in the griffiths-avery-mccarty experiments, how was it determined that it was dna that transformed rough bacteria into smooth?

Answers

Answer 1

Answer: In the Griffiths-Avery-McCarty experiments, it was determined that it was DNA that transformed rough bacteria into smooth by the experiments of Oswald Avery and his colleagues.

Avery, Colin MacLeod, and Maclyn McCarty's experiments were based on Griffith's finding that R strain bacteria could be transformed into S strain bacteria by dead heat-killed S strain bacteria. It was important to establish the mechanism of transformation, as well as the substance that carried the genetic material responsible for the transformation.

To achieve this goal, Avery, MacLeod, and McCarty executed a series of experiments that were designed to identify the material responsible for the transformation. They first divided the S strain into its main biochemical components, including RNA, DNA, proteins, and lipids, and then treated the R strain with each of these components. The transformation only occurred when the R strain was treated with DNA.

The R strain, on the other hand, did not transform when treated with RNA, protein, or lipid. The researchers came to the conclusion that DNA is the hereditary substance that transmits genetic information from one generation of organisms to the next. This discovery altered the study of genetics and advanced research on DNA's function and structure.



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following ingestion of mushrooms found growing in his garden, a man develops symptoms of oliguria, lethargy, and edema. many renal tubular epithelial (rte) cells are observed in his urinalysis. this is a case of:

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Ingestion of mushrooms found growing in his garden, a man develops symptoms of oliguria, lethargy, and edema. many renal tubular epithelial (rte) cells are observed in his urinalysis. This is a case of: Mycetism.

Mycetism is the poisoning that occurs when toxic substances from certain mushrooms are ingested. Symptoms of mycetism may include oliguria (decreased urination), lethargy (extreme tiredness), and edema (swelling due to fluid buildup).

Urinalysis may reveal an increased number of renal tubular epithelial (RTE) cells, which are cells that line the inside of the kidney tubules. Treatment may involve supportive care, gastric lavage (stomach pumping), and/or antifungal drugs. It is important to recognize and avoid toxic mushrooms in the future to avoid this type of poisoning.

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describe the zones of the epiphyseal plate and their functions, and the significance of the epiphyseal line.

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The epiphyseal plate, also known as the growth plate, is composed of four zones: the resting zone, the proliferative zone, the hypertrophic zone, and the calcified zone. The epiphyseal line, or growth line, is the division between the epiphyseal plate and the diaphysis and is where all growth stops.

The resting zone is the first zone in the epiphyseal plate and is located at the epiphyseal side of the plate. It contains cells that are inactive but can divide to form more chondrocytes, which are essential for the formation of bone and cartilage.

The proliferative zone is the second zone and is the site of cell division and growth.

The hypertrophic zone is the third zone and is the site of most growth. It is also the site of most of the extracellular matrix mineralization, as chondrocytes in this zone produce high levels of collagen and other matrix proteins.

The calcified zone is the fourth and last zone and is composed of cells that are no longer able to divide or grow. It contains mature, mineralized cartilage.

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Which of the following are responsible for sending messages from the
midbrain to the cerebrum?
A. Sensory neurons
B. Interneurons
C. Hormones
D. Motor neurons

Answers

Answer:A. Sensory neurons

Explanation:

>> We know that, the he Sensory neurons conduct signals from sensory organs to the CNS.

>> The Sensory Neurons arise from the dorsal root ganglion which are specialized clusters present at the dorsal roots of the spinal cord.

>> The Sensory neurons lack distinct axons and dendrites.

>> The soma of the sensory neurons possesses a nucleus and other cell organelles.

>> A synaptic junction with second-order sensory neurons is formed as the central branch extends from soma to the posterior horn of the spinal cord.

The functions of sensory neurons are :

>> Its the Controlling the Heartbeat and Blood Circulation

>> The sensory receptors in the blood vessels are responsible for registering blood pressure.

>> The Sensory neurons can be found in the aorta carotid arteries pulmonary artery capillaries in the adrenal gland and the tissues of the heart itself from where the signals are sent to the medulla and thus the help in controlling BP and blood circulation.

