Answer:
1. To calculate the frequency of the brown allele, count the number of BROWN ALLELES and divide by the total number of alleles in this population
2. In this beetle population, the number of brown alleles is 8
3. In this beetle population, the total number of alleles for color is 20
4. The frequency of the brown allele in this beetle population is .4
5. The frequency of the green allele in this beetle population is .6
A person has entered a hunger strike. After 60 days of starvation, this person is on the edge of dying. What is likely to be the cause of death?a. too many ketone bodies in the blood, which causes ketoacidosisb. lack of oxygen in the body for oxidative metabolismc. shut down of gluconeogenesis in the liverd. the breakdown of essential proteins of the brain and hearte. exhaustion of fat from adipose tissue
Answer: option D - the breakdown of essential proteins of the brain and heart
Explanation:
During starvation (food deprivation), the body cells begins to utilize the glycogen reserves in the liver and muscle. Then, it switches to the fats (stored in the adipose tissues).
If starvation persists, the body turns to the proteins for energy derivation. At this stage, the person's DYING STATE is usually visible because the muscles of the body shrinks greatly.
Note that DEATH set in when essential proteins of the BRAIN and HEART are metabolized to produce energy due to the vital functions both organs do in the overall body processes.
Which could be an example of an Adaptive Radiation?
A. a burst of new species in a new habitat where selective pressure is different from an ancestral habitat, like has been noticed for "Darwin’s Finches" The Cichlid Fishes and numerous examples from the fossil record.
B. HUMAN domestication of wild plants into crop plants due to selectively breeding plants with desirable characteristics over several hundred thousand years
C. Descent with modification and natural selection
Answer:
The correct answer will be option-A
Explanation:
Adaptive radiation is a process of speciation by which organisms get diversified into many forms very rapidly from an ancestral species.
The process takes place when the species gets into a new environment and new niches which results in morphological and physiological traits.
In the given question, the various forms of Darwin’s Finches on the same island is the result of adaptive radiation as the different forms of the beak of Finches is the result of different niches and the different food provided by nature.
Thus, option-A is the correct answer.
Answer:
A
Explanation:
Adaptional radiation- The comparatively quick evolution of the many species from one common ascendant, often happens when an organism enters a new area with lots of ecological opportunities.
EXAMPLE: "Darwin Finches".
It can also happen due to mass extinction or when new traits evolve.
Ex: Mammals after the extinction of the dinosaur.
9. Consider the need for accuracy in the flow of information from DNA to RNA to functional protein molecules. What are the different levels of stringency in replication, transcription and translation? What kinds of mechanism improve the accuracy of these processes?
Answer:
The central dogma of the molecular biology explains the flow of information from the replication of the DNA to the expression of proteins. This central dogma must be error free so that correct information can be transferred. The error in this process can cause life threatening situation. So, the phases of proof reading and rectification is required in this process.
From the DNA replication to the transcription of the RNA and final translation into the proteins requires proof reading mechanisms and many enzymes with the non enzymatic processes. The enzymes of transcription, replication and translation shows high degree of proof reading and works in efficient way. The incorporation of the amino acids, post translation modification makes the information flow as correct as possible.
When the antibiotic valinomycin is added to actively respiring mitochondria, several things happen: the yield of ATP decreases, the rate of O2 consumption increases, heat is released, and the pH gradient across the inner mitochondrial membrane increases. Explain these observations by listing the causal order of events that occur beginning with the addition of valinomycin to the respiring mitochondria and ending with the release of heat. Note: some events listed do not occur and should not be placed Valinomycin added The rate of ATP synthesis decreases The rate of electron transfer and oxygen consumption increases The pH gradient across the The valinomycin-K complex moves K ion out of the mitochondrial matrix. The electrical potential across the The rate of ATP hydrolysis increases The electrical potential across the mitochondrial membrane decreases. Valinomycin binds K" ion. The valinomycin-K' complex moves into the mitochondrial matrix. The pH t across the Heat released
Valinomycin binds to K+ ions, disrupting the potassium gradient and causing a decrease in ATP production. Increased efforts to maintain energy result in enhanced electron transport and oxygen consumption, along with heat generation.
When valinomycin is added to actively respiring mitochondria, it complexes with potassium ions (K+) and facilitates their transfer across the mitochondrial membrane. This disrupts the osmotic pressure and affects the proton gradient. This sequence of events begins with valinomycin binding to K+ ions on the surface of the mitochondria and forming the K+-valinomycin complex. The complex becomes highly soluble in the nonpolar interior of the membrane, allowing it to cross the membrane. Once inside, K+ is released, and the valinomycin is free to return and bind to another K+ ion, continually disrupting the normal potassium gradient. This action increases the permeability of the membrane to K+, leading to the dissipation of the electrical potential across the membrane. This reduction in electrical potential affects the proton gradient, which in turn impairs the ATP synthase activity, leading to a decrease in ATP production. Without the efficient generation of ATP, the rate of electron transport and oxygen consumption increases as the mitochondria try to maintain energy production, while the dysfunctional gradient causes the proton pumps to work harder. The extra work results in the generation of heat. Meanwhile, the pH gradient across the inner mitochondrial membrane increases due to the altered balance between proton influx and efflux.
