Answer:
when u do ittttt
Explanation:
26. predict the elect of the loss of forests on cellular respiration in other organisms
HELP NEEDED ASAP! 30 POINTS AND BRAINLYEST TO WHOEVER ANSWERS WITH THE BEST ANSWER! What are 3 examples of gene therapies that are used to treat diseases?
Answer:
Viruses used in gene therapy are retroviruses, adenoviruses, adeno-associated virus, herpes simplex, and vaccinia. Retroviruses are most commonly used because they can incorporate their genetic material into the host cell’s DNA, thus changing the genetic component of that cell.
Explanation:
Gene therapy is a medical field which focuses on the genetic modification of cells to produce a therapeutic effect or the treatment of disease by repairing or reconstructing defective genetic material. The first attempt at modifying human DNA was performed in 1980 by Martin Cline, but the first successful nuclear gene transfer in humans, approved by the National Institutes of Health, was performed in May 1989. The first therapeutic use of gene transfer as well as the first direct insertion of human DNA into the nuclear genome was performed by French Anderson in a trial starting in September 1990. It is thought to be able to cure many genetic disorders or treat them over time.
Gene augmentation therapy.
Gene inhibition therapy.
Killing of specific cells.
I think this is what you want but i'm only 14 and in 10th so sry if its not
what is the process of producing energy in living organisms
Answer:
cellular respiration
Explanation:
how many cans of green beans in green bean casserole
Match each hazardous waste with its matching description.
Words:
-asbestos
-mercury
-cadmium
-lead
Definitions:
A) Substance from thermostats known to cause reproductive issues
B) Substance known to cause lung cancer, found in older landfills
C) Substance found in computer monitors that causes brain damage
D) Hazardous cancer agent found in the batteries of many devices
Answer:
B is to asbestos
A is to lead
D is to cadmium
C is to mercury
if i could get a brainly answer i would appreciate
how are the 5 levels (cell tissue organ system and organism) of organization in living things similar to the levels of organization in a school building?
Answer:
student_teacher_principal_proprietor_minister of education
What is the structure of the forebrain that connects the two hemispheres of cerebrum?
Answer:
c. Corpus callosum
Explanation: is correct
Answer:
corpus callosum
Explanation:
right on test
IQ is determined by:(1 point)
multiple genes only
both a single gene and environmental factors
a single gene only
multiple genes and environmental factors
IQ is determined by ; ( D ) multiple genes and environmental factors
IQ ( intelligence quotient ) is influenced by more than a single gene ( i.e. multiple genes ) and this makes IQ a polygenic trait found in organisms such as humans.
IQ is not only determined by multiple traits but is strongly influenced by environmental factors just like the multiple genes. IQ measures the intellectual ability of humans ( i.e ability to read and think critically outside the box and also think logically . )
Hence we can conclude that IQ is determined by multiple genes and environmental factors
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Answer:
multiple genes and environmental factors
Explanation:
_____ cells are long to carry impulses quickly from one place to another
Answer:
Neuron cells
Explanation:
Neurons use axons to carry nerve impulses
Answer:
i thinks they are called neurons/nerve cells
Explanation:
Genes located near one another on the same chromosome are often inherited together. These are called ________.
Answer: Linked Genes
Explanation:
The _______ horn is the region at the rear of the spinal cord that receives inputs from receptors in the skin.
Answer:
Dorsal
Explanation:
What are organelles.
Organelles are specialized structures that perform various jobs inside cells.
What organisms does lactic acid fermentation occur in.
Answer:
It occurs in Yogurt Bacteria
A partial marine food web is shown.
image
Which of the following lists only organisms that are secondary consumers in this food web?
A.
euphausilds, sweeps, water fleas, sea urchins
B.
crabs, pilchards, blennies, snappers
C.
blennies, sweeps, tiger sharks, dolphins
D.
bull kelp, sweeps, water fleas, dinoflagellates
Crabs, pilchards, blennies and snappers are the examples of secondary consumers.
