Answer:
The Answer is B.
Explanation:
The process by which a zygote becomes a blastocyst is cleavage division.Thus, option "B" is correct.
What is fertilization and example?A method of reproduction that entagled the union of male and female germ cell (each with a single, haploid set of chromosomes) to create a diploid cell called zygote.
Thus, option "B" is correct, A group of blastomeres produced by cleavage divisions is known as pre-embryo.
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All of the following would be visible on a karyotype except
aneuploidy.
polysomy.
trisomy.
homologous chromosomes.
Answer: I think aneuploidy id wait for someone else to confirm this
Explanation:
All of the following would be visible on a karyotype except is aneuploidy
How to identify karyotype?The normal karyotype of the species is described as 46, may XX or XY, in which the first number of chromosomes is followed by a comma indicating the chromosomes. Thus, 46, XX corresponds to the karyotype of a woman; and 46, XY represents a male karyotype.
Aneuploidy is the presence of an abnormal number of chromosomes in a cell so this is the exception.
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One of the procedures for cloning foreign DNA segments takes advantage of restriction endonucleases such as HindIII that produce complementary single strand ends (sticky ends). These enzymes produce identical sticky ends on cleaved foreign DNA and on the vector DNA into which you insert your DNA. Assume you have inserted your favorite gene into the HindIII site of the Bluescript cloning vector with ligase, you amplified it in E.coli, and isolated a large quantity of the plasmid with the insert. How do you excise your favorite gene from the plasmid?
Answer:
Restriction endonucleases are utilized to cut the deoxyribonucleic acid at a specific Restriction site. HindIII is an example of Restriction endonuclease. The Restriction site of HindIII is as per the following:
5'A*AGCTT3'
3'TTCGA*A5' (* represents restriction site)
The Restriction endonuclease cuts the particular gene from the foreign DNA and produces numerous fragments with the sticky ends. The part of the outside DNA which contains the gene of intrigue or a particular gene is ligated with the plasmid.
On the off chance that foreign DNA and plasmid DNA are separated by the same Restriction endonuclease, they will create sticky ends integral to one another. In this way, the segment of foreign DNA can be ligated to the plasmid DNA by the assistance of DNA ligase at the integral sticky end to shape a recombinant DNA.
Additionally, by the assistance of the same Restriction endonuclease for example HindIII for this situation, the recombinant plasmid can be treated with a similar Restriction endonuclease to extract the ideal gene or gene of intrigue.
HELP ASAP DUE NOW PLSSS HELP 10 points and brainlist Which arrows show matter moving from a producer to an omnivore? Select all that apply.
Answer:
my best guess is number answer 2 and answer 4
because crows love to eat desert woodrats and for the last one i watched it in national geo, its like a cycle grass hopper died by a mouse dies by a rattle snake.
Which is the pair of the enzyme activities most significantly affected by glucagon- and insulin-dependent phosphorylation and dephosphorylation of 6-phosphofructo-2-kinase/fructose-2,6-bisphosphatase (PFKFB)?
a. hexokinas ::::: glucose-6-phosphatase
b. phosphofructokinase ::::: Fructose-1,6-bisphosphatase
c. pyruvate kinase :::::: pyruvate carboxylase
d. glycogen synthase ::::: glycogen phosphorylase
e. None of these
Answer:
b. Phosphofructokinase; Fructose-1,6-bisphosphatase
Explanation:
Glucagon is a hormone which functions to initiate gluconeogenesis whwreas insulin functions to initiate glycolysis.
Fructose-2,6-bisphosphate (F26BP) is an allosteric regulator of the glycolytic and gluconeogenic enzymes phosphofructokinase-1 (PFK-1) and fructose-1,6-bisphosphatase (FBPase-1).
It stimulates glycolysis and inhibits gluconeogenesis by binding to its allosteric site on phosphofructokinase-1 reducing its affinity for the allosteric inhibitors ATP and citrate .
It is formed by phosphorylation of fructose-6-phosphate catalyzed by phosphofructokinase-2 and is broken down by fructose-2,6-bisphosphatase which are two separate enzymatic activities of a single, bifunctional protein (PFK-2/FBPase-2).