Fish is high in omega-3 fatty acids, which are most beneficial for the __________ as mentioned in the video. Fish is high in omega-3 fatty acids, which are most beneficial for the __________ as mentioned in the video.
1. immune system
2. kidneys
3. heart
4. digestive tract
Answer: option 3) Heart
Explanation:
Omega-3 fatty acids are an integral part of cell membranes throughout the body and affect the function of the cell receptors in these membranes.
They provide the starting point for making hormones that regulate blood clotting, contraction and relaxation of artery walls, and inflammation.
Thus, they are most beneficial to the HEART
which of these is not a basic function of a cell
obtaining energy
storing energy
destroying energy
releasing energy
Answer: destroying energy
Explanation:
The cell is the basic unit of life. It consists of several components such as
Nucleus,
Mitochondria: it helps in obtaining energy through lipid oxidation within its layer
Cytosol: It helps in releasing energy from glucose breakdown (glycolysis)
Vacuoles, endoplasmic reticulum also help in storing energy in cells of adipose tissues.
But, NONE of the cell components helps in destroying energy, since energy is needed and usually well-regulated.
Most consumers have a wide variety of choices in terms of the food products they choose to buy. These decisions can be influenced by factors like food advertising or the labels on the products themselves.Choose the correct statements about marketing and labeling.Select all that apply.__Organic foods are free of all pesticides.__All free-range animals have unlimited access to the outdoors.__Most food advertising is for processed foods and packaged foods.__Organic foods have higher levels of nutrients than their conventional counterparts.__The term "natural" on food labels has no standardized definition.
Not all organic foods are free from pesticides and 'free-range' does not necessarily mean unlimited outdoor access. Most food advertising is for processed and packaged foods, organic foods do not always have more nutrients and the term 'natural' has no standardized definition in food labeling.
Explanation:It's true that many consumer decisions related to food purchases are influenced by marketing and labeling. However, not all the statements listed are accurate. Organic foods are not necessarily free of all pesticides. Sometimes, organic farmers use naturally-derived pesticides over synthetic ones but this doesn't mean they are pesticide-free. Similarly, while free-range animals should have access to the outdoors, it doesn't necessarily mean that they have unlimited access.
Furthermore, it's a fact that most food advertising is indeed aimed at processed and packaged foods. This could be because they have a longer shelf-life compared to fresh produce, making them more appealing to grocery stores. As for nutrient levels, organic foods do not always have higher levels of nutrients than their conventional counterparts. The nutrient content can vary greatly depending on a variety of factors, including how the food is grown and harvested.
Finally, the term 'natural' on food labels indeed has no standardized definition. Therefore, it can be used loosely by companies to market their products.
Learn more about Food Marketing and Labeling here:https://brainly.com/question/15846552
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What are the conditions that lead to a "beer gut" due to the excess consumption of alcohol? NADH produced from the metabolism of ethanol stimulates the citric acid cycle for glucose-derived acetyl CoA. Excess ethanol metabolism leads to an accumulation of NADH that inhibits fatty acid metabolism. The processing of acetate in the liver becomes inefficient leading to a pH imbalance in liver cells, reducing enzyme efficiency in general. NADH inhibits ketone body formation, stimulating glucose rather than fatty acid metabolism. NADH stimulates citric acid cycle enzymes that stimulates glucose-derived acetyl CoA metabolism.
Answer: Excess ethanol metabolism leads to an accumulation of NADH that inhibits fatty acid metabolism.
Explanation:
Fatty acid metabolism is often activated by limited or absence of NADH, however ethanol (the main constituent of most alcoholic drinks) on metabolism yields several molecules of NADH which rather stimulates belly/abdominal fat production often referred to as "beer gut".
The data from a replicated experiment are very different from the data obtained during the initial experiment. Which statements identify possible reasons for the difference in the data? Check all that apply
The initial experiment was not a valid experiment.
The replicated experiment was performed incorrectly.
The replicated experiment used a different laboratory.
The initial experiment involved more trials and data.
The replicated experiment changed the variables being tested.
The initial experiment was not a valid experiment is the statement which identifies possible reasons for the difference in the data.
Explanation:
The replicated experiment however is crucial but doesn't satisfy the condition of justifying the validity of a scientific theory or hypothesis. However, the initial or repeated experiment are more satisfying in producing logical result from the experiment. It involves repeated methods of a particular experiment with more and more trials and observations of more data.
