Answer:
it is probably b.
Explanation:
hope this helps :)
For traits controlled by a single genetic locus, dominant phenotypes are expected to occur at a frequency of approximately 75%, in a randomly mating population.
true or false and why
Answer:
True.
Explanation:
The dominant phenotypes are expected to occur at a frequency of approximately 75%, in a population because of its dominant nature which allows them to appear in the physical structure of the organisms. In most of the offspring, the dominant allele shows their characteristics in the population whereas the recessive allele are hidden in that organisms and we can't see them in the physical appearance of the organisms.
3. If the solute concentration in the water is high (hypertonic solution), does water move
into or out of the paramecium?
Answer:
its flows out
Explanation:
i hope ill help
:)
How many alleles are typically present in a genotype?
Answer:A genotype is the combination of two alleles, one received from each parent. The expression of a genotype is called the phenotype and the specific combination of the two alleles (the genotype) influences the physical expression (the phenotype) of the physical trait that the alleles carry information for.
Explanation: so 2
17. When a person looks up into the sky
during new moon, what fraction of the
bright surface of the moon do we see?
Answer:
i dont really know why and how im here but im here
Explanation:
Where are diploid cells found ?
A. only in reproductive organs
B. only in twin siblings
C. through out the body
D. only in the lungs
Answer:
c
Explanation:
diploid cells are cell that have 2 genetic alleles, one from your mom and one from dad, so all your cells are diploid, except haploid cells in the rep system.
The skin cells of an organism are analyzed and cells identified in G1 and have an average of 12 picograms of DNA. In the same organism a cell from gonadal tissues is identified and that single cell contains 6 picograms of DNA. Given this information, which of the following is true about the single gonadal tissue cell with 6pg of DNA?
1. The cell is at the beginning of Prophase I.
2. The cell has already completed meiosis.
3. The cell is not dividing (is in interphase).
4. The cell contains unduplicated chromosomes.
Answer:
El Rasho Macuin
Explanation:
The statement which is true about the single gonadal tissue cell with 6 picograms of DNA is that the cell has already completed meiosis. So, the correct option is B.
What do you mean by DNA?It is a complex molecule that contains all the information necessary to build and maintain an organism. It stands for Deoxyribonucleic acid. It is considered the genetic material in most organisms.
The skin cells in the G1 phase have an average of 12 picograms of DNA. As the amount of DNA in the G1 phase is 2C, hence 2C = 12 pg of DNA.
Now in the same organism, a cell from gonadal tissue is identified and that single cell contains 6 pg of DNA, which means, the amount of DNA is 6 pg that is C. Therefore, after the completion of meiosis, the cell contains half the amount of DNA as compared to the parent cell in G1.
So, Amount of DNA = C = 6 pg.
Therefore, the statement which is true about the single gonadal tissue cell with 6 picograms of DNA is that the cell has already completed meiosis.
To learn more about the Cell cycle, refer to the link:
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In fruit flies, curved wings (w) are recessive to the dominant straight wings (W), and ebony body (b) is recessive to the dominant gray body (B). A fly that has curved wings and ebony body is crossed with a fly that has straight wings and gray body. Of the offspring there were approximately equal numbers of straight gray, straight ebony, curved gray, curved ebony. What are the genotypes of the parents and offspring
Answer:
The correct answer =
parents : WwBb x wwbb
offspring :
WwBb, Wwbb, wwBb, and wwbb
Explanation:
It is given that the straight wings (W) is dominant over curved wings (w) and the ebony body (b) is recessive to the dominant gray body (B) and there is a cross between curved wings and ebony body and straight wings and gray body is made. The phenotypic ratio is: 1:1:1:1 for the four different phenotypes straight gray, straight ebony, curved gray, curved ebony.
It is not given the genotype of parents, however, for the curved and ebony fly, it is clear that both alleles are recessive as recessive alleles only appear in absence of the dominant allele of the trait.
For another parent, the possible genotypes can be WWBB or WwBb, however in the case of WWBB the phenotypic ratio would not be equal therefore,
crossing WwBb x wwbb
This can be represented in the form of a punnett square as below:
wb wb wb wb
WB WwBb WwBb WwBb WwBb Straight wing, Gray body
Wb Wwbb Wwbb Wwbb Wwbb Straight wing, Ebony body
wB wwBb wwBb wwBb wwBb Curved wing, Gray body
wb wwbb wwbb wwbb wwbb Curved wing, Ebony Body
The phenotypic ratio is: 1:1:1:1 for the four different phenotypes
Why is the 3-dimensional shape of a macromolecule (ex: protein) important to know from the perspective of understanding
the molecular basis of human disease? 2 pts
A/ Mutations in genes that cause a change in amino acid primary sequence can cause a change in the protein's function in a
cell. For example, it can cause the protein to mislocalize; the protein is now recruited to a specific site in the cell, like the
nucleus, where it can cause cellular dysfunction.
B/ If the protein is an enzyme, the 3-dimensional shape of the active site can be altered and decrease the rate of a reaction that
is catalyzed by that enzyme.
C/ If the protein is an enzyme, the 3-dimensional shape of the active site can be altered and increase the rate of a reaction that
is catalyzed by that enzyme.
D/ Only options A and B may be true.
F/Antions A, B and C may be true.