>> The Taste receptor cells on our tongues form a group of 50 to 150.

>> These cells respond to the chemicals present in the food and thus the form taste buds which help us in differentiating among the food items of different tastes.

Answer:

Interneurons

Explanation:

took the quiz

PLSSSS HELP IF YOU TURLY KNOW THISSS

Answers

Which type of cloud is very close to the earth's surface?

Fog

The altostartus clouds are found in the upper troposphere

The cirrus clouds are found in the troposphere

The cumulonimbus clouds are found in the lower troposphere...

What are the main functions of the ear? Please respond in 1-2 complete sentences
using your best grammar.

Answers

Hearing, Balance and equilibrium: The ear is also very important for keeping your balance and equilibrium, which is important for your posture, movement, and sense of where you are in space.

Pressure regulation: The Eustachian tube, which connects the middle ear to the back of the throat, is opened and closed by the ear. This helps keep the pressure in the middle ear at the right level.

Protection: Hair and wax line the ear canal, which helps keep dust, dirt, and other foreign particles from getting into the ear's delicate structures.

Temperature regulation: When the temperature outside changes, the ear responds by widening or narrowing the blood vessels in the ear.

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the red portion of the human lip: question 12 options: integumentary lip. has no facial markings. must be treated by hypodermic tissue building in every case. mucous membrane.

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The red portion of the human lip is known as the mucous membrane. It does not have any facial markings and must be treated by hypodermic tissue building in every case.

What is the mucous membrane?

The mucous membrane is a layer of tissue that lines various parts of the body's openings and cavities that are in contact with the outside environment. It is a moist membrane that secretes mucus, a slimy substance that assists in trapping germs and other foreign substances, as well as keeping the surface moist.

The red portion of the human lip: Mucous membrane. The red portion of the human lip is the mucous membrane. The mucous membrane of the lips is often known as the vermilion zone. It is a transition zone between the skin and the mucous membrane.

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6. What type of body blan do sponges have?
A. Cephalization
B.asymmetry
C.bilateral symmetry
D.radial symmetry

Answers

The Option B is correct. The body blan  have the assymetry types of sponges in them

What are sponges?

Sponges (Phylum Porifera) are a type of aquatic animal that lack true tissues and organs. They are considered the simplest of all animals and do not have a body plan based on symmetry.

Instead, sponges exhibit a type of symmetry known as "asymmetry," which means they have no plane of symmetry or any organization of body parts around a central axis. Therefore, the correct answer is B. asymmetry.

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the presence of a rug or its metabolites in cells, tissue, organs, or other edible products of an animal is referred to as a

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The presence of a rug or its metabolites in cells, tissue, organs, or other edible products of an animal is referred to as a Drug Residue.

A drug residue is any medication that remains in animal tissues, fluids, or edible goods at the time of slaughter or when an animal is harvested or when an animal is given to the owner for consumption.

Drug residues are defined as any compound found in animal tissue, edible animal products, or animal feed, including their metabolites, which are unapproved for use in food animals or are used at higher doses, routes of administration, or withdrawal times than allowed in official labeling.

The presence of a drug residue in an animal's body, as well as the amount of that drug residue, can be influenced by various factors, including the animal's health status, dosage, route of administration, withdrawal times, and the presence of other drug residues.

Drug residues may persist in animal tissues, fluids, and edible goods long after the drug has been administered to the animal. The accumulation of drug residues in animal tissues and products raises health concerns for humans who eat the products.