However, the replicate experiments are individual experiments of a specific hypothesis which fails to experiment are individual experiments of a specific hypothesis which fails to produce a logical result.
Answer:
The initial experiment was not a valid experiment.
The replicated experiment was performed incorrectly.
The initial experiment involved more trials and data.
The replicated experiment changed the variables being tested.
Explanation:
A scientific experiment is performed to test if the predicated hypothesis can be proved or not.
The scientific experiment involves the repetition of the experiment by the same experimenter and the replication of the experiment which could be performed by the same experimenter or another experimenter to test if the results of the initial experiment are valid or not.
In the given question, the results of the initial experiment and the replicated experiment are different which could be due to various reasons:
The initial experiment was not valid - if the initial experiment performed is not valid. The replicated experiment was performed incorrectly- the replication procedure differs from the initial experiment or there could be some error while experimenting. The initial experiment involved more trials and data- the initial experiments could have involved more replicas, trials, and data. The replicated experiment changed the variables being tested - the variables of this two experiment could differ.What is the chance that the 4th child from the couple indicated by the arrow will have a girl who suffers from the disease?
Explanation:
Tho i can see no arrow but there is probability of 75% that a fourth child suffers the disease of the parent
All of the following can reduce crop water use, except for which one? Group of answer choices Applying irrigation water at times of day when evaporation is low. Leaving the soil surface bare to improve infiltration. Selecting crop varieties that are well-suited to the local climate. Using drip irrigation techniques to apply water only where plants can use it.
Answer:
Leaving the soil surface bare to improve infiltration.
Explanation:
All of the following can reduce crop water use, except Leaving the soil surface bare to improve infiltration.
Leaving the soil bare will do no good to soil's water retention capacity rather it would decrease significantly. This happen because of water runoffs as there would no crops to hold water and the top fertile layer of the soil will runoff along with water.
Answer:
Explanation:
Leaving the soil surface bare for infiltration. When the soil surface is bare it exposes the it's layers to direct sunlight which absorbs all the water content present in the soil and thus is harmful for the soil and crop
The visible colored portion of the eye is the:
a. cornea.b. pupil.c. sclera.d. iris.
Answer:
D. IRIS
Explanation:
Generally in humans and also in most mammals and birds, the iris is a very small circular structure in the eye, responsible for controlling the diameter and size of the pupil and also with the amount of light reaching the retina.
finallly eye color is also known to be defined by that of the iris
Your friend is a pioneer in ES cell research. In her research, she uses an ES cell line that originated from an inbred strain of laboratory mice called FG426. She has just figured out methods that allow her to grow an entire liver from an ES cell and has successfully grown 10 livers. She demonstrates that the newly grown livers are functional by successfully transplanting one of the new livers into a FG426 laboratory mouse. You are particularly excited about this, because you have a sick pet mouse, Squeaky. You are very attached to Squeaky, as you found him when you were out camping in New Hampshire. Unfortunately, Squeaky has developed liver disease and will not live much longer without a liver transplant. After you see your friend on TV talking about her new method for growing mouse livers, you immediately grab your cell phone to ask her whether Squeaky could have one of the newly grown livers. Just as you are about to dial your friend, you remember something you learned in cell biology and realize that instead, you should ask your friend about possibly using therapeutic cloning for Squeaky’s benefit.
A. Why do you think that one of the newly grown livers may not work in Squeaky?
B.Explain how using iPS cells could solve this problem.
Answer:
Explanation:
a)Organ transplantation requires that the donor organism and recipient be genetically close so that the graft or transplant will not be attacked by the immune system of the recipient leading to rejection and damage. Squeaky is likely to be made up of a different genetic configuration compared to laboratory inbred FG426 mouse
b) ips (induced pluripotent stem cell) on the other hand can benefit squeaky since the cells are somatic cells such as B cells, Keratinocytes, neuronal progenitors cells, kidney and muscles gotten from the donor that are reprogrammed by reactivating silent genes through fusing of another different cell such as ES (embryonic stem cell) and introduction of some transcriptional factors such oct4, sox2,kf4, and k-myc leading to transcriptional activity and DNA methylation. This induced pluripotent stem cells can be grown into organ that can be transplanted to the recipient who was initially the donor of the reprogrammed somatic cells. Because it is from the host, the transplanted organ is not likely to trigger immune response compare to those grown from ES from other bred.
Any object struck at its natural frequency will vibrate violently. This is known as resonance. Applications of this include earthquakes knocking down building and
A) rainbows created in air.
B) microwaves created in ovens.
C) light being created by the sun.
D) sound being created in music instruments.
Answer: option D) sound being created in music instruments
Explanation:
Resonance occurs when an object close in contact to a vibrating body begins to vibrate at the same frequency of the vibrating body.
