Answer:
Gas
Explanation:
Gas has the most kinetic energy.
Gas has the most energy than the solid ones.
Three states of matter
1: solid
2: liquid
3: Gas
Hope this helps.
how do cells in a multicelluar organism become specialized
Answer:
Explanation:
Each type of cell in a multicellular organism has slightly different DNA, allowing the different types of cells to specialize. ... The genes necessary for different types of cells to function segregate into different cell types in a multicellular organism.
Which is a characteristic of pseudoscience?
Answer:
Pseudoscience is often characterized by contradictory, exaggerated or unfalsifiable claims; reliance on confirmation bias rather than rigorous attempts at refutation; lack of openness to evaluation by other experts; absence of systematic practices when developing hypotheses; and continued adherence long after the pseudoscientific hypotheses have been experimentally discredited.
Explanation:
Under certain circumstances, potassium ions ( K+K+ ) in a cell will move across the cell membrane from the inside to the outside. The potential inside the cell is −70.5 mV−70.5 mV and the potential outside the cell is zero. What is the change in the electrical potential energy of a single potassium ion as it moves across the membrane?
Answer:
the change in the electrical potential energy of a single potassium ion as it moves across the membrane is 1.12941 × 10⁻²⁰ J
Explanation:
The formula used in determining the change in the electrical potential energy can be expressed by the relation.
Charge in electrical Potential energy = [tex]q ( V_2 - V_1)[/tex]
where q = the potassium ion
q = 18 electrons and 19 protons
q = 19-18 = 1 electron
1 electron = 1.602 × 10⁻¹⁹ Coulombs
Thus;
the potential inside the cell is −70.5 mV
= -70.5 × 10⁻³ V
∴ Charge in electrical Potential energy = [tex]q ( V_2 - V_1)[/tex]
Charge in electrical Potential energy = [tex]1.602*10^{-19} ( 0 - (-70.5*10^{-3})[/tex]
Charge in electrical Potential energy = 1.12941 × 10⁻²⁰ J
Thus; the change in the electrical potential energy of a single potassium ion as it moves across the membrane is 1.12941 × 10⁻²⁰ J
Which substance reduces the level of acid in blood?
sodium
bicarbonate
urea
nitrogen
Answer:
Bicarbonate
Explanation:
You have a strain of bacteria that is defective for DNA glycosylase activity. You expose one culture of this strain to EMS and the other to cisplatin (a chemotherapy drug that creates bulky adducts on purine bases). You notice that the cisplatin is much more toxic to this mutant strain of bacteria than is the EMS. Explain your observation.
Answer:
Bacteria under successive rounds of selection may acquire genetic resistance to the treatment with mutagenic compounds including ethyl methanesulfonate (EMS).
Explanation:
Artificial selection is a process known to generate resistance of the organisms under treatment. For example, this type of selection has been reported to generate resistance to the treatment with different chemical compounds including mutagenic chemical compounds (in this case EMS), antibiotic drugs, etc.
A farmer is raising rabbits. The average body weight in his population is 3kg. The farmer selects the 10 largest rabbits in the population, whose average body weight is 4kg, and interbreeds them. If the heritability of body weight in the rabbit population is 0.7, what is the expected body weight among the offspring of the selected rabbits?
Answer:
The expected weight increase that is expected of offspring of selected parents is 0.7 kg
Explanation:
Given from the question:
The average body weight in his population of rabbits = 3 kg
heritability of body weight in the rabbit population (h^2)=0.7
Average body weight of the 10 selected largest rabbit =4 kg
(R) = response to selection
selection differential (S)= Average body weight of the 10 selected largest rabbit - The average body weight in his population of rabbits
selection differential (S)= 4-3=1 kg
But heritability of body weight in the rabbit population (h^2)=0.7
Then, Heritability of body weight in the rabbit population (h^2) can be calculated using the formula below;
Heritability of body weight in the rabbit population (h^2)=
response to selection (R) / selection differential (S)
response to selection (R) = Heritability of body weight in the rabbit population (h^2) × selection differential (S)
R= h^2 x S= 0.7 x 1kg= 0.7 kg
The expected weight increase that is expected of offspring of selected parents is 0.7 kg
Average weight of offspring = The average body weight in his population of rabbits + heritability of body weight in the rabbit population (h^2)
Average weight of offspring= 3 kg + 0.7 kg = 3.7kg
Therefore, the Average weight of offspring is 3.7kg and The expected weight increase that is expected of offspring of selected parents is 0.7 kg
A man with type AB blood is married to a woman with type O blood. They have two biological children, and one adopted child. The children’s blood types are: A, B, and O. Which child was adopted?