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In the same mouse species, a third unlinked gene (gene C/c) also has an epistatic effect on fur color. The presence of the dominant allele C (for color), allows the A/a and B/b genes to be expressed normally. The presence of two recessive alleles (cc), on the other hand, prevents any pigment from being formed, resulting in an albino (white) mouse.Matchthe phenotypes on the labels at left to the genotypes listed below. Labels can be used once, more than once, or not at all.agoutisolid colorsolid coloragouti blackalbinoAaBbccAaBBCCAabbccAAbbCcaaBbCcAABBcc

Answers

The phenotype "agouti" would be matched with the genotype AaBb, "solid color" with the genotype AaBB or Aabb, "black" with the genotype AABB or AABb, and "albino" with the genotype cc. This is because the presence of the gene C/c (epistasis) determines the fur color of the mouse, and the genotypes above show the different combinations of alleles. If two recessive alleles (cc) are present, it will result in an albino (white) mouse.

Explanation:
Physical characteristics like the fur color of a mouse are determined by the combination of genes in the organism's DNA. Epistasis is a phenomenon in which the expression of one gene affects the expression of another gene. When an organism reproduces, genes are inherited by offspring from their parents. In the context of this problem, the genes involved in determining fur color are A/a, B/b, and C/c. C is the gene that has an epistatic effect on fur color.

Here, are the matched genotypes with phenotypes: AaBbcc - agouti solid colorAaBBCC - solid colorAgouti black - AAbbCc, AaBbCcAlbino - aabbcc, aabbCc, aabbCC, aaBbcc, aaBbCc, aaBBcc.The label agouti solid color matches with the genotype AaBbcc. The solid color matches the genotype AaBBCC. The label agouti black matches with the genotypes AAbbCc and AaBbCc. The label albino matches with the genotypes aabbcc, aabbCc, aabbCC, aaBbcc, aaBbCc, and aaBBcc.

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what is the main function of the ribosomes in the cell? multiple choice to break down proteins into individual amino acids to provide strength and structural support for the cell membrane to form the nuclear envelope to synthesize proteins to synthesize dna

Answers

Answer:

break down proteins into individual amino acids

Explanation:

if dna contains the code for making proteins, wherein the structure of the double helix do you think the code is found?

Answers

DNA contains the code for making proteins. The code in DNA is found in the structure of the double helix in several different ways.

The double helix structure is composed of two strands of nucleotides that are linked together by hydrogen bonds. The code is found in the sequence of nucleotides along each strand of the double helix. The sequence of nucleotides is what determines the genetic code. The genetic code is read in groups of three nucleotides called codons. Each codon codes for a specific amino acid, which is then used to build proteins. In addition to the sequence of nucleotides, the code is also found in the way that the double helix is folded and coiled. The three-dimensional structure of the double helix determines which parts of the DNA are accessible and which parts are not. This, in turn, determines which genes are expressed and which are not. The double helix structure of DNA is a complex structure that contains the code for making proteins in many different ways.

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type iii hypersensitivity is caused by soluble antigen-antibody complexes that avoid being phagocytized by macrophages. true false g

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Type III hypersensitivity is caused by soluble antigen-antibody complexes that avoid being phagocytized by macrophages. This statement is true.

What is type III hypersensitivity?

Type III hypersensitivity occurs when a large amount of antigen enters the body and combines with an antibody, forming an insoluble complex. These are difficult to eliminate, and they begin to settle in the tissues, particularly those with a low blood supply and a high concentration of protein. They elicit an inflammatory response and, as a result, the release of proteases, hydrolases, and complement factors is increased.These immune complexes can become stuck in blood vessels or other organs, resulting in symptoms such as joint pain, fever, and rash. These symptoms usually manifest in the tissues where the complexes are deposited.

What are the causes of type III hypersensitivity?

The causative agents of Type III hypersensitivity are usually proteins, such as serum proteins or microbial proteins, that combine with specific antibodies to form circulating immune complexes. If the immune complexes become deposited in the blood vessels, they can result in vasculitis, inflammation, and subsequent tissue damage. Type III hypersensitivity is responsible for diseases like systemic lupus erythematosus, rheumatoid arthritis, and serum sickness.

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on the cellular level, how is gastrulation accomplished in echinoderms, amphibians, and birds? in general terms what does gastrulation accomplish?

Answers

Gastrulation in echinoderms, amphibians, and birds is accomplished through the invagination of different cells.