Usually each object has a THRESHOLD frequency which must be equaled or exceeded before it is set into vibration.
Therefore, this is the PRINCIPLE behind MELODY i.e how sound is being created in music instruments.
What is the function of the lacZ gene?
a. This gene encodes an enzyme, galactoside permease, which transports lactose into the cell.
b. This gene encodes an enzyme, b-galactosidase, that cleaves lactose into two glucose molecules.
c. This gene encodes an enzyme, b-galactosidase, which cleaves lactose into glucose and galactose.
d. This gene encodes the repressor of the lac operon.
Answer:
the answer is c
Explanation:
The lacZ gene encodes an enzyme called β-galactosidase, which is responsible for splitting lactose (a disaccharide) into readily usable glucose and galactose (monosaccharides). The lacY gene encodes a membrane protein called lactose permease, which is a transmembrane "pump" that allows the cell to import lactose.
The lacZ gene encodes the enzyme β-galactosidase, which breaks down lactose into glucose and galactose, playing a crucial role in lactose metabolism in bacteria such as E. coli.
Explanation:The function of the lacZ gene is to encode an enzyme called β-galactosidase, which is responsible for cleaving the disaccharide lactose into its monosaccharide components, glucose and galactose. This process is an essential step in the metabolism of lactose by bacteria such as E. coli. When lactose is present in the environment, E. coli can use the lac operon system to express the lacZ gene, along with other genes necessary for the utilization of lactose as a carbon source, such as the lacY gene for lactose permease and the lacA gene for transacetylase.
What are some ways to prevent wildfires?
Ways to prevent wildfires can include:
If you see new fires, report it if its not getting taken care of.Don't use fireworks in areas with lots of trees/bushes.When you smoke cigarettes, don't throw them on the ground.When you are using flammable liquids, please be cautious with it.When you are using a campfire, put it out before you leave.Best of Luck!
Answer:
Report unattended fires. ...Extinguish fire pits and campfires when done. ...Don't throw lit cigarettes out of your moving car. ...Use caution when using flammable liquids. ...Pay attention to local ordinances for trash burning. ...Only use fireworks in clear areas with no woods nearby.Explanation:
hope this helps you
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a cell found in the root of the plant would most likely lack what cell structure.
Mitochondria
Cell wall
Cell membrane
Chloroplasts
Answer:
The cells found in the root of the plant normally lacks chloroplasts, as roots do not perform photosynthesis. The main function of chloroplast is photosynthesis.
Explanation:
The plant cell contains nucleus, cell wall, cell membrane, mitochondria, chloroplasts, vacuole, endoplasmic reticulum, ribosomes etc. Chloroplasts contain green pigments which give green color to the plants. The main function of chloroplast is photosynthesis. During photosynthesis, the plants produce glucose and release oxygen by using carbon dioxide from the air, and absorbing water and nutrients from the soil. Using the glucose in the plants and oxygen, mitochondria produce energy.
The root system of the plants absorb water and nutrients from the soil. The absorbed water and nutrients are then transported to various parts of the plant to carry out processes like photosynthesis, maintaining turgor pressure etc. They have different types of cell which perform some specific functions. These cells also contain cell wall, cell membrane and mitochondria. Mitochondria provides energy for the active transport of water and nutrients. But chloroplasts are absent in root cells as they do not perform photosynthesis.
What are the functions of the three major forms of RNA (ribosomal RNA, messenger RNA, and transfer RNA)?
Answer:
Ribosomal RNA: Structural part of ribosomes
Messenger RNA: Carry genetic information from DNA to proteins
Transfer RNA (tRNA): Transport amino acids to protein synthesizing complex.
Explanation:
Ribosomes are made up of ribosomal RNA (rRNA) and proteins. The catalytic activity for the formation of peptide bonds between amino acids during protein synthesis resides the RNA of ribosomes.
Messenger RNA (mRNA) is formed by the process of transcription during which the nucleotide sequence of the template DNA strand is copied into that of the RNA. The mRNA serves as a template for protein synthesis. The nucleotide sequence of mRNA is read in the form of genetic codes to specify the amino acid sequence of a protein. In this way, the genetic information stored in DNA is carried to the proteins.
During the process of protein synthesis, tRNAs carry amino acids to the mRNA-ribosome complex so that the amino acids are incorporated into the polypeptide. For the purpose, there is a tRNA with a specific anticodon sequence for a particular amino acid.
The Correlated trait exercise shows that :
A) when two characters are correlated, selection always favors larger values of both or smaller values of both
B) When two characters are correlated, one of them can evolve toward higher values even if smaller values are better for survival
C) when two characters are correlated, selection is always strong on both or weak on both
D) when two characters are correlated, the optimal values and selection strengths rise and fall together
Answer:The correlated trait exercise shows that when two characters are correlated, the optimal values and selection strengths rise and fall together.