Phosphorylation of PFK-2/FBPase-2 due to glucagon release enhances its FBPase-2 activity thereby stimulating gluconeogenesis and inhibiting glycolysis.
Dephosphsorylation of PFK-2/FBPase-2 due to insulin release enhances its PFK-2 activity, thereby stimulating glycolysis and inhibiting gluconeogenesis.
Describe the sizes of the sun, moon and Earth models given to the participants. Before embarking on the actual activity, what did the participants learn as a result of the size of the models?
Answer:
(A). (1).SUN AND EARTH= the distance between the sun and the earth is about 0.5 miles.
(2). MOON AND SUN: the distance between the moon and the sun is about 400m.
(B). Kindly check Explanation.
Explanation:
PART A:
(GIVEN TO THE PARTICIPANTS) DESCRIPTION OF THE DISTANCES SIZES OF;
(1).SUN AND EARTH= the distance between the sun and the earth is about 0.5 miles.
(2). MOON AND SUN: the distance between the moon and the sun is about 400m.
PART B = what did the participants learn as a result of the size of the models?
The participants learnt if they so wish to make the size of the moon to be equal to that of the sun, the people or the participants have to move or go back because initially to them(participants) they thought that if are able to move the moon to the sun, it will be equal.
help asap giving branlist and 50 points!!!!!!!!!
Answer:
[tex]\boxed{Option \ 2}[/tex]
Explanation:
=> In the first graph, we come to know that winters are colder and summers are hotter.
=> In the second graph, winters have more precipitation than summer.
=> On the whole, Winters have lower temperature and more precipitation than summers. ← [Option 2]
=> Summers have high temperatures and less precipitation.
Answer:
[tex]\boxed{\sf Option \ 2}[/tex]
Explanation:
We can see in the first graph that the winter months have a lower temperature compared to the summer months. The summer months have higher temperatures.
In the second graph, the winter months have much higher average precipitation compared to the summer months. The summer months have lower average precipitation.
Winters have lower temperatures and more precipitation than summers.
Which of the following is not a benefit of Genetic Engineering? A. Enhancing the desirable traits of organisms B. Helping to reduce world hunger C. Enhancing the effectiveness of the new DNA by spreading it to other plants D. Helping to treat disease
Answer:
b
Explanation:
because all the rest are about genetics
The one that is not a benefit of genetic engineering is helping to reduce world hunger. The correct option is B.
What is genetic engineering?Genetic engineering is the method of modifying an individual's DNA through the use of research lab innovations.
This could be as simple as changing a single base pair, erasing a region of DNA, or inserting a new DNA segment.
Scientists can use genetic modification to transfer preferred genes from one plant or animal to another.
Genes may also be transferred from one animal to another. GMOs are another term for genetically engineered organisms.
Techniques for genetic engineering have produced the creation of medically valuable things such as human insulin, human growth hormone, and hepatitis B vaccine.
It is also involved in the development of genetically modified organisms such as disease-resistant plants.
Thus, the correct option is B.
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Which best explains the evolution of gymnosperm plants? They evolved at the same time as monocots and dicots. They evolved after the seeded flowering plants. They evolved after the seedless vascular plants. They evolved before the seedless nonvascular plants.
Answer:
They evolved after the seedless vascular plants.
Explanation:
The gymnosperms were the first seed plants to have evolved some about 382.7 million to 358.9 million years ago in the Devonian period. They were the ancestor of the seed plants like the heterosporous, earlier gymnosperms were pteridosperms or seed ferns. The vascular plant's particular ferns consist of tissues that don't make flowers or seeds but have unicellular pores.Answer:
the answer is C- They evolved after the seedless vascular plants.
Explanation:
How does mitosis introduce genetic variation into offspring
Answer:
it is when a domminant allele mixes with a recesive allele creating a new pattern or genetic variation such as spot colors of a dog
How can protists exhibit both animal-like and plant-like characteristics?