In general, gastrulation is the process that reorganizes cells to form the three germ layers, which are necessary for the further development of an organism.

Gastrulation is the process in which cells rearrange to form the three germ layers: the ectoderm, mesoderm, and endoderm.

In echinoderms, gastrulation is accomplished through the process of archenteron formation, which is when the mesoderm forms from the invagination of cells from the surface of the embryo.

In amphibians, gastrulation is accomplished through blastopore closure, which is when the opening at the blastula stage of the embryo closes.

In birds, gastrulation is accomplished through the formation of the primitive streak, which is when the ectoderm folds and inwards to form a groove-like structure.


In summary, gastrulation is the first step of morphogenesis, the development of form and structure, which will determine the shape of the organism. The three germ layers will further differentiate and develop into the organs, tissues, and cells that make up the organism.

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problem 5: in an alaskan village of inuit indians, an inordinate number of cats have 6 toes on each foot. the trait of polydactyly (many digits) is caused by a dominant allele. if 22% of the cats have 6 digits per foot, what is the allele frequency of this dominant allele in this population of cats?

Answers

The allele frequency of the polydactyly (many digits) trait in the population of cats in the Alaskan village of Inuit Indians is 0.22 (22%).

Polydactyly is caused by a dominant allele, meaning that the allele is expressed in the organism even when the organism only has one copy of it.
This means that in the population of cats, 22% of them are expressing the trait, indicating that 22% of the cats have one or two copies of the dominant allele for polydactyly.

In order for the cats to have this trait, at least one of their parents must have the same dominant allele, meaning that the parents of the cats expressing the trait must have a combined allele frequency of 0.22 (22%) or more.
The allele frequency of 0.22 (22%) is then passed on to the offspring of the cats expressing the trait, meaning that the cats expressing the trait must have a combined allele frequency of 0.22 (22%) or more.

This means that 22% of the cats in the population have either one or two copies of the dominant allele for polydactyly.

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a gardener would like to grow a lemon tree from a lemon. what is the first thing he should do?

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If a gardener wants to grow a lemon tree from a lemon, the first thing he should do is to remove the seeds from the lemon to germinate.

A gardener who wants to grow a lemon tree from a lemon should follow a series of steps. These steps are as follows:

Step 1: Remove the seeds from the lemon. The seeds should be washed and cleaned with water. The gardener should be careful not to damage the seeds.

Step 2: Prepare the soil. The soil should be well-draining, rich in nutrients, and have a pH of 5.5 to 6.5. The gardener can mix sand, perlite, and vermiculite to the soil to increase its drainage.

Step 3: Plant the seeds. The gardener should plant the seeds about 1 inch deep into the soil. The soil should be moist but not waterlogged.

Step 4: Cover the pot with a plastic bag or a plastic wrap to create a greenhouse effect.

Step 5: Place the pot in a warm and sunny location. The temperature should be around 70 degrees Fahrenheit.

Step 6: Water the soil regularly. The soil should be kept moist but not waterlogged.

Step 7: Wait for the seeds to germinate. It may take a few weeks to a few months for the seeds to germinate.

Step 8: Once the seedlings have grown big enough, they can be transplanted into a bigger pot. The plant should be kept in a warm and sunny location. The soil should be kept moist but not waterlogged.

Step 9: The lemon tree should be fertilized with a citrus fertilizer every two weeks during the growing season.

Step 10: The lemon tree should be pruned regularly to remove dead, damaged, or diseased branches.

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imagine a condition where the vessels that carry blood between the lungs and the body tissues were permeable to oxygen. what would you expect to observe relative to the normal condition of low permeability to oxygen in the vessels that carry blood from the lungs to the tissues?

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If the vessels between the lungs and body tissues were permeable to oxygen, there will be a decrease in the oxygen supply to the body tissues.

Normally, oxygen-poor blood from the body tissues flows into the right side of the heart, and is then pumped to the lungs where it picks up oxygen and releases carbon dioxide. The oxygen-rich blood then flows back to the left side of the heart, where it is pumped out to the body tissues to supply oxygen to the cells.