Explanation: Correlation refers to statistical (linear) relationship between two random variables. When traits are correlated, change in one is associated with change in the other.
Correlation coefficient (c.c.) measures strength of association between two variables in the same individual or experiment. It can range from -1 to +1. C.c. can be positive, negative or weak.
1. A positive c.c. means that an increase in one variable is associated with an increase in the other variable.
2. A negative c.c. means that an increase in one variable is associated with a decrease in the other.
3. A c.c. near zero indicates a weak relationship between the variables.
Correlation can be represented by scatter plot as shown in the attached image.
Option B: When two traits are correlated, one can evolve toward higher values even if smaller values are better for survival.
Option B states that when two characters are correlated, one of them can evolve toward higher values even if smaller values are better for survival. This implies that selection pressures can favor changes in one trait even if those changes are not directly beneficial for survival. This phenomenon, known as evolutionary constraint or evolutionary trade-off, occurs when traits are genetically linked or when the same genetic mechanisms control multiple traits. As a result, evolutionary changes in one trait may be constrained by the genetic correlation with another trait, leading to seemingly suboptimal evolutionary outcomes. Therefore, Option B provides a nuanced understanding of how correlated traits can evolve.
Saccharomyces cerevisiae is a diploid yeast species that can reproduce either sexually or asexually. An experiment was performed to induce mitotically dividing S. cerevisiae cells in G2 to undergo meiosis. Which of the following best describes the steps these cells will follow to form gametes?
A. The first division will result in crossing over between homologous chromosomes, and the second division will reduce the original number of chromosomes by half in the daughter cells.
B. The first division will reduce the number of chromosomes by half for each daughter cell, and the second division will result in each daughter cell having one-fourth of the original number of chromosomes.
C. The first division will move single chromatids to each daughter cell, and the second division will double the number of chromosomes in each daughter cell.
D. The first division will reduce the number of chromosomes by half for each daughter cell, and the second division will move single chromatids to each daughter cell.
The statement that best describes the steps these cells will follow to form gametes is "the first division will reduce the number of chromosomes by half for each daughter cell, and the second division will move single chromatids to each daughter cell"
MEIOSIS:
Meiosis is the process by which a single cell produces four daughter cells with a reduced number of chromosomes (by half). Meiosis occurs in two step division process namely: meiosis I and meiosis II. In meiosis I, homologous chromosomes are separated while in meiosis II, sister chromatids are separated. Meiosis I reduces the chromosome number by half i.e. from diploid (2n) to haploid (n). Therefore, in the experiment that was performed to induce mitotically dividing S. cerevisiae cells in G2 to undergo meiosis, the first division will reduce the number of chromosomes by half for each daughter cell, and the second division will move single chromatids to each daughter cell.Learn more at: https://brainly.com/question/12327654?referrer=searchResults
Meiosis is a reduction division in which number of chromosomes in the daughter cells are halved. In mitosis, the number of chromosomes in the daughter and parent cell are similar after cell division. The statement which describes the correct sequence is:
The first division will reduce the number of chromosomes by half for each daughter cell, and the second division will move single chromatids to each daughter cell.The yeasts are the unicellular microorganisms that can reproduce sexually or asexually. During the sexual reproduction, the number of chromosomes in the in the daughter cells are halved. Some of the features of meiosis are:
Meiosis in yeasts produces four haploid daughter cells, known as gametes. Meiosis takes place in two steps, meiosis I and meiosis II. In meiosis I, the homologous chromosomes are separated, whereas, in the meiosis II separation of sister chromatids occur. Meiosis I is the reduction division in which diploid cell produces haploid cells. Thus, the experiment performed on yeast to induce mitosis will result in the cells undergoing meiosis during G1 phase. The first division will lead to reduction in the chromosome numbers by half and the second division will move single chromatids to each daughter cell.Therefore, option D is correct.
To know more about cell division, refer to the following link:
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In humans, hemophilia is caused by an X-linked recessive gene. A woman who is a nonbleeder - that is, she does not display the blood-clotting irregularities associated with hemophilia - had a father who was a hemophiliac.
She marries a nonbleeder, and they plan to have children.
Calculate the probability of hemophilia in the female and male offspring.
Answer:
one normal girl
One normal boy
One carrier girl
One colour blind boy
Explanation:
Haemophilia is a sex linked feature as it is transmitted with the chromosomes determining sex.
In humans,the male has XY and the female has XX.the Y chromosome is usually genetically empty.so as mall received his Y chromosomes from his father,he cannot inherit his father's sex linked traits .but women can as they receive and X chromosomes from their fathers.so a cross between a carrier woman Cc and a normal man C result in;
XX--CC
XX--Cc
XY--C
XY--c
As you continue to cross the offspring resulting from the previous question with albinos, what will occur in the growing population?