Crabs, pilchards, blennies and snappers are considered as secondary consumers because they feed on primary consumer which feed on producers such as plants and algae. Secondary consumer are those organisms which feed on herbivorous organisms or primary consumers.
They can feed on producers due to different digestive system so we can conclude that crabs, pilchards, blennies and snappers are the examples of secondary consumers.
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In a particular diploid organism, somatic cells have 24 chromosomes. How many pairs of homologous chromosomes would be present in the gametes of that organism
Answer:
12
Explanation:
It says 'pairs'.
how much atp is gained at the end of cellular respiration
Which sequence of dna bases would pair with this partial strand cat tca ctg?.
A) Identify a signaling molecule from the model present. Explain how receptors play a role in cell differentiation.
B) Identify the dependent variable and two controls the experimenters used when conducting this experiment.
C) Evaluate if the number of Variant 1-Type cells with mating projections was significantly different from those of the Wild Type. Use chi-square analysis.
D) Scientists propose that a mutation has occurred that either changed the mating pheromone or receptor site on the Variant 1-Type yeast cells. Predict where the mutation occurred. Justify your prediction with evidence from the experiment and scientific reasoning, based on your knowledge of cell-signaling pathways.
Answer:
A) A signaling molecule from the model shown is the pheromone. The pheromone binds to the receptor to create the cellular response in yeast to stop growth and produce shmoo. Shmoo is a nodule that allows the yeast cells to join together. Receptors play a role in cell differentiation, because the signaling molecules bind to the receptor in order to produce a response. When the pheromone binds to the receptor, a series of steps are followed in the transduction pathway in order to create shmoo, a differentiation in the cell. Without the receptor, the signaling pheromone would not be able to trigger the transduction pathway that ultimately results in the differentiation of the cell. Only signaling molecules with a specific shape and size can bind to a specialized receptor and cause a cellular response. Different yeast cell types may have varying receptors, affecting the ability of each pheromone to bind to the receptor to create a mating differentiation and, therefore, the rate of mating.
B) The dependent variable of the experiment is the number of cells that differentiated. The number of cells that differentiated depended on the type of yeast exposed to the pheromones. In this experiment, the experimenters controlled the sample size and the application of the pheromones. Each treated group consisted of 1,000 cells and was given the same concentration of pheromones. The same three pheromones, Wild Type-created, Variant 1-created, and Variant 2-created, were also used for each yeast cell type. These controls allowed the experimenters to observe how yeast cell type affects the rate of mating without the influence of other factors that could have skewed the results.
C) Chi-square analysis can be used to determine if the number of Variant 1-Type cells with mating differentiations significantly differed from those of the Wild Type. The Variant 1-Type yeast cells are being compared to the Wild Type cells, so the Variant-1 Type cells are the observed data and the Wild Type cells represent the expected data. In order to find the chi-square value, the square of the difference between the observed and expected values divided by the expected value must be calculated for each category. For the Wild Type-created pheromone type, there were 450 differentiated cells in the Wild Type cells and 203 in the Variant 1-Type cells. By using these values in the formula, a value of approximately 135.58 results. There were 606 differentiated Wild Type cells and 411 differentiated Variant 1-Type cells in the groups exposed to the Variant 1-created pheromone, showing a value of about 62.75. The value for the Variant 2-created pheromone category can be calculated as 16.82, with 50 differentiated Wild Type yeast cells and 21 differentiated Variant 1-Type cells. Then, these values are added to find the final chi-square value, 215.15, which can be compared to a critical chi-square value to determine the significance of the difference. The critical value with a 95% confidence for three categories is 5.99. The calculated chi-square value is far greater than the critical value, showing a significant variation between the number of cells with mating projections in the Wild Type and Variant 1-Type yeast cells. This also rejects the null hypothesis that there is not an important variation in the values, supporting the alternative hypothesis that a factor is affecting the rates of mating in Variant 1-Type yeast cells.
