Organisms can adapt
A.to other organisms
B. To their physical surroundings
C. To selective breeding
D. Both a and b
Answer:
ta and b both
hope it will help you
Explanation:
b. to their physical surroundings
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Describe the characteristics of the keratinocyte membrane that restrict water movement across the membrane. If isolated keratinocytes are placed in a hypotonic solution for 24 hours, explain why there are likely to be far fewer cells at the end of this time than there were at the start.
Answer:
It has four layers and attach with each other with desmosome.
Explanation:
The keratinocytes are arranged in four different layers which are the stratum basale, stratum spinosum, stratum granulosum, and stratum corneum. These cells are attached with desmosome which is a membrane protein. The presence of these four layers made a barrier against damage done by heat, UV radiation, water loss, pathogens etc. There are fewer cells at the end of the time as compared to the start because may be of the water pressure on the membrane or these cells for a long time.
The digestive system of most birds includes a structure called the gizzard. The
gizzard crushes and grinds food, often with the help of small stones that the bird
has swallowed.
Which of the following structures in the human digestive system has a function most
similar to that of a bird's gizzard?
F esophagus
G large intestine
H small intestine
J teeth
Answer:
the small intestine
Explanation:
the small intestine breaks down food and export it to the large into the large intestine.
the large intestine converts the food into turd aka poop.
A test tube is filled with water and a green plant called Elodea. When the plant is exposed to sunlight, it begins to produce bubbles. What is the most likely conclusion for the identity of the gas in the bubbles? Explain.
Answer:
The correct answer is - oxygen.
Explanation:
The experiment of a test tube and elodea plant in which a plant of elodea is placed in a test tube filled with water and exposed to the sunlight and one is placed in the dark. The plant that is exposed to the sunlight starts forming a gas that can be seen as bubbles in the test tube.
This experiment shows the process of photosynthesis that converts sunlight into glucose with the help of carbon dioxide and water and produces oxygen as a byproduct that releases from the plants.
Thus, the gas that forms in the test tube in the form of bubble is oxygen which is released in the process of photosynthesis.
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Answer:
The correct answers are:
1. A. eukaryote 2. B prokaryote 3. A. animals 4. B bacteria
Explanation:
The cell illustrated in A is a eukaryotic cell as it has a membrane-bound nucleus and organelles. The nucleus contains chromatin in the denser area in the center of the nucleus called the nucleolus. An example of this type of cell is animals or plants.
The other cell is a prokaryotic cell as it has no cell organelle or true nucleus. It has a denser area in cell plasm called nucleoid that has chromosome present. The example of these cells are bacteria or archeae.
QUESTION 12
neuronal pool that might rely on fatigue to end its impulse propagation is a:
A. converging circuit
B. diverging circuit
C. oscillating circuit
D. parallel after-discharge circuit
E. circuit breaker
At the end of Meiosis II, each daughter cell has 23 single-stranded chromosomes.
True
OR
False
Answer:
y the end of meiosis, the resulting reproductive cells, or gametes, each have 23 genetically unique chromosomes. ... Each daughter cell is haploid, because it has half the number of chromosomes as the original parent cell.
Explanation:
30 chromosomes
Each daughter cell will have 30 chromosomes. At the end of meiosis II, each cell (i.e., gamete) would have half the original number of chromosomes, that is, 15 chromosomes. 2.
From a diploid germ cell 2n =24 double stranded chromosomes, meiosis will produce four gamete haploid cells n =12 single stranded chromosomes.
Meiosis produces four with half because double stranded chromosomes represents DNA replication. Twice as much DNA as a diploid cell needs.
From a diploid germ cell 2n =24 double stranded chromosomes, mitosis will produce two diploid daughter cells 2n =24 single stranded chromosomes for the same reason.
DNA replication doubles the genome before cell division. Sister chromatids of double stranded chromosomes represent the replicated DNA. During cell division, sister chromatids of double stranded chromosomes are separated into different cells.
Meiosis has two stages, the first separating homologous pairs of double stranded chromosomes, and then the second stage separating sister chromatids into four haploid (n) gametes.
5.) What is the process by which the egg is released called?
6.) When does ovulation occur?
Answer:
5.) What is the process by which the egg is released called?
A.) Ovulation
6.) When does ovulation occur?
A.) Ovulation typically occurs about 14 days before the start of the next menstrual period.
-TheUnknownScientist
Will give brainliest
Below are two sequences of a segment of DNA.
Normal sequence CAT TAC ACC
Mutated sequence CTT TAC ACC
Which type of mutation has occurred?
Choose 1 answer:
A deletion is the removal of a nucleotide from a DNA sequence.