Answer selections:
a) the percentage of agouti will increase
b) the percentage of chinchillas will increase
c) the percentage of both dominant traits (agouti and chinchilla) will increase
d) the percentage of albinos will increase
*any explanation with answer is welcome for best understanding*
Answer:
D. The percentage of albinos will increase.
Explanation:
In the growing population it is certain that the percentage of albinos will surely increase because it is continuous breeding involving albinos.
Answer:
The percentage of albinos will increase.
6. Before direct proof of DNA being the genetic material, several lines of indirect evidence had suggested DNA as the appropriate macromolecule. Which ONE of the following statements supports the hypothesis that DNA is the genetic material?
a. RNA is more stable than DNA or protein.
b. Molecular composition of DNA is variable in different cells of an organism, whereas the molecular composition of RNA and protein does not change in different cells of an organism.
c. DNA consists of only four bases while proteins are comprised of twenty different amino acids.
d. protein content is most abundant in cells just prior to division.
e. Gametes have 1/2 the amount of DNA as somatic cells.
Answer:
e. Gametes have 1/2 the amount of DNA as somatic cells.
Explanation:
a. RNA is more stable than DNA or protein
-the reverese is the case, DNA has double helix structure as a result of phosphodiester bonds in the sugar-phosphate backbone plus 2 or 3 Hydrogen bonds between the nitrogenous bases.
DNA is more stable
b. Molecular composition of DNA is variable in different cells of an organism, whereas the molecular composition of RNA and protein does not change in different cells of an organism.
with the exception of gametes which contain half the amount of DNA necessary, each Cells have equal amount of DNA when compared to variable amount in cells RNA and proteins due to the different degree of expression and cell differentiation
c. DNA consists of only four bases while proteins are comprised of twenty different aminoacids.
This evidence lacks basis of why DNA is the genetic material.
d. Protein content is most abundant in cells just prior to division.
This evidence lacks basis of why DNA stands as the genetic material in individual organisms
e. Gametes have 1/2 the amount of DNA as somatic cells.
-this Argument is firmly supported since half of parent DNA is donated to its offspring forming the total genome of a somatic cell
Which of the following best describes the change in kdr genotype frequencies over time in A. gambiae?a)From pre-2006 to post-2006, the frequency of the r/r genotype increased slightly while the frequencies of +/+ and +/r genotypes decreased dramatically.b)From pre-2006 to post-2006, the frequency of the r/r genotype increased dramatically while the frequencies of +/+ and +/r genotypes decreased dramatically.c)From pre-2006 to post-2006, the frequency of the r/r genotype increased dramatically while the frequencies of +/+ and +/r genotypes remained relatively constant.d)From pre-2006 to post-2006, there was little change to the r/r, +/+, and +/r genotype frequencies.
Answer:
b)From pre-2006 to post-2006, the frequency of the r/r genotype increased dramatically while the frequencies of +/+ and +/r genotypes decreased dramatically.
Final answer:
The frequency of the resistant r/r genotype in Anopheles gambiae mosquitoes increased dramatically post-2006, while the susceptible +/+ and +/r genotypes decreased dramatically, indicating evolution due to insecticide selection pressure.
Explanation:
Population genetics examines the changes in allele frequencies and genotype frequencies in a population over time. According to the provided reference, the change in kdr genotype frequencies in Anopheles gambiae mosquitoes indicates that the frequency of the resistant allele (denoted as 'r') and the corresponding homozygous genotype (r/r) increased dramatically post-2006, which suggests that the use of anti-vector interventions may have selected for this resistance allele at a higher rate. On the contrary, the frequencies of the susceptible homozygous genotype (+/+) and heterozygous genotype (+/r) decreased dramatically over this time period.
The correct description of these changes would most likely be that from pre-2006 to post-2006, the frequency of the r/r genotype increased dramatically while the frequencies of +/+ and +/r genotypes decreased dramatically. This reflects a population undergoing evolution due to selection pressure from the use of insecticides, as alleles that confer resistance to these chemicals are favored and become more common within the population.
Which statements describe telomeres and the actions of telomerase?
a) Telomerase activity is turned off in most human cells, causing the telomeres to gradually shorten as the individual ages.
b) Telomerase contains a G‑rich template sequence that adds a C‑rich DNA sequence to the lagging strand of the chromosome.
c) Telomerase consists of an RNA template and a protein component that adds complementary nucleotides to the 3′ end of DNA.
d) Inappropriate activation of telomerase can result in cellular immortality, one of the cellular changes implicated in the development of cancer.
e)A telomere forms a loop with a section of the complementary DNA strand, protecting the chromosome from degradation by endonucleases.