D) The significant variation between the data values could have resulted from a mutation in the Variant 1-Type cells. This mutation likely changed the receptor site of the cell by affecting its shape. Without the proper shape of specialized receptor sites, the pheromones are inhibited from binding to the receptor. When signals bind to receptors, the signal is received and a sequence of changes occurs throughout the transduction pathway in order to produce a response. Since pheromones cannot bind to the receptor sites to produce a response as easily, the overall cellular response of differentiation cannot be produced as often. The mutation resulted in the inability for signals in the pheromones to be received and communicate the correct response. Therefore, the mutation in the receptor site Variant 1-Type cells explains the significant variation in the values between the Variant 1-Type cells and the Wild Type cells. The data in the experiment shows that different pheromones resulted in differing amounts of cell differentiation in the Variant 1-Type yeast cells. This further suggests that the mutation affected the receptor site, not the pheromones, as the pheromones could still bind in some cells and the differences in the pheromones were not lost. The data indicates that the receptor sites of Variant 1-Type cells were changed by a mutation, creating a significant difference between the number of differentiated cells in the Wild Type and Variant 1-Type yeast cells.
Ligands are called signaling molecules because they bind to receptors and carries information.
A) The signaling molecule in the model is the pheromone. It binds to the receptor to generate a cellular response in the yeast system. It inhibits the growth of yeast cells and secretes shmoo.
Receptors are important in cell differentiation as they bind with receptors and create shmoo via the transduction pathway.
In the absence of a receptor, the transduction pathway will not occur and shmoo will not be produced.
Signaling molecules having specific shapes and sizes can only bind to receptors.
B) In the above experiment, a dependent variable is the number of cells differentiated. The number of cells differentiated depends on its exposure to pheromones.
The application of pheromones and the sample size of cells were in control by the experimenters. The yeast cell types were exposed to Variant 1-created, Variant 2-created and Wild Type-created pheromones equally.
This helped the researchers in determining the rate of mating without any influence of other factors.
C) Chi-square can be used to evaluate the number of variant 1 type cells with that from wild type.
The observed data includes variant type 1 cells and the expected data includes the wild type cells.
[tex]\rm Chi - square = \dfrac{( Observed - Expected \:values)^{2}}{ \:Expected \:values}[/tex]
The required value with a 95% certainty for the three types is 5.99. The calculated chi-square value is greater than that of critical values.
This shows the difference in the mating of variant and wild type varieties. It also repudiates the null hypothesis.
D) The mutation likely occurred on the receptor site of the variant type that changed the shape of the site binding.
The change in the receptor site will inhibit the pheromones from binding. This will affect the cell differentiation and transduction pathway.
Therefore, mutation on the receptor site indicated the variant and the wild type have different cell differentiation and mating rates. The mutation is responsible for the varied data and not pheromones.
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what is gene expression
Answer:
Gene expression is the process the cell uses to produce the molecule it needs by reading the genetic code written in the DNA.
Explanation:
To do this, the cell interprets the genetic code, and for each group of three letters it adds one of the 20 different amino acids that are the basic units needed to build proteins.
How does evaporative cooling work biology.
spermiogenesis begins with a ________ and ends with a ________.
Answer:
Answer:
germ cell and four gametes
Explanation:
Spermiogenesis begins with a germ cell and ends with four gametes.
Based on figure 2-2, how many electrons does an atom of Aluminum contain? (assume all atoms are neutral)
Answer:
see exp.
Explanation:
Aluminum has 13 electrons in a neutral atom. The 1S2, 2S2 and 2P6 shells are full, and so is the 3S2. The 3P1 shell has one electron in it (in Al's neutral atom) and that gives aluminum one lone valence electron and two "semi-valence" electrons which can be loaned out. :D
pls join clan NaGa
While you’re waiting for the lactase solution to cool, read the directions on the test strips. The test strips in the Edmentum lab kit will react to glucose within a few seconds. If you use different strips, the reaction time may vary. Now follow these steps to test the solutions. Record your data in the answer space.