Answer:
The adopted child is the one with type O blood.
Explanation:
Available data:
A man with type AB blood is married to a woman with type O blood. They have two biological children and one adopted child.The man´s genotype for blood type is AB (Codominant type of blood) and the woman´s genotype for blood type is OO, as this type of blood is expressed by two recessive alleles.
Cross:
Parental) AB x OO
Gametes) A B O O
Punnet square) A B
O AO BO
O AO BO
F1) Half of the biological progeny will be AO and the other half BO.
There are no possibilities that any of the biological child could be OO. This means that the adopted child´s blood type is O and the biological children´s blood type are A and B.
The part of the atom that carries a positive charge is the _____, and the part that carries a neutral charge is the _____
Answer:
The first blank is proton and the second is neutron
Explanation:
Transfer RNA molecules pick up amino acids
which are free in the cytoplasm and carry them to
what features of meiosis are important in sexual reproduction
Answer:
The daughter cells are identical to the mother cell in shape and size but different in chromosome number. The daughter cells are haploid. Recombination and segregation take place in meiosis. The process occurs in the reproductive organs and results in the formation of gametes.
Explanation:
During sexual reproduction, meiosis generates genetic variation in offspring because the process randomly shuffles genes across chromosomes and then randomly separates half of those chromosomes into each gamete. The two gametes then randomly fuse to form a new organism.
Which is a method of making many copies of DNA?
O A. Nuclear transfer
B. Gel electrophoresis
C. Reverse transcriptase
D. Polymerase chain reaction
Answer:
D. Polymerase chain reaction
Explanation:
PCR is very percise; used to amplify or copy a specific DNA target from a mixture of DNA molecules by:
two short DNA sequences (primers) are bind to start and end the DNA targetperform the PCR; DNA templates is added to a tube that contains primers, free nucleotides, and the enzyme (DNA polymearse)it is mixed in a PCR machine that increases and decreases the temperature of the samplemixture is heated to seperate the double strnaded DNA templated into single strandsmixture is cooled so the primers can bind to the DNA templatethe DNA polymerase will synthesize to new strands of DNA from the primersduring synthesis, the end of the first cycle, the DNA molecule will consist of a new and old DNA strandPCR continues with added cycles to repeat the stepsnewly synthesized DNA segements serve as templates in later cycles; which can be copied over and over.The resource competition model of sympatric speciation requires: A. divergent/disruptive selection driving character displacement. B. directional selection driving character displacement. C. evolution of reproductive isolation D. Both A & C E. Both B & C
Answer:
The correct answer is option D, that is, both disruptive selection and evolution of reproductive isolation.
Explanation:
The phenomenon of the evolution of a novel species from a thriving ancestral species and at the same time both continue to live in a similar geographical area is termed as sympatric speciation. Due to reproductive isolation of the population of species that were thriving in a similar habitat, the phenomenon of sympatric speciation results.
On the other hand, sympatric speciation also results due to disruptive selection as in the process the extreme kind of individuals within the population are favored over the hybrids.
What makes a molecule a carbohydrate?
which of the following represents a testcross
A- t? x tt
B- XX x XY
C-T? x Tt
D- T? x TT
Answer:
The option should be like this:
A- T? x tt
Explanation:
In an inheritance pattern in which a particular allele is exhibiting doiminance over the other, the dominant phenotype can either be homozygous or heterozygous. Hence, it is impossible to tell the genotype of a dominant phenotype by merely observing it. A cross called TEST CROSS is used.
A test cross is a cross used to determine the actual genotype of a dominant phenotype i.e. whether it is homozygous or heterozygous by crossing tye unknown genotype with a homozygous recessive parent. If all of the offspring of this cross have a dominant phenotype, it means the dominant parent is HOMOZOGOUS but if offsprings with an equal ratio of dominant and recessive phenotypes are produced, it means the dominant parent is HETEROZYGOUS.
In this case, the dominant phenotype whose genotype is unknown is (T*) while the recessive parent will be (tt).
Answer:
a
Explanation:
Application of biotechnology in Agriculture
Answer:
Biotechnology can be used to engineer GMOs, or genetically modified organisms. These GMOs can be resistant to viruses, diseases, etc. that make them more desirable. One example is virus-resistant squash, a GMO that is resistant to viruses that often destroy fields of the squash. This GMO yields much more fruit and saves farmers' energy and resources. Biotechnology has been very useful in the agriculture industry for reasons such as the ones listed above, it can prevent crop decay and many more issues by engineering GMOs.