Final answer:
Correct statements about telomeres and telomerase include telomerase activity being generally off in most human cells (a), telomerase containing an RNA template and a protein component adding to the 3' DNA end (c), and inappropriate activation of telomerase contributing to cellular immortality and cancer (d).
Explanation:
Which statements describe telomeres and the actions of telomerase? The statements that accurately describe telomeres and the actions of telomerase are as follows:
a) Telomerase activity is turned off in most human cells, causing the telomeres to gradually shorten as the individual ages.
c) Telomerase consists of an RNA template and a protein component that adds complementary nucleotides to the 3′ end of DNA.
d) Inappropriate activation of telomerase can result in cellular immortality, one of the cellular changes implicated in the development of cancer.
Statement b) is incorrect because telomerase adds a G-rich sequence to the telomere, not a C-rich sequence. Statement e) describes the formation of a T-loop by the telomere but does not pertain directly to how telomerase functions.
Which statement describes double fertilization in an angiosperm?
a. Two egg cells are fertilized within each ovule.
b. A sperm nucleus fuses with an egg cell and a sperm nucleus fuses with polar nuclei.
c. Two sperm are required for fertilization of one egg cell.
d. A flower can engage in both self-pollination and cross pollination.
e. One sperm fertilizes the egg while two polar nuclei fuse with a second egg.
Answer:b. A sperm nucleus fuses with an egg cell and a sperm nucleus fuses with polar nuclei.
Explanation:
This mode of fertlization occurs only in angioseprms .One of the sperms forms 2n as zygote.
The second sperm fuses with the two polar nuclei to form 3n as endosperm.
The importance of these are:
1. Restoration of diploid condition with the fusion of haploid male and female gametes.
2.it makes plants to recieve stimulus based on which ovary develops into a fruit and shows thst it is ovules that develops into seeds.
Answer:
B: A sperm nucleus fuses with an egg cell and a sperm nucleus fuses with polar nuclei.
Explanation:
In double fertilization in angiosperms, one sperm fertilizes the egg to form the 2n zygote, while the other sperm fuses with two polar nuclei to form the 3n endosperm. After fertilization, embryonic development begins
Toemas’s syndrome is a autosomal dominant disease, homozygous lethal, seen at a frequency of 1 in 10,000 cats. Felines with this disease have a very strong desire to have their backs scratched. Assume the locus is in Hardy Weinberg Equilibrium.
A male cat with Toemas's syndrome and an unaffected female cat have a newborn kitten. What is the probability that the kitten is male AND he has Toemas’s syndrome? Assume mating is at random.
(Hint: recall your work on matings for two independent loci!)
Answer: 50% or 25% according to the genotype of the male cat.
Explanation:
The female cat is unaffected, and since this is an autosomal dominant diease, it means this cat does not have any affected alleles. Thereby, its genotype is aa. This is because dominant alleles show their effect even if the organisms only has one copy of the allele. The male cat is affected with this syndrome so its genotype is either AA or Aa. For this disease to manifest itself, only one affected allele is needed. Although both alleles can be affected so the individual can be homozygous dominant or heterozygous.
To find this we have to do a punnett square. Here, we use the gametes produced by each individual. Gametes are sex cells, egg or sperm, which only have one allele of each gene.
Assuming that the cross is between aa and AA, the gametes produced by "aa" are "a", and the gametes produced by "AA" are "A" So we are having a cross between an "a" gamete and a "A" gamete. 100% of the offspring will be Aa so it is a heterozygous, which will develope this syndrome. And, males have one Y chromosome and one X chromosomes, thereby the chances of having a male kitten is 50%. Then, to conclude, there is a 50% chances of having a male kitten with this syndrome.
And, assuming that the cross is between aa and Aa, the gametes produced by "Aa" are "A" but also "a" So we are habing a cross between an "a" gamete and "A" or "a" gamete (See picture) In this example, there is a 50% chances of having a kitten with the syndrome, because only one genotype produced (Aa) has the affected allele. And, as we said before, since there is a 50% chances of having a male kitten, we can said that there is a 25% chances of having an affected male. Because to calculate both probabilities so that they occur simultaneously, it is necessary to multiply them. That is, 0.5 x 0.5 = 0.25 x 100 = 25%.
In the datura plant, purple flower color is controlled by a dominant allele P. White flowers are found in plants homozygous for the recessive allele p. Suppose that a purple-flowered datura plant with an unknown genotype is self-fertilized and that its progeny are 28 purple-flowered plants and 10 white-flowered plants.