(Choice B)
B
Nonsense mutation
(Choice C)
C
Insertion mutation
(Choice D)
D
Substitution mutation
Answer:
D
Explanation:
CAT= CTT
A was substituted by T
It didn't stop, can't be B
It didn't add more letters to the sequence, can't be C
is the earth flat or not
Answer:
its not flat
Explanation:
Answer:
Hello There!!
Explanation:
The Earth is not flat but it is round.
hope this helps,have a great day!!
~Pinky~
Rabbits in most areas have a brown coat. The arctic hare has a bright white coat in
winter. How does the arctic hare's white color help it survive?
Answer:
Indeed, rabbits, in most of the planet, tend to have their fur colored brown, copper or other similar shades. Now, in the Arctic regions, these animals do not have this color, but a white one, similar to the environment that surrounds them. This is so because these animals, for evolutionary reasons, have adapted their fur to the snowy environment that surrounds them in the Arctic, in order to avoid being detected by potential predators, protecting themselves and trying to ensure their survival.
What is the importance of mitosis for unicellular organisms?
A) asexual reproduction
B) sexual reproduction
C) repair of damaged tissue
D) growth
Answer:
A) asexual reproduction
Explanation:
Mitosis plays an important part in the life cycle of most living things, though to varying extents. In unicellular organisms such as bacteria, mitosis is a type of asexual reproduction, making identical copies of a single cell. In multicellular organisms, mitosis produces more cells for growth and repair.
Answer:
hello again
Explanation:
motosis play an important
Please me help fast!!
Answer: you can just press sumbit for plato
Explanation:
s
Predation and parasitism are similar because in both predation and parasitism -
both organisms are harmed in some way.
one organism is harmed while the other benefits.
one organism is neither harmed nor benefits.
both organisms receive a benefit.
Answer: on organism is harmed while to the other benefits
Explanation:
Predation and parasitism are similar because, in both predation and parasitism, one organism is harmed while the other benefits. Therefore, option B is correct.
What are predation and parasitism?The act of eating all or a portion of the body of another organism, the prey, is known as predation. Herbivory is the term for predatory behavior in which the prey is a plant.
When one creature, the parasite, feeds on another, the host, hurting or even killing the host, the connection is said to be parasitism. The parasite stays within or on the host's body. Tapeworms, fleas, and barnacles are a few examples of parasites.
Thus, predation and parasitism are similar because, in both predation and parasitism, one organism is harmed while the other benefits. Therefore, option B is correct.
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Cell cycle is divided into two basic phases. What are they??
take 50 points
Fruiting Bodies of Fungi The recognizable forms of fungal reproductive bodies, or fruiting bodies, encountered in nature vary structurally. Common forms of fungal fruiting bodies include trumpet, club fungus, bracket, cup fungus, gill fungus, bolete, puff ball, and stinkhorn. Close observation of fruiting body anatomy can lend clues during identification of a species. Differing species of fungi can appear similar, and it is never advised to eat a wild mushroom because poisonous varieties can resemble edible ones.
1. Fungal fruiting bodies, or reproductive bodies, have different structural forms.
a. true
b. false
2. Many experiments require student-supplied items, such as _____.
Answer to Question 1:
True
Explanation
As indicated in the passage, fruiting bodies of fungi differ structurally. However, different species can appear to be similar and often misleading thus creating the risk of food poisoning.
Question 2:
Many experiments require student-supplied items, such as _____.
1. Chicken
2. Fresh flowers
3. Distilled water
4. All of the Above.
Answer to Question 2:
The correct answer is 3.
Explanation
Water comes in varying degrees of purity. When water that contains other substrates or impurities is used for an experiment, it distorts the results and creates errors. So the scientist's ability to determine the root of a problem becomes enervated.
Using distilled water impurities removes the possibility of such an error. Distilled water is also used for cleaning laboratory equipment.
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what type of organism is Eospirifer?
Answer:
Its a fossil that is in the section of Rhynchonellata
• 1. What is the difference between a gamete and a zygote?
• 2. Are gametes haploid or diploid? Are zygotes haploid or diploid?
• 3. What occurs at fertilization?
• 4. Define diploid and haploid.
• 5. What characterizes sexual reproduction?
• 6. What characterizes asexual reproduction?
• 7. What type of reproduction occurs in prokaryotes?
• 8. What is somatic tissue? What is germ-line tissue?
• 9. What do somatic cell lines and germ-cell producing cell lines have in common? What makes them different?
• 10. What happens in each stage of Meiosis I and Meiosis I1?
• 11. How does each stage of Meiosis compare to Mitosis?
• 12. Describe three ways that meiosis contributes to genetic diversity in a species?
