modified by Rivers?
A. The pathogen must be present in every case of the disease.
B. The pathogen must be grown in pure culture.
C. The pathogen must cause the same disease when introduced into experimental animal hosts.
D. The pathogen must be recovered from the diseased experimental lab animals.
Answer:
B. The pathogen must be grown in pure culture.
Explanation:
Virus, and also some bacteria can not be cultured in "lifeless media". In the case of bacteria, we refer to the obligate intracellular ones, and all virus. Thus, this postulate can not be fulfilled in all cases. Virus can be propagated on tissue cultures, but certainly this was not considered by Koch. Moreover, these methods often result in spontaneous genetic alterations, which leads to adaptation and even attenuation in animal models of disease, so, the cultured virus perhaps does not induce the same disease when introduced to another specimen.
A woman reports a racing heart that began while she was playing softball 2 hours ago. Her ECG shows a rapid and regular heart rate, and she is given an intravenous bolus of a drug that activates G protein–coupled receptors. Which drug mechanism may lead to termination of the arrhythmia?
Answer: increase reflux of potassium
Explanation:
The woman possibly has supraventricular tachycardia which may precipitate by a premature atrial deplorization. Adenosine is administered to her which will bind and activate the G-protein coupled receptor that has adenosine binding site. This activation leads to the efflux of potassium and the hyperpolarization of supraventricular tissue.
The drug type used in this case should be beta blocker drug who has G protein as transduction unit where beta 1 or beta 2 adrenergic receptor is blocked.
The characters of "seed color" and "seed shape" showed independent assortment in Mendel's experiments because
A. the genes for these two characters were very close together on the same chromosome
B. these two characters were the product of the same gene
C. the genes for these two characters were on different chromosomes
D. the genes for both characters were sex-linked
E. either answer A or B could be correct
Answer:
C. the genes for these two characters were on different chromosomes
Explanation:
Mendel conducted several experiments with peas to demonstrate the mechanisms of heredity. He decided to use peas because they self-fertilize, have a short life cycle, are easily grown and produce large numbers of offspring. Peas, in addition to the advantages already mentioned, have simple characteristics to observe, such as the color and shape of the seed. Mendel analyzed seven of these characteristics that had no intermediates: seed texture, seed color, seed coat color, pod shape, pod color, flower position, and plant height.
Through his study, Mendel concluded that the characters "seed color" and "seed shape" showed an independent variety. This is because the genes related to these two traits are distant from each other in the chromosome and therefore have no dependence between them.
Complex I, also called NADH dehydrogenase, is an entry point for electrons to the electron transport chain. Complex I transfers electrons from NADH to Q. The reduction of Q is linked to the active transport of protons out of the matrix. How many protons does complex I pump into the intermembrane space?
Final answer:
Complex I, or NADH dehydrogenase, of the electron transport chain pumps four protons into the intermembrane space for every two electrons it receives from NADH.
Explanation:
Complex I, also known as NADH dehydrogenase, is a crucial component of the electron transport chain in cellular respiration. Its role is to transfer electrons from NADH to coenzyme Q (ubiquinone), facilitating the conversion of ubiquinone to ubiquinol. Through the process of electron transfer, Complex I is able to pump protons from the mitochondrial matrix into the intermembrane space, thus contributing to the proton gradient used for ATP synthesis.
Importantly, Complex I is responsible for the translocation of four protons across the inner mitochondrial membrane for every two electrons it receives from NADH. This active transport of protons is crucial for the establishment and maintenance of the proton motive force that drives ATP synthesis via ATP synthase.
What is the advantage of having Complexes I, III, and IV associated with one another in the form of a respirasome?
a. It facilitates the energetically unfavorable electron transfer reactions carried out by Complexes I, III, and IV.
b. A respirasome composed of Complexes I, III, and IV allows the efficient transfer of electrons from to oxygen.
c. It enables the rapid transfer of electrons, thereby increasing the efficiency of proton pumping.
d. The respirasome allows and to interact directly with the inner mitochondrial membrane.
Answer:
The correct answer is c. It enables the rapid transfer of electrons, thereby increasing the efficiency of proton pumping.
Explanation:
Having three complexes will increase efficiency due to the fact that these complexes are connected, being more efficient in proton pumping.
The visceral and parietal pleural membranes are normally held together by
A. loose connective tissue.
B. dense connective tissue.
C. decreased pressure in the pleural cavity.
D. surfactant in the pleural cavity.
E. elastic connective tissue.
Answer:
C) decreased pressure in the pleural cavity.
Explanation:
The pleura is a membrane that surrounds the lung and consists of two membranes, the parietal and the visceral, which are continuous in the hilum region of the lung. The parietal and visceral membranes delimit, for each lung, an independent cavity fully lined by the mesothelium. Typically, this pleural cavity is virtual, containing only a thin layer of lubricating fluid (pleural fluid), allowing smooth sliding of these two membranes during respiration, preventing friction between the visceral and parietal mesothelium.
The visceral and parietal pleural membranes are usually held together due to decreased pressure in the pleural cavity.
The visceral and parietal pleural membranes are held together by the decreased pressure in the pleural cavity, enabling smooth lung function with minimal friction.