If the vessels between the lungs and body tissues were permeable to oxygen, oxygen-rich blood from the lungs would flow into the right side of the heart, mix with oxygen-poor blood from the body tissues, and then be pumped out to the body tissues.

This would result in a reduced delivery of oxygen to the tissues, as some of the oxygen-rich blood from the lungs would bypass the body tissues and flow back to the lungs.

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what could you infer if scientists discover that south america split from africa well before the evolution of the common ancestor of all modern primates?

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This would mean that the common ancestor of all modern primates did not exist in South America at the time of the split and that the primates in South America evolved from a different lineage.

Evolution explained.

If scientists discover that South America split from Africa well before the evolution of the common ancestor of all modern primates, it would suggest that the primates in South America evolved independently from those in Africa. This would mean that the common ancestor of all modern primates did not exist in South America at the time of the split and that the primates in South America evolved from a different lineage.

This finding could have important implications for our understanding of primate evolution and biogeography. It would suggest that primates have a more complex evolutionary history than previously thought, and that their distribution and diversification may have been influenced by a variety of factors, including continental drift, climate change, and ecological interactions.

Additionally, it could help to explain some of the unique characteristics of the primates in South America, such as the presence of the platyrrhine dental formula (2133) and the absence of a number of primate groups that are found in other parts of the world. It could also shed light on the processes that drove the evolution of primates in both South America and Africa, and how these processes may have influenced the diversification of primates more broadly.

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if pure water and a solution containing a nonpenetrating solute are separated by a membrane that is permeable only to water, what would occur?

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Water will diffuse by osmosis toward the side with the solute, until stopped by opposing hydrostatic pressure.

If pure water and a solution containing a nonpenetrating solute are separated by a membrane that is permeable only to water, osmosis will occur.

Osmosis is the movement of water molecules across a membrane in order to equalize the solute concentration on either side. As the solute molecules are unable to pass through the membrane, only the water molecules are allowed to pass. This results in the transfer of water molecules from the pure water to the solution containing a nonpenetrating solute, thus increasing the solute concentration on the pure water side and decreasing the concentration on the other side. In the end, equilibrium is achieved and the water molecules will stop moving.

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s you read your textbook, note the similarities and differences between the different land biomes and aquatic ecosystems. there will be more than 1 biome that fits into each feature, and each biome can be used more than once. record your work in the table.

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These are environments found in water, either freshwater or marine. Examples include lakes, rivers, estuaries, and coral reefs.

What kind of environment found in water?

As I cannot view the specific textbook or table you are using, I will provide general information about the similarities and differences between land biomes and aquatic ecosystems. Please refer to your textbook and adjust the information accordingly.

Land biomes: These are large regions defined by their climate, vegetation, and animal life. Some examples include forests, grasslands, and deserts.

Similarities: Land biomes share features such as soil type, precipitation levels, and temperature ranges. They also contain diverse plant and animal life adapted to the specific conditions.
- Differences: Land biomes differ in climate, vegetation, and animal life. For example, forests are characterized by a high density of trees, while grasslands have predominantly grasses and deserts have little vegetation.

Aquatic ecosystems: These are environments found in water, either freshwater or marine. Examples include lakes, rivers, estuaries, and coral reefs.

Similarities: Aquatic ecosystems share features such as water depth, salinity, and temperature. They also contain diverse aquatic plants and animal life adapted to the specific conditions.

Differences: Aquatic ecosystems differ in the type of water (freshwater or marine), water movement, and available sunlight. For example, lakes are still bodies of freshwater, while rivers have flowing freshwater. Estuaries are where freshwater meets marine water, and coral reefs are marine ecosystems with high biodiversity.

To record your work in the table, you can list each biome and aquatic ecosystem, then note their similarities and differences based on the features mentioned above. Please refer to your textbook for specific examples and more detailed information.

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Which of the following is NOT found in saliva? A) urea and uric acid. B) electrolytes. C) lysozyme. D) protease. D) protease.