Answer:
Explanation:
A gamete is a haploid reproductive cell that fuses with another gamete of opposite strain to form a diploid
zygote.
A zygote is a cell formed by the fusion of the male and female gamete before it undergoes division.
gametes are haploid while a zygote is diploid.
There is exchange of genetic material during sexual reproduction and fusion of gametes. Meiosis is also involved.
The opposite occurs in asexual reproduction and mitosis is involved instead.
Asexual reproduction is common in prokaryotes eg bacteria.
How do landfills impact groundwater? A. litter
B. Sharp metal and glass get in the soil.
C. Liquids leak into the ground.
D. Landfills do not impact groundwater.
Answer: c. Liquids leak into the ground
Explanation:
Where do most transform boundaries occur?
Answer:
QUESTION:
Where do most transform boundaries occur?
ANSWER:
Transform Plate Boundaries are locations where two plates slide past one another. The fracture zone that forms a transform plate boundary is known as a transform fault. Most transform faults are found in the ocean basin and connect offsets in the mid-ocean ridges.
Explanation:
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What is the function of the raphe in diatoms?
a.
defense
b.
photosynthesis
c.
capturing food
d.
locomotion
Answer:
Locomotion
Explanation:
I am not sure but it will be helpful. Thank You !!!
Starting with the European settlers, humans have introduced earthworms from Europe and Asia into North American forests. These introductions continue through the transport of soil that contains non-native earthworms, such as during construction and through the release of non-native earthworms used for fishing. The effects of non-native earthworms are especially large in forests that did not have any native earthworms. For example, forests of the Great Lakes region did not previously have earthworms until humans introduced them. When non-native earthworms are introduced, the thick layer of leaf litter disappears quickly, thereby altering biogeochemical cycles. You conduct an experiment to compare the nitrogen cycle in soils with and without non-native earthworms. Predict the results of your experiment.
Answer:
The soils with earthworms will show a faster rate of ammonification
Explanation:
Originally, the nitrogen released by organisms when they die (or excrete waste products) is organic nitrogen, e.i., amino acids and nitrogenous bases in DNA. Ammonification is the property that decomposer organisms have to mineralize organic nitrogen in order to produce inorganic nitrogen in the form of ammonium (NH4+). In agroecosystems, earthworm activity can increase the ammonification rate. For example, Lumbricus terrestris is an invasive earthworm (which is native to Europe) that has been shown to increase both ammonification and nitrification (nitrate production) rates in the soil of different ecosystems.
Which can be used to prevent initial entry of bacteria into the body?
Answer:
safeguard
Explanation:
safeguard sabon na skalam
Andrew did a mutant screen with his favorite bacterium that he used for marine phage hunting. He identified three base-substitution mutations on the bacterial recA gene:
Mutant A: A base substitution occurred, which changed T to C, and consequentially abolished transcription of the recA gene.
Mutant B: A base substitution, which changed codon 2 from ACC to ACT, which did not change encoded amino acid.
Mutant C: A base substitution occurred, which changed codon 3 from TCA to a stop codon TGA.
Based on the above info, which of the mutants would still have functional RecA protein?
a. Mutant B
b. Mutant A
c. Mutant C
Answer:
a. Mutant B
Explanation:
This question is describing a kind of mutation called SUBSTITUTION MUTATION, in which one or more nucleotide base is replaced by another in a sequence. In this case, the base-substitution mutations occcurs in the bacterial recA gene sequence.
However, base substitutions can be grouped into three based on the different results it yields namely: missense mutation, nonsense mutation, silent mutation.
- In Mutant A: A base substitution occurred, which changed T to C, and consequentially abolished transcription of the recA gene. This is an example of missense mutation because it changes the sequence.
- In Mutant B: A base substitution, which changed codon 2 from ACC to ACT, which did not change encoded amino acid. This is an example of silent mutation because it does not alter the sequence of the gene.
- In Mutant C: A base substitution occurred, which changed codon 3 from TCA to a stop codon TGA. This is an example of nonsense mutation because it leads to a STOP CODON.
According to the question, MUTANT B would still have functional RecA protein because it doesn't change the encoded amino acid sequence, which will still lead to the production of the RecA protein.
m/forms/d/e/1FAIpQLSf1siktOoZRkggNo-AzHPw62091WglmEyyJt30t8yy3LbzpVQ/viewform
15. When two similar unicellular organisms fuse and exchange nuclear materials,
it is known as
O A. Haploidy
B. Conjugation
O .