Explanation:The visceral and parietal pleural membranes are normally held together by decreased pressure in the pleural cavity. The pleural cavity, which surrounds the lungs, contains a small amount of fluid that allows the visceral and parietal layers of the pleura to slide effortlessly over each other with minimal friction, aiding in lung movement during breathing. This fluid and the resulting suction-like effect keep these membranes close together.
Even though she has been taking lithium for a year, you review some teaching about drug therapy with S.T. Which statement by S.T. reveals a need for further education? a. "I will call my doctor if I have severe vomiting or diarrhea." b. "I need to be careful because lithium is addictive."c. "I take the lithium tablets with meals." d. I will keep my appointments to have my drug levels checked.
Answer:
The answer is letter A.
Explanation:
Because vomiting and diarrhea are patological signs of lithium's intoxication. Which means ST is taking higher doses than she should.
In certain bacteria, an oval shape (O) is dominant over round (o) and thick cell walls (T) are dominant over thin (t). Show a cross between a heterozygous oval, thick cell walled bacteria with a round, thin cell walled bacteria. What are the phenotypes of the F1 and F2 offspring
Answer:
Phenotypes –
Oval Thick - OoTt
Oval Thin - Oott
Round Thick - ooTt
Round thin - oott
Explanation:
Given -
Allele "O" is dominant over allele "o"
and allele for thick cell wall "T" is dominant over thin celled wall "t"
Genotype of heterozygous oval, thick cell walled bacteria is OoTt
Genotype of round, thin cell walled bacteria is oott
The cross between the two parents will produce following offspring
OT Ot oT ot
ot OoTt Oott ooTt oott
ot OoTt Oott ooTt oott
ot OoTt Oott ooTt oott
ot OoTt Oott ooTt oott
Phenotypes –
Oval Thick - OoTt
Oval Thin - Oott
Round Thick - ooTt
Round thin - oott
Final answer:
The cross between a heterozygous oval, thick cell-walled bacteria and a round, thin cell-walled bacteria will result in all F1 offspring being oval and thick-walled. The F2 offspring will have a 3:1 phenotype ratio for both cell shape and wall thickness if complete dominance is assumed, or a 1:2:1 ratio for cell wall thickness if incomplete dominance occurs.
Explanation:
Genetics of Bacterial Shapes and Cell Walls
In regard to the genetics of bacterial characteristics, when crossing a heterozygous oval-shaped bacteria with thick cell walls (genotype OT) with a round, thin cell-walled bacteria (genotype oo tt), we must consider the dominance of the alleles. The dominant traits are oval shapes (O) over round (o), and thick cell walls (T) over thin (t). A Punnett square can be used to predict the genotypes of the F1 offspring, which would all be heterozygous oval-shaped with intermediate cell wall thickness (Oo Tt), assuming incomplete dominance for cell wall trait. The F1 phenotype would exhibit the dominant traits of oval shape and thick cell wall.
For the F2 generation, we would cross two F1 heterozygotes (Oo Tt x Oo Tt), which will produce offspring with varying combinations of these traits by following the principles of Mendelian genetics. The F2 phenotype ratio for oval and round would be 3:1 and for thick and thin cell walls also 3:1 if cell wall thickness shows complete dominance. If there is incomplete dominance for cell wall thickness, we would observe a 1:2:1 ratio of thick, intermediate, and thin cell walls.
One of the genes that controls color vision is found on the X chromosome. The dominant allele results in normal color vision, and the recessive allele causes colorblindness. Ginny has normal color vision and is heterozygous, but Harry is colorblind. If Ginny and Harry have three children, what is the probability that at least one of them is a colorblind son?A. 0 B. 1/8 C. 9/64 D. 3/8 E. 27/64.
Answer:
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The gene "X" codes for the membrane-associated protein X. It is a receptor used in signal transduction. When a ligand binds the receptor protein, it activates a signal transduction pathway in the cell. A mutant allele of this gene, X0, encodes a hyperactive X protein, which has the same size as the wild type and associates with the plasma membrane like the wild type, but is always active: unlike the normal protein, it activates the signal transduction pathway whether or a ligand is bound to it. Interestingly, X/X0 heterozygotes and X0/X0 homozygotes both have shorter legs and arms. With respect to the function of the X gene, the X0 allele can be considered:
(A) haploinsufficient.
(B) dominant to the Xallele.
(C) codominant to the X allele.
(D) haplosufficient.
(E) incompletely dominant to the Xallele.
(F) recessive to the X allele.
Answer:
(B) dominant to the X allele.
Explanation:
Take a look at it, from the question; the heterozygotes and the homozygotes mutants have the same phenotypes (shorter legs and arms), i.e X/XO and XO/XO have identical phenotype.
Also, the XO produces the mutant protein (encoding a hyperactive X protein), which is always active in order for the signal transduction to take place in the cells with XO protein on the plasma membrane. This same signal is developed in the transduction pathway with the second messenger in the cytosol so that both the X/XO and XO/XO will have the same phenotype. As such, we can thereby conclude that the XO allele is dominant to the X allele.