Answers

Proteases enzyme is not found in saliva , hence option 'D' is correct

The natural execration occurs from salivary gland, thus it accounts for high concentration of urea and uric acid found in saliva. Since the amount of creatinine production is consonant in 24 hours , uric acid and urea -to- creatinine ratio are better to clarify the changes of this compound concentration in saliva . Therefore option A is incorrect.

The main inorganic components are sodium , potassium, chloride, calcium, phosphate , and bicarbonate , all contributing to the ionic strength of saliva. Therefore option B is incorrect.

As an important part of the non specific immune defense mechanism , lysozyme is an important component of antibacterial in saliva. Therefore option C is incorrect.

Proteases are released by pancreas into the proximal small intestine ,where the mix with proteins already denatured by gastric secretion's and break down into amino acids. Therefore option "D" is correct.

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meiosis divides one cell into four cells, but the resulting cells have half the amount of dna as compared to the original cell. how do you think this is possible?

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During meiosis, one cell is divided into four cells, but the resulting cells have half the amount of DNA as compared to the original cell. This is because of the two cell divisions, meiosis I and meiosis II, that occur during meiosis.

During meiosis I, homologous chromosomes separate, resulting in two cells with half the number of chromosomes as the original cell.

During meiosis II, sister chromatids separate, resulting in four cells, each with half the number of chromosomes as the original cell.

In other words, the resulting cells have half the amount of DNA because meiosis results in four cells, each containing half the number of chromosomes and, therefore, half the amount of DNA as the original cell.

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procaine (novocaine) is metabolized primarily by the group of answer choices liver. lungs. plasma. kidneys.

Answers

Answer: plasma

Explanation:

the thyroid gland has primary responsibility for the fight or flight stress response. true or false?

Answers

Answer:

In this example of literal language, the writer means to explain exactly what is written: that he or she chose to ride the bus because of the heavy rain. Figurative language is used to mean something other than what is written, something symbolic, suggested, or implied.

Explanation:

what was the first disease shown to be bacterial in origin? what was the first disease shown to be bacterial in origin? cholera malaria yellow fever tuberculosis anthrax

Answers

The first disease shown to be bacterial in origin was cholera. It is characterized by diarrhea, vomiting, and dehydration

Cholera is an acute gastrointestinal infection caused by the bacteria Vibrio cholera, which is found in contaminated water or food. In 1854, John Snow, an English physician, concluded that cholera was spread through water contaminated with feces, leading to the first scientific demonstration that a disease was caused by bacteria. This realization was an important milestone in the history of medicine, as it showed that diseases were caused by microorganisms and could be prevented and treated by controlling their environment. Cholera remains an important disease, especially in developing countries, where sanitation is often poor and water-borne diseases are common.

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what is the specific receptor site on the host cell that the virus needs to attach and infect?

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The specific receptor site on the host cell that the virus needs to attach and infect is the cell surface receptor.

A cell surface receptor is a protein that spans the plasma membrane of a cell and acts as a signal transducer that recognizes extracellular molecules and stimulates an intracellular response.

This response could involve changing the membrane potential or an intracellular signaling pathway. The virus's attachment to a host cell is dependent on the presence of specific host cell receptors. The virus uses these receptors to enter host cells and replicate, causing disease.

Many viruses bind to specific proteins on the cell surface of the host, while others bind to glycoproteins or glycolipids. For example, the flu virus binds to sialic acid molecules on the surface of host cells, while the human immunodeficiency virus (HIV) binds to the CD4 receptor and the chemokine receptor.

The binding of a virus to a cell surface receptor is often the first step in viral infection. Once the virus binds to the receptor, it triggers a series of events that result in the virus entering the cell and taking over its machinery to replicate itself.



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a pea plant is homozygous dominant for seed shape and seed color and produces round, yellow seeds. what genotype is possible for the offspring?