A heterotrophic and chlorophyll protist
Food is ingested by protists in three forms. We produce, eat and digest their own organic molecules. Meat ingestion or english bacteria ingests protists. The cell wall and cell membrane are stretched to create an alimentary vacuolum around the foodstuff. Enzymes extract the food inside the food vacuole. At the other side, absorbent protists consume food molecules through their cell membrane by diffusion. In decomposition, absorbent protists play a crucial role. They are assumed to be essential decomponents. Light energy is used to make your own food by big farmers including photosynthetic protists.
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Brainliest would be great!
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With all care,
07x12!
Glycogen Synthase Kinase:_______.
A. Will only phosphorylate glycogen synthase (GS) after GS has been phosphorylated by casein kinase.
B. Phosphorylates glycogen phosphorylase
C. Phosphorlyates Casein Kinase.
D. Is activated by the insulin signaling pathway.
Answer:
A. Will only phosphorylate glycogen synthase (GS) after GS has been phosphorylated by casein kinase.
Explanation:
Glycogen synthase kinase (GSK) is a type of enzyme capable of phosphorylating serine and threonine residues on the glycogen synthase (GS). The insulin hormone activates the synthesis of glycogen by dephosphorylating GS, but it is a mechanism that depends on the previous phosphorylation of GS by Casein Kinase (CK). GSK is also known to phosphorylate the insulin receptor (IRS), and thus controlling glucose metabolism.
1. Hypothesize in what type of environments it would be advantageous for a protist to be autotrophic, heterotrophic, or mixotrophic.
Answer:
See below
Explanation:
Protists fall into the category of eukaryotes. They are diverse in nature and typically are found to be single celled organisms containing very organized structures.They are classified based on what is their source of nutrition.
Autotrophic - capable of photosynthesis to produce their own food, so an environment where sunlight, CO2 and water is available.
Heterotrophic - feed on other organisms , an environment where prey is plenty and very limited or no access to sunlight.
Mixotrophic - a combination of the above two, an environment with all of the aforementioned factors present.
Which of the following statements best describes communication style versatility?
The early bird gets the worm.
You can t judge a book by its cover.
I cannot control you, but I can control myself
alf a loa better than none
Answer:
I cannot control you but I can control myself.
Explanation:
Communication style versality is the ability to identify and understand way or taste of communication and adapt to it in order to make people open and receptive to it there by creating intimate relationships with people.
The features of communication style versality is I cannot control you but I can control myself in the sense that I will have to understand different styles and communication preference, then adapt to it by controlling myself, in me controlling myself people will tend to be open and receptive to it and intimate relationships will be built. I can't control other people but I will have to control or work on myself.
Mike is studying the phylogenetic tree for big cats. A big cat phylogenetic tree is shown. A bracket goes to clouded leopard and to another bracket. That bracket goes to 2 additional brackets. One bracket goes to tiger and snow leopard. The other bracket goes to janguar and another bracket. The other bracket has lion and leopard. He made several statements about the tree. Which statement is not true? The leopard is more closely related to the jaguar than to the snow leopard. The leopard is more closely related to the tiger than to the clouded leopard. The leopard is more closely related to the snow leopard than to the clouded leopard. The leopard is more closely related to the tiger than to the snow leopard.
Answer:
The leopard is more closely related to the snow leopard than to the clouded leopard.
Explanation:
Hope it helps.
Answer:
C
Explanation:
The Answer is C my fellow Brainly users
Bryozoans have several different forms. The branching form is often mistaken for a type of coral. Which is true and can be used to differentiate bryozoans from corals?
Answer:
Bryozoans belong to phylum bryozoa
while corals belong to phylum cnidaria.
Bryozoans are more advanced than corals.
bryozoans have an anus while corals do not have an anus
bryozoans tentacles don’t sting like corals.
Which feature separates humans from other primates? smaller brain size bipedalism use of tools complex social skills
Answer:
Complex social skills
Explanation:
Answer:
Complex social skills.
Explanation:
Humans actually have larger brains than other primates, so smaller brain size is incorrect.
Other primates, such as chimpanzees, orangutans, and gorillas, all have the ability to use tools, so that is incorrect.
Some primates, such as chimps and baboons, do stand on their hind legs at times, so bipedalism does not separate humans from other primates.
So, complex social skills separate humans from primates.
Hope this helps!