The mature ovary of a plant is a _____.
· pistil
· flower
· fruit
· seed
Answer:
The matured ovary of a plant is a seed.
Question 1
The gaining of electrons from an atom or molecule is an example of:
reduction
a.
b.
adhesion
O c. capillarity
d.
cohesion
oxidation
e.
Answer:
a. reduction
Explanation:
The reduction is an example of the gaining of electrons from an atom or molecule or the loss of oxygen atom from a molecule.
Oxidation and reduction always occur together so in oxidation electrons loss by one substance is fulfil by gaining electrons in reduction process.
For example: The copper ion form copper by gaining electrons and undergoing reduction.
Hence, the correct answer is a. Reduction.
1. Which of the following statements explains how N-formylmethionine (fMet) is incorporated only in response to the 5'-AUG initiation codon in E. coli, given that methionine has only one codon?
A. Methionine is only ever found as the first amino acid in a polypeptide
B. There is only one tRNA with the CAU anticodon.
C. Only Met-tRNAmCl can bind to a specific ribosomal entry site.
D. The N-formyl group on methionine prevents fMet from entering interior positions in a polypeptide.
E. fMet is always incorporated at interior AUG sites and then deformylated after translation.
2. Which of the following is not true of the second step of translational elongation in bacteria?
A. Puromycin inhibits translation by mimicking the 3' end of an aminoacyl-tRNA
B. The carboxyl group oxygen of the A-site amino acid is the nucleophile in peptide bond formation.
C. A peptide bond is formed between the two amino acids bound by the tRNAs in the A and P sites.
D. The growing polypeptide chain is transferred to the tRNA in the A site.
E. The a-amino group of the A-site amino acid attacks the carbonyl carbon of the P-site amino acid.
3. Which of the following is not a function of the release factors of bacterial translation termination'?
A. mediating the hydrolysis of the terminal peptidyl-tRNA bond
B. assisting in the initial folding of the nascent polypeptide chain
C. releasing the last uncharged tRNA from the P site
D. recognition of a stop codon
E. occupation of the ribosomal A site
4. Which of the following is NOT a possible outcome of the CRISPWCas9 system?
A. disruption of a gene
B. repair of a gene
C. deletion of a chromosome
D. blocking or enhancing gene expression
E. all of the above are possible
5. Which of the following is NOT true of the reaction catalyzed by RNA polymerases?
A. Template DNA is copied in the 3' to 5' direction.
B. The 3'-hydroxyl group of the growing RNA strand is the nucleophile.
C. The reaction is magnesium ion dependent.
D. U residues are incorporated into the RNA to pair with T residues of the template.
E. The RNA molecule is synthesized in the 5' to 3' direction.
6. The sequence of the messenger RNA molecule (primary transcript) synthesized from a DNA molecule with a template strand having the sequence 5'-ATGCTACCGTTA-3' is:
A. 5 '-UACGAUGGCAAU-3'
B. 5 '-UAACGGUAGCAU-3'
C. 5'-AUGCUACCGUUA-3'
D. 5'-AUUGCCAUCGUA-3'
E. None of these choices is correct
Answer: D. The N-formyl group on methionine prevents fMet from entering interior positions in a polypeptide.
Answer:
1) Option D (The N-formyl group on methionine prevents fMet from entering interior positions in a polypeptide)2) Option B (The carboxyl group oxygen of the A-site amino acid is the nucleophile in peptide bond formation)3) Option B (assisting in the initial folding of the nascent polypeptide chain)4) Option A (disruption of a gene)5) Option A (Template DNA is copied in the 3' to 5' direction)6) Option A (5 '-UACGAUGGCAAU-3')Explanation:
1) Option D(The N-formyl group on methionine prevents fMet from entering interior positions in a polypeptide)
2) Option B (The carboxyl group oxygen of the A-site amino acid is the nucleophile in peptide bond formation) is not responsible for the second part of the translation elongation in bacteria.
3) Option B (assisting in the initial folding of the nascent polypeptide chain) This is because the folding of the protein does not takes place in the ribosome. Afetr the binding of the release factor into the A site. the nascent peptide is released from the ribosomal complex. This nascent peptide is now present in the primary state and folding occur in secondary or tertiary state of the protein
4) Option A (disruption of a gene)
5) Option A (Template DNA is copied in the 3' to 5' direction) is wrong and not copied by the RNA polymerase. RNA polymerasze copy the Coding strands of the DNA in 3' to 5' direction
6) Option A (5 '-UACGAUGGCAAU-3') During transcrition of the RNA from the template DNA.is done by the enzyme RNA polymerase.