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Genotype possible for the offspring when a pea plant produces round, yellow seeds: RRYY, RRYy, RrYY, RrYy

Pea plants have a variety of seed shapes and colors. These characters are managed by alleles, alternative versions of a gene. A pea plant is homozygous dominant for seed shape and seed color and produces round, yellow seeds. Let's determine the genotype possible for the offspring. It is homozygous dominant for both seed color and seed shape, therefore the genotype must be RRYY. R represents the round seed, and Y represents the yellow seed. It means the two alleles (genes) on each pair are the same and dominant. Round (R) is dominant over wrinkled (r) and yellow (Y) is dominant over green (y). Therefore, the offspring must contain either RRYY, RRYy, RrYY, RrYy.

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what was the control group in this study? a the transplanted population in the killifish pools b the transplanted population in the pike-cichlid pools c the source population in the killifish pools d the source population in the pike-cichlid pools

Answers

In an ecological study involving killifish and pike-cichlid pools, the control group is the source population in the pike-cichlid pools as it did not receive any intervention in the study.

In a study, the control group refers to the group that does not receive any treatment or intervention and is used as a comparison to the experimental group. In this scenario, the source population in the pike-cichlid pools is the control group as it did not receive any intervention in the study. The study is not mentioned in the question, but based on the options provided, it is likely an ecological study involving killifish and pike-cichlid pools. The transplanted population is most likely the experimental group. The source population in the killifish pools and the source population in the pike-cichlid pools are both control groups that did not receive any intervention in the study.

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transport of a solute across a membrane where the solute is going up its concentration gradient and using protein carriers driven by the expenditure of chemical energy, is known as

Answers

Transport of a solute across a membrane where the solute is going up its concentration gradient and using protein carriers driven by the expenditure of chemical energy is known as active transport.

What is active transport?

Active transport is the movement of molecules against the concentration gradient, which means moving from lower to higher concentrations. It involves a direct energy source (ATP) to drive the movement of molecules. The active transport method involves the use of protein pumps to move molecules across the cell membrane. These pumps can help move molecules, including sodium, calcium, and potassium, against the concentration gradient, which allows the cell to regulate what enters and exits. During active transport, the cell must use energy in the form of ATP to transport the molecules.

In summary, the transport of a solute across a membrane, where the solute is going up its concentration gradient and using protein carriers driven by the expenditure of chemical energy, is known as active transport. Active transport requires energy, which is provided by the hydrolysis of ATP. Active transport is necessary because it allows the cell to maintain its internal environment despite the external environment's changes.

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7. Which of the following organisms could be considered a primary
consumer and a secondary consumer according to the food web?
A. Fox
B. Snake
C. Caterpillar
D. Mouse

Answers

i think it’d be a mouse because they’re omnivores

Help with my biology please

Answers

Carbohydrates are composed of monosaccharides, proteins are composed of amino acids, and nucleic acids are composed of nucleotides.

What are the elements present and the building blocks in carbohydrates, proteins, and nucleic acids?

Carbohydrates, proteins, and nucleic acids are three major classes of biomolecules that are essential for life.

Here are the elements present and the building blocks of each:

Carbohydrates:

Carbohydrates are organic molecules that contain carbon, hydrogen, and oxygen in the ratio of 1:2:1. The building blocks of carbohydrates are monosaccharides, which are simple sugars that cannot be broken down into smaller molecules. Examples of monosaccharides include glucose, fructose, and galactose.

Proteins:

Proteins are complex molecules that are made up of amino acids. Amino acids contain carbon, hydrogen, oxygen, nitrogen, and sometimes sulfur. There are 20 different types of amino acids, and they are joined together by peptide bonds to form polypeptide chains, which fold into specific three-dimensional structures to form proteins.

Nucleic acids:

Nucleic acids are macromolecules that store and transmit genetic information. They are composed of nucleotides, which are made up of a nitrogenous base, a sugar, and a phosphate group. The four nitrogenous bases in DNA are adenine, guanine, cytosine, and thymine, while in RNA, uracil replaces thymine. The sugar in DNA is deoxyribose, while in RNA, it is ribose. The nucleotides are joined together by phosphodiester bonds to form a linear chain called a polynucleotide.

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