Predict what will happen as more time passes? (To Earth)
Where near your home do you think the highest level of photosynthesis is happening? Think about
where in your area photosynthetic organisms are growing.
Answer:
Depends where the individual lives.
If the person lives in a rural area near a forest, photosynthesis would most likely occur in the forest.
If the person lives in an urban area in a city, a park or garden would have the highest level of photosynthesis occurring.
Answer:
There is a wonderful big park near my house with a lot of trees and plants, and I believe there is where the most photosynthesis occurs.
Explanation:
One characteristic of fungi is that they have a cell wall made of chitin. What are two other characteristics of fungi? Produce spores and heterotrophic Contain membrane-bound nucleus and photosynthetic Heterotrophic and produce seeds Absorb food and photosynthetic
Answer:
Produce spores and are heterotrophic
Explanation:
All animals and fungi are heterotrophic and fungi produce spores which is how they reproduce.
Answer:
Produce spores and heterotrophic
The karyotype reveals that the aliens have a total of 200 chromosomes in each somatic cell. Assuming this alien reproduces sexually, how many chromosomes would its gametes have?
Answer:
100
Explanation:
in sexually reproduction gametes are formed by meiosis where number of chromosome is reduced by half(haploid gametes)
Which of the following processes requires energy from the cell?
A.
Osmosis
B.
Active transport
C.
Diffusion
D.
Passive transport
The following processes requires energy from the cell - B. active transport
Active transport is used by cells to get required molecules such as glucose and amino acids. It requires energy in form of adenosine triphosphate (ATP) is known as primary active transport.
Active Transport works against the concentration gradient and therefore needs energy.Passive transport such as Diffusion and Osmosis, from areas of high concentration to areas of low concentration without the need for energyThus, the following processes requires energy from the cell - B. active transport.
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Which feature separates humans from other primates? smaller brain size bipedalism use of tools complex social skills
Answer:
Bipedalism.
Explanation:
Bipedalism is a human's ability in which they can use their two hind legs to run or walk without the help of their front limbs, unlike other primates.
Answer:
Bipedalism
Explanation:
i took the test and failed
2 Diseases of a cassava
The overall outcome of the activities carried out under the CaCESA initiative
will improve cassava production and productivity. The effectiveness of
program is based on five axes:
- provision of healthy cassava cuttings to vulnerable families;
- the organization of extensive awareness and communication campaigns;
- national and regional coordination of actors in the production chain and
cassava post-production;
- the improvement of cultivation, processing and conservation practices
cassava;
- control and management of the spread of diseases, whether they are transmitted
insects and / or infected cassava cuttings.
To achieve these results, it will be necessary to build capacity at different
levels. The objectives will be achieved through the implementation of appropriate activities
Explanation:
Answer:
Explanation:
Oidium manihotis
Root Rot Pathogens
Classify the following organisms into their respective kingdoms (i) Yeast (ii) Penicillium (iii) Rhizobium (iv) Mushroom (v) Amoeba (vi) fish
Answer:
(i) Yeast- Kingdom Fungi
(ii) Penicillium- Kingdom Fungi
(iii) Rhizobium- Kingdom Eubacteria
(iv) Mushroom- Kingdom Fungi
(v) Amoeba- Kingdom Protista
(vi) fish- Kingdom Animalia
Explanation:
Living organisms were classified into a large group called KINGDOM, which represent the largest and most generic grouping consisting of organisms that share a few similarities. The kingdoms that living organisms were classified into are as follows: Plantae, Animalia, Fungi, Protista, Eubacteria, Archaebacteria, etc.
Based on the question, the mentioned organisms are classified into the following kingdoms:
(I) Yeast: Yeast is a unicellular microbe classified under the Kingdom Fungi due to possession of chitin in its cell wall and other similar features with members of kingdom Fungi.
ii) Penicillium- Penicillium is a genus classified under the Kingdom Fungi.
(iii) Rhizobium- Rhizobium is a genus of prokaryotic organism (lack membrane-bound nucleus and organnelles) classified under the Kingdom Eubacteria.
(iv) Mushroom- Mushrooms are saprophytic (heterotrophic) species of organisms classified under the Kingdom Fungi.