C. Meiosis
D. Nuclear growth
When the gene encoding a certain cytoskeleton protein is deleted, the resulting mutant cells round up and do not form their normal appendages. These mutants can be rescued when a gene encoding an N-terminal green fluorescent protein (GFP) fusion of the protein is expressed, but not when a gene encoding a C-terminal GFP fusion is expressed. Which fusion protein is appropriate to use in studying cellular localization and activity
Answer: N-terminal green fluorescent protein (GFP)
Explanation:
Protein synthesis takes place in ribosomes located in the cell cytoplasm, in which amino acids are transported by transfer RNA corresponding to each amino acid to the messenger RNA where they bind in the appropriate position to form new proteins. The messenger RNA has a sequence of nucleotides that are translated into protein, as each codon (set of three amino acids) codes for one amino acid.
Genes are the storage units of genetic information, so they are segments of DNA that contain the information on how the cells of the organism should function. Each gene codes for a protein, so if a gene is damaged or absent, the protein will not be obtained. In this case, mutant cells with a deleted gene, will round up and wont form their normal appendages. This mutation can be rescued or repaired with a gene that encodes an N-terminal green fluorescent protein (GFP) fusion of the protein. The N-terminal end is the amino-terminal end and it refers to the end of a protein that ends with an amino acid that has a free amino group. The C-terminus or carboxyl-terminus is the end of a protein that ends in a carboxyl group. The convention for writing peptides is to place the C-terminal end to the right and write the sequence from the N- to the C-terminal end. So, when an N-terminal green fluorescent protein (GFP) fusion of the protein is expressed, the gene will produce the protein with a GFP tag.
GFP is Green Fluorescent Protein, and it is a protein produced by the jellyfish Aequorea victoria that emits bioluminescence in the green region of the visible spectrum. When a gene is fused to another gene (at either the N- or C-terminus, although in this case it is the N-terminus), the entire messenger RNA is translated together as if it were a single fused protein. Thereby, since the protein will be produce with a GFP tag, it can be seen under the microscope and it will be apropiate to use in studying cellular localization and activity.
There are some considerations that this problem does not question, such as that there must be a start codon for protein synthesis and a stop codon, and this stop codon must not be in the middle of the gene or between the gene and the GFP tag.
A particular species of bacteria has a "a generation time" of 20 minutes. That is, every 20 minutes a single bacterium completes prokaryotic fission -- one bacterium is now two bacterial cells. This morning at 6:00 a.m. you made a sandwich and packed your lunchbox. What you were not aware of was that there were 10 bacteria cells of this species on the dollop of mayonnaise that you spread across your bread. At 12:00 p.m. (no on) You sit down and eat your entire sandwich! How many bacteria cells did you have for lunch? (Hint: use the formula 2axb x C = numbers of cells, where "a" - number of times per hour the bacteria reproduces, "5" the number of hours the bacteria had to reproduce, and "C" = the number of bacteria at the start).
Answer:
Number of bacteria cells = 2621440 cells
Explanation:
The generation time of a bacteria specie is the time it takes for a population of bacteria to double in number. The generation time for most common bacteria is quite short, about 20 minutes to 1 hour.
Bacteria reproduce asexually by binary fission, in which the parent cell divides into two to produce two identical daughter cells.
In the given question, the generation time of the bacterial species is 20 minutes.
The time that have passed from between the time the sandwich was packed at 6:00 am to lunch time by 12:00 pm is 6 hours.
Using the given formula in the hint: number of cells = 2^(ab) × C
Where a = number of times per hour the bacteria reproduces; b = number of hours; C = number of bacteria at the start
a = 60 minutes/ 20 minutes = 3; b = 6; C = 10
Number of cells = 2^(3 × 6) × 10
Number of cells = 2621440 cells
The technology time for the maximum not unusual place bacteria is pretty short, approximately 20 mins to one hour. The number of bacteria cells did have for lunch =2621440 cells.
Bacteria reproduce via way of means of binary fission, wherein the discern molecular divides into a time of the bacterial species is 20 mins.The time which has handed from among the time the sandwich became packed at 6:00 am to lunchtime via way of means of 12:00 pm is 6 hours.Using the given formulation withinside the hint: range of cells =2^ (ab)×C.Where a = range of instances in step with hour the reproduces; b = nu of hours; C = number of bacteria on the start.a = 60 mins/ 20 mins = 3; b = 6; C =10.The number of bacteria cells did have for lunch =2621440 cells.What is bacterial cel?Bacterial cells are exceptional from plant and animal cells. Bacteria are prokaryotes, because of this that they haven't any nucleus. A bacterial includes Capsule.
Thus it is clearly explained above that the number of bacteria cells did have for lunch =2621440 cells.
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what name is given to an enzyme which catalyses' the breakdown of proteins?
Answer:
Proteases
Explanation:
Pro - protein
eases - enzyme