Milk and water solution: Fill the first test tube one fourth full of milk. Fill the small graduated cylinder with water and gently add it to the milk in the test tube until the test tube is half full. Use the stirrer to thoroughly mix the solution. Then insert the test strip for 10 to 20 seconds. Look at the test strip, and record whether it changed color. Wash the stirrer.
Milk and lactase enzyme solution: Fill the second test tube one fourth with milk and one fourth with the lactase solution. Use the stirrer to thoroughly mix the solution. Insert the test strip for 10 to 20 seconds, and record whether it changed color. Wash the stirrer.
Milk an
Which of the following is an advantage of sexual reproduction?
A. Produce identical copies of an organism.
B. Produce natural variation to adapt to environmental change.
C. Quickly populate a new habitat.
D. Reproduce very quickly.
Answer:
B
Explanation:
Produce natural variation to adapt to environmental change is an advantage of sexual reproduction. Correct Option is 2.
The advantage of sexual reproduction is that it produces genetic variation within a population. During sexual reproduction, offspring inherit a combination of genes from both parents, leading to unique genetic combinations. This genetic diversity results in natural variation among individuals within a population.
This variation is advantageous because it allows a population to better adapt to changing environmental conditions. In a changing environment, some individuals may possess traits that are better suited to the new conditions, making them more likely to survive and reproduce. Through natural selection, these advantageous traits can become more prevalent in the population over time, enhancing the population's overall ability to adapt and thrive.
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What are the two phenomenon that results in drastic weather change in the world?
Answers:
Hurricane or Tornado
Explanation:
Whenever a Hurricane or Tornado occurs it can cause drastic change in weather causing floods, wind, and damage to houses, buildings, and even people and animals.
How long does the eclipse of an earth-like planet take?
A teacher showed this animal to students on a field trip. Which tool will the students use to record the animal's behavior? A. Graduated cylinder B. Notebook C. Tape measure D. Hand lens
Answer:
note book
Explanation:
nothing else would work
Una parte de la radiación
infrarroja es absorbida
por los gases de efecto
invernadero. ¿En qué capa
de la atmósfera tiene lugar
este fenómeno?
La troposfera es la capa de la atmósfera donde una parte de la radiación infrarroja es absorbida por los gases de efecto invernadero.
En la capa de la troposfera, que es la capa más baja de la atmósfera, hay vapor de agua, dióxido de carbono, metano y algunos otros gases que son responsables de la absorción de la radiación infrarroja. Parte de la radiación infrarroja se escapa al espacio, pero una parte es detenida y absorbida por los gases de efecto invernadero presentes en la atmósfera.
Esta absorción de radiación infrarroja por los gases de efecto invernadero contribuye a un aumento de la temperatura de la superficie de la tierra y de la atmósfera. Entonces, podemos concluir que la troposfera es la capa de la atmósfera donde una parte de la radiación infrarroja es absorbida por los gases de efecto invernadero.
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Which two structures engage to create a muscle contraction?.
Muscle contraction occurs when actin connects with myosin.
Muscle contraction occurs thanks to an action potential, which transmits a stimulus to the interior of the muscles.
For fiber tension to occur, the myosin head must bind with actin, slip, release, and then connect with another actin.This electrical signal allows the actin and myosin proteins to move, thereby generating contraction of the muscles.Therefore, we can conclude that myosin is a protein found in muscle fibers in order to bind to actin protein and cause muscle contraction.
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what is this i tried doing it but i just dont get it at all 7.5 + n + 9.63.
Answer:
+17.13
Explanation:
brainliest please
Solving?
Answer:
n+17.13
Explanation:
Add the numbers
7.5+n+9.63
17.13+n
Rearrange terms
17.13+n
n+17.13
Solution
n+17.13