This RNA pol. copy the all the neucleotide from the coding strand of the DNA but replace Thymine by Uracil.
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https://brainly.com/question/16834143
Which describes the greenhouse effect
2. Why is variety in offspring so important?
Answer:
Important to Darwinian Evolution
Explanation:
Some traits in the offspring will be more useful than others. Because of Natural Selection, the offspring with the more favored traits are more likely to reproduce and pass on their genes.
In what phase do the organelles disperse out into the daughter cells, and the cell prepares for cytokinesis?
1. Prophase
2. Interphase
3. Mitosis
4. Telophase
alapogos tortoises have a number of shell variations, shown right. What are the advantages of choosing a system examining the beaks of finches instead of the shells of tortoises? Check all that apply. Birds have shorter lifespans, making generations shorter. The bird population is likely much greater, allowing for more data. Tortoises move too slowly for natural selection to influence shell phenotype. Birds are easier to track and monitor.
The question lacks essential part that is options to answer, however the options are given:
1. Birds have shorter lifespans, making generations shorter.
2. The bird population is likely much greater, allowing for more data.
3. Tortoises move too slowly for natural selection to influence shell phenotype.
4. Birds are easier to track and monitor.
Answer:
The correct options are : 1, 2, and 4.
Explanation:
For the accuracy and dependability of any experiment, various observations must be as extensive as possible. It decreases the impact of any observation or any event which may have occurred by some coincidence or mistake.
Therefore, finches were smarter to record observation as they were accessible in enormous numbers when compared with the turtle.
Likewise, the life cycle of birds is less when contrasted with the turtle which empowered the researchers to observe more generations.
In conclusion, the birds were anything but difficult to catch and track when contrasted with the turtle. Turtle is a land and water proficient which made it hard to follow them.
Answer:
noe. . g.aa.aaaaaaaaaaaaaaaaas
Explanation:
You conduct a series of two-point mapping crosses involving five genes located on chromosome III and obtain the following percentage of recombinant offspring:
h-cu eyg-h eyg-cu rai-eyg rai-h rai-cu th-cu th-h th-eyg th-rai
23.5% 9.0% 14.5% 18.5% 9.5% 33.0% 6.8% 16.7% 7.7% 26.2%
Considering that rai is near the end of chromosome III, construct a map of these genes. Drag gene names to blue targets and map distances to pink targets.
a. 6.8
b. 7.7
c. 9.0
d. 9.5
1. cu
2. eyg
3. h
4. th
1. The genes for miniature wings (m) and garnet eyes (g) are approximately 8 map units apart on chromosome 1 in Drosophila. Phenotypically wild-type females (m+g / mg+) were mated to miniature-winged males with garnet eyes (mg / mg). Which of the following phenotypic classes reflect offspring that were generated as a result of a crossover event?
a. miniature wings,
b. wild-type eyesminiature wings,
c. garnet eyeswild-type wings,
d. garnet eyeswild-type wings,
e. wild-type eyes
2. If 800 offspring were produced from the cross, in what numbers would you expect the following phenotypes?
a. wild-type wings, wild-type eyes :_____.
b. miniature wings, wild-type eyes :_____.
c. wild-type wings, garnet eyes :_____.
d. miniature wings, garnet eyes:_____.
Answer:
I am not a doctor
Explanation:
I am sorry
Round seeds and yellow seed color are dominant to wrinkled seeds and green seed color. Which Punnett square represents the correct way to set up the cross RRYy ´ RrYY? A 4-column Punnett square has 4 rows. The columns are labeled Upper R Upper Y, Upper R y, r Upper Y, r y. The rows are labeled Upper R Upper Y, Upper R y, r Upper y, r y. A 4-column Punnett square has 4 rows. The columns are labeled Upper R Upper Y, Upper R y, r Upper Y, r y. The rows are labeled Upper R Upper Y, Upper R y, Upper R Upper y, Upper R y. A 4-column Punnett square has 4 rows. The columns are labeled Upper R Upper Y, Upper R Upper Y, r Upper Y, r Upper Y. The rows are labeled Upper R Upper Y, Upper R y, Upper R Upper Y, Upper R y. A 4-column Punnett square has 4 rows. The columns are labeled Upper R Upper Y, Upper R Upper Y, r Upper Y, r Upper Y. The rows are labeled Upper R Upper Y, Upper R Upper Y, Upper R Upper Y, Upper R Upper y.
Answer:
A 4-column Punnett square has 4 rows. The columns are labeled Upper R Upper Y, Upper R Upper Y, r Upper Y, r Upper Y. The rows are labeled Upper R Upper Y, Upper R y, Upper R Upper Y, Upper R y.