(v) Amoeba- Amoeba is a genus of unicellular eukaryotes classified under the Kingdom Protista.
(vi) fish- Fishes are organisms classified under the Kingdom Animalia
About how much energy is transferred from a primary consumer to a secondary consumer?
90% of the original energy from the Sun
10% of the original energy from the Sun
10% of the primary consumer's energy
O 1% of the primary consumer's energy
wi
Answer:
90 percent of the original energy from the sun
Answer:
10% of the primary consumer's energy
Explanation:
Somatic cells in elephants have 56 chromosomes. How many chromosomes would be carried by the gametes produced by elephants
Answer:
As somatic cells are diploid (2n) 2n=56
gametes receive only haploid set so, n=28
The number of chromosomes carried by the gametes produced by elephants would be : 28
Given that The elephants posses 56 chromosomes in its somatic cells and somatic cells are known as diploid cells ( i.e. 2n ) while the gametes will only carry a haploid set of chromosome ( i.e. n ).
Determining the number of chromosomes contained in the gametes
since ; 2n ( chromosomes of a somatic cell ) = 56
∴ n ( chromosomes contained in a gamete ) = 56 / 2 = 28
Hence we can conclude that The number of chromosomes carried by the gametes produced by elephants would be : 28.
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Which of the following are roles carbohydrates play in biological systems? Group of answer choices the basis of most of the organic matter on our planet (cellulose, chitin, etc.)
Answer: Here are the completed options
fuels
components of DNA and RNA
signalling
forming barriers between organelles
digesting enzymes
breaking up lipids for digestion
providing structure
The correct options are
most of the organic matter on our planet (cellulose, chitin, etc.)
Carbohydrates act as Body fuels
providing structure
Explanation:
Carbohydrates are biological molecules and are constituents of starches, sugars, glucose e.t.c and are made up of carbon, hydrogen and oxygen which are the major constituents elements of the most organic matter on the planet.
Carbohydrates is a body fuel that act as the major energy source in the body. It gives and provide energy in the body needed for biological activities.
In plants, carbohydrates make up part of the cellulose, giving plants strength and structure. carbohydrates is to form a structure called the glycocalyx.
Observations of organisms include descriptions of why the organism has certain traits. descriptions of the number, size, and arrangement of parts of the organism. conclusions made about the characteristics of an organism. conclusions made based on an organism’s DNA.
Answer:
The correct statement is descriptions of the size, number and arrangement of parts of the organism.
Explanation:
The act of obtaining information from a suitable primary source is termed as an observation. It is the very initial step in the array of steps that comprises a scientific method and it incorporates the recording of data regarding an observed natural process.
Observation permits the researchers to inquire regarding the apposite scientific questions, to produce an applicable hypothesis, and to set up significant experiments, which will provide suitable answers to the queries asked earlier.
Answer:
B. descriptions of the number, size, and arrangement of parts of the organism.
Explanation:
took the test, hope this helps.
characters of a fertile soil
Answer:
There are so many characteristics of fertile soil like: It is rich in nutrients necessary for plant nourish ment. It includes nitrogen , phosphorus and potassium. It consists adequate minerals such as boron, chlorine, cobalt, iron, magnese, sulphur etc.Explanation:
Hope it helps you.The genomic DNA of an organism has a base composition of 40% C-G base pairs and 60% A-T base pairs. Assuming a random sequence of bases, what is the expected frequency of the Tsp E1 restriction sites 5'-CCGG-3'?
Answer:
The correct answer is (0.2)^4.
Explanation:
Based on the given information, the given genomic DNA is A-T rich, and the base pair composition is C-G = 40 % and A-T = 60%. Therefore, as per Chargaff's rule, Each C and G will possess 20%, and each A and T will possess 30%.
Therefore, the probability of each base will be,
C = 20/100 = 0.2
G = 20/100 = 0.2
A = 30/100 = 0.3
T = 30/100 = 0.3
Now the expected frequency of the Tsp E1 restriction sites 5'-CCGG-3' will be 0.2 * 0.2 * 0.2 * 0.2 = (0.2)^4.