Part A
Use the results of the self-fertilization to determine the genotype of the original purple-flowered plant.
a. Pp
b. PP
c. pp
Part B
If one of the purple-flowered progeny plants is selected at random and self-fertilized, what is the probability it will breed true?
a. 1/3
b. 1/4
c. 3/4
d. 2/3
Answer:
Part A: Pp
Part B: 1/3
Explanation:
A: The purple flower color is regulated by dominant allele "P". Since self-fertilization of the purple-flowered plant with unknown genotype gives both purple and white-flowered progeny in almost 3:1 ratio, the parent plant was heterozygous dominant for the trait. Its genotype is "Pp".
Segregation of recessive and dominant alleles during gamete formation leads to the production of two types of gametes with either "P" or "p" alleles. The random fusion of these gametes produces progeny in following phenotype ratio= 3 purple: 1 white.
B. Out of the 3 purple-flowered progeny, 1 is homozygous dominant (PP). Therefore, the probability of the purple-flowered progeny to breed true is 1/3.
Final answer:
The genotype of the original purple-flowered datura plant is determined to be heterozygous (Pp) based on the ratios of purple to white flowered progeny. The probability that a randomly selected purple-flowered progeny will breed true for the purple color is 1/4, meaning there is a 25% chance it is homozygous dominant (PP).
Explanation:
Part A: Genotype Determination
To determine the genotype of the original purple-flowered datura plant, we can look at the progeny from its self-fertilization. Given that the purple flower color is dominant and represented as allele P and the white color is recessive and represented as allele p, we can infer the genotype of the parent plant based on the ratio of purple to white flowered plants among its progeny. The progeny consists of 28 purple-flowered plants and 10 white-flowered plants. This suggests that the parent plant is not a homozygous dominant (PP) because we expect all progeny to have purple flowers in that scenario. Since white flowers (pp) are a recessive trait, they can only appear in the offspring if the parent carries the recessive allele. Therefore, the parent plant must be heterozygous Pp.
Part B: Probability of True Breeding
The probability that a randomly selected purple-flowered progeny plant will breed true (be homozygous dominant PP) can be inferred from the parent's genotype. Since the parent plant is heterozygous Pp, its possible gametes are P or p. With self-fertilization, there is a 1 in 4 chance (or 25%) of getting PP, a 1 in 4 chance of pp, and a 2 in 4 chance of Pp. The probability that the purple progeny plant selected is a homozygote (PP) is the same as finding a homozygous dominant offspring in the self-fertilization, which is 1/4 or option 'b'. The other two-thirds will be heterozygous purple (Pp).
In 1906 Harden and Young, in a series of classic studies on the fermentation of glucose to ethanol and CO2 by extracts of brewer's yeast, made the following observations.(A) Inorganic phosphate was essential to fermentation; when the supply of phosphate was exhausted, fermentation ceased before all the glucose was used.(B) During fermentation under these conditions, ethanol, CO2, and a sugar phosphate accumulated.(C) When arsenate was substituted for phosphate, no sugar phosphate acumulated, but the fermentation proceeded until all the glucose was converted to ethanol and CO2.Answer the following questions.1. Which enzyme of glycolysis requires inorganic phosphate and therefore stops when no phosphate is available?(a) glyceraldehyde 3-phosphate dehydrogenase(b) phosphoglycerate mutase(c) phosphofructokinase-1(d) phosphoglycerate kinase
Answer:
(a) glyceraldehyde 3-phosphate dehydrogenase.
Explanation:
Glyceraldehyde 3-phosphate dehydrogenase is the enzyme which requires inorganic phosphate (Pi) essentially for it's functioning. It catalyzes the conversion of glyceraldehyde 3-phosphate into 1,3-bisphosphoglyceric acid (1,3 BPG) during glycolysis.
All other enzymes of glycolysis use ATP for phosphorylation of their substrate but this is the only enzyme which rather uses an inorganic phosphate to phosphorylate it's substrate.
A RNA differs from DNA in all EXCEPT which of the following ways?
A. the presence of uracil
B. the 5'-3' orientation of the polynucleotide strand
C. the sugar molecule
D. the number of different functions performed
Answer: B. the 5'-3' orientation of the polynucleotide strand .
The complimentary pairing of DNA molecules ensures 5' 3' and 3'[ 5 ' arrangement for correct pairing of purines and pyrimidine (Adenine E with thymine and cytosine with guanine) in opposite direction. .
RNA only contains five prime and three prime only, but DNA contains both 5'3' and 3'5'.
RNA differs from DNA in the presence of uracil, the sugar molecule, and its various functions. However, both RNA and DNA share the 5'-3' orientation of the polynucleotide strand.
Explanation:The RNA differs from DNA in several ways, including the presence of uracil, the type of sugar molecule it contains, and its various functions. However, the 5'-3' orientation of the polynucleotide strand is a feature that RNA shares with DNA. Both RNA and DNA have this orientation, which describes the direction that the molecules are read and synthesized in cellular processes such as DNA replication and transcription.
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