Explanation:
This question involves a dihybrid cross i.e. two distinct genes; one coding for seed shape and the other for seed colour. The alleles for round seeds (R) and yellow seeds (Y) are dominant over the alleles for wrinkled seeds (r) and green seeds (y) in the first and second gene respectively.
In a cross between a RRYy and RrYY parent, each will undergo meiosis and produce the following combinations of gametes;
RRYy- RY, Ry, RY and Ry
RrYY- RY, RY, rY, and rY
These gametes will be used to construct a punnet square, which is a square-shaped diagram used to determine the possible genotypes that will be produced by two parents. The punnet square has 4 columns and 4 rows. The four possible gametes of one parent are filled into the four columns and the four possible gametes of the other parent are filled into the four rows (see attached image).
Hence, if the gametes of parent RrYY are filled into the four columns. The columns will labelled:
RY, RY, rY and rY
Also, if the gametes of parent RRYy are filled into the four rows. The rows will be labelled: RY, Ry, RY, and Ry.
This makes the third option correct.
Answer:
its c on edge 2020
on the top row it says
RY RY rY rY
on the side row it says
RY Ry RY Ry
Explanation:
i took the test
What is occurring when a transport protein uses energy to pump a solute against its gradient across a
cell membrane
Answer:
Active transportation
Explanation:
Active transportation is happening because energy is need to pump against the gradient. Whenever energy is used the transportation is active.
nteractions between organisms and their environment impact the organism’s overall population. The jaguar Panthera onca is the largest cat in North America. It is found in areas across the Southwest, including Arizona, New Mexico, and Texas. It is a carnivore that has powerful jaws and sharp teeth and preys on fish, turtles, tapirs, and many smaller mammals. Which shows the relationship between the jaguar and turtles?
This question is incomplete because the options are missing, here is the question and the options:
Interactions between organisms and their environment impact the organism’s overall population. The jaguar Panthera onca is the largest cat in North America. It is found in areas across the Southwest, including Arizona, New Mexico, and Texas. It is a carnivore that has powerful jaws and sharp teeth and preys on fish, turtles, tapirs, and many smaller mammals. Which shows the relationship between the jaguar and turtles?
A. Adaptation principle
B. Competitive exclusion principle
C. Predator-prey relationship
D. Niche principle
The correct answer is C. Predator-prey relationship
Explanation:
In a prey-predator relationship, one organism which is the predator feeds on another animal known as the prey. Additionally, predators hunt and kill their prey, which is essential for predators to obtain nutrients and survive. Besides this, predators can hunt and feed on only one type of animal or hunt and feed on different types of animals. This relationship occurs between the jaguar and the turtle because the jaguar hunts, kills, and eats the turtles, which means jaguars hunt and kill other animals to get nutrients. Also, in the case of the jaguar, this hunts different animals including fish, tapirs, turtles, and small mammals.
Answer:
predator-prey relationship
Explanation:
edge
How could a tropical storm affect you personally?
Answer:
it can stress you out because of the danger you're in
What are two examples of extreme weather on exoplanets?
Answer: rock rain on Corot-7b and diamond rain on saturn
Explanation:
during drought years on the galapagos,small easily eaten seeds become rare, leaving mostly large, hard-cased seeds that only birds with large beaks can eat. if drought continues for several years, measurable changes to the beak size of the galapagos bird population can be observed. which of the following identifies the changes to the beak size of the galapagos bird population what would most likely be expected as a result of natural selection? a) more small-beaked birds dying than larger beaked birds. the offspring produced in later generations have a higher percentage of birds with large beaks. b) small birds gaining larger beaks by exercising their mouth parts c) small birds mutating their beak genes resulting in later-generation offspring having larger beaks d) small birds anticipating the long drought and eating more to gain weight, resulting in them growing larger beaks
Answer:
A) More small-beaked birds dying than larger beaked birds
Explanation:
Natural selection will cause the small-beaked birds to die, while large-beaked birds will begin producing more birds with that gene. This causes large-beaked birds to become dominant, and less and less small-beaked birds will survive.
Hope this helps!
Answer:
A.
Explanation:
Considering that the seeds have changed to where larger birds can only consume them (or birds with larger beaks, anyways..) then smaller beaked birds would die or as a result of the changing in the food supply.If the food supply revolves around having a stronger or larger beak, then birds with larger beaks are more likely to flourish.
Which is a density-independent factor in controlling a population? A. Floods B. Access to light C. Water supply D. Predators