Answer:
A, B and C may be true
Explanation:
The three-dimensional structure (3D) of a protein, also known as tertiary structure, is critical to its function. In general, the 3D protein structure consists of alpha-helix and beta-sheets (secondary structures) associated through disordered coiled-coil regions. Genetic mutations may cause modifications in the 3D protein structure by modifying amino acids that are added to the growing polypeptide chain during the process of translation. In consequence, the relationships among amino acids (i.e., the intermolecular attractive forces that hold them together) may also be altered by these mutations. In general, mutations that generate amino acid changes at the active site of the enzyme will have a deleterious effect, thereby decreasing or inactivating protein function. However, there are situations where a mutation may be beneficial and can eventually increase the rate of reaction of a protein.
Match the following prokaryotic classes with differing modes of metabolism with the appropriate descriptions - obligate anaerobes - Facultative anaerobes - Aerotolerant anaerobes - Obligate aerobes A. can shift their metabolism between anaerobic and aerobic modes B. Oxygen-sensitive, can only use anaerobic mode of nutrition C. unable to survive for extended periods in the absence of oxygen. They require oxygen for cellular respiration. D. cannot conduct cellular respiration but are not damaged by oxygen when it is present
Answer:
- Obligate anaerobes: B
- Facultative anaerobes: A
- Aerotolerant anaerobes: D
- Obligate aerobes: C
Explanation:
Obligate anaerobes are microorganisms that dead in normal atmospheric oxygen (O2) levels (around 21%). Some examples of obligate anaerobes are procaryotic bacteria such as Clostridium, Fusobacterium, Actinomyces and Peptostreptococcus. Facultative anaerobes are organisms that can grow without O2 and produce energy by fermentation, but they can produce more energy (ATP) by the aerobic respiration mechanism. Facultative anaerobes include both prokaryotic (e.g., Escherichia coli) and eukaryotic (e.g., Saccharomyces cerevisiae) organisms. Aerotolerant anaerobes are microorganisms that can live in O2 atmospheric conditions, but they use fermentation as cellular mechanism to produce energy (ATP). Some examples of aerotolerant anaerobes are lactobacilli and streptococci (bacteria) found in the oral microbiota. Finally, obligate aerobes are organisms that can grow only in normal O2 conditions and metabolize foods through cellular respiration, a process more energetically efficient than fermentation. An example of an obligate aerobe is the human species.
what is a cell membrane and its functions
Answer:
The cell membrane is a strip of membrane around the cell and it keeps bad things from entering the cell
Explanation:
Mary received a flu vaccine. Why didn't the vaccine make her sick?
Mary didn't get sick from the flu vaccine because it contains the smaller portion of the flu which isn't strong enough to infect you. Moreover, your white blood cells attack the portion of the flu while the memory cells take note of the flu, so when it comes next time, it won't affect your body because it knows how to fight the virus.
Hope this helps!
During cell reproduction, inherited information is carried by.
genes
bacteria
cytoplasm
nuclei
Answer:
During cell reproduction, inherited information is carried by:
A. nuclei
B. bacteria
C. cytoplasm
D. genes
The answer is D genes
Explanation:
When two parents join to form a new individual-
Offspring will NOT be identical to one parent.
It will have traits from both parents
Both A and C
Offspring will be identical to one parent.
Help please!!
Answer:
"B"
Explanation:
It will have both traits from both parents.
The combination of fibers and ground substance in fluid connective tissue is known as
A) ground substance
B) fat
C) cytosol
D) plasma
BIOLOGY CIRCULATORY SYSTEM!! WILL NAME BRAINLIEST IF YOU ANSWER CORRECTLY!!
Answer:
C, G, F, I, H, D, A, E, B
They are in order from top to bottom
Evergreen broadleaf trees, along with abundant rain, warm temperatures, and consistent day length, characterize the _ biome.
Answer:
Rainforest biome (tropical rainforest)
Explanation:
The characteristics of the tropical rainforest biome are:
Located in humid and warm regions of equatorial climate.Medium temperature among 18 and 30 ºCAnnual precipitation is around 1000 and 8000 millimeters. Precipitation levels are never inferior to 1500 mm. There is no dry season, or if there is, it is too short. There is no hydric deficit. Fanerophic species dominate in this biome. Evergreen tall trees might reach up to 50 meters in height. They develop dense foliation and branches, becoming a refuge and source of food for many species. Vegetation does not need to develop strategies for surviving the dry season. Instead, they need to develop strategies for competing for light and nutrients. We can find different species of plants such as epiphytes, lianas, stranglers, aerial-rooted pants, tabular-rooted plants, among other vegetation forms. Tropical and subtropical rainforests might differentiate from each other in latitude and seasonality. Tropical rainforests might be found in latitudes of the Capricorn and Cancer tropics. From cero to ten degrees latitude. They are in America, Africa, and Asia continents. It is the biome with the greatest biodiversity on the planet. Rainforests have an extraordinary richness of animals and plants.
Drag the tiles to the correct boxes to complete the pairs.
Identify the types of inheritance based on the expression of alleles in the organism.
Answer:
Presence of many alleles for the trait to be expressed --> Multiple
Presence of two alleles for the trait to be expressed --> recessive
Presence of a single allele for the trait to be expressed --> dominant
Answer:
Presence of two alleles for the trait to be expressed: recessive
Presence of many alleles for the trait to be expressed: multiple
Presence of a single allele for the trait to be expressed: dominant
Explanation:
Edmentum
what do you call a soft ,heated mass of crushed leaves and other substances that spread over the skin to treat swelling or pain ?
a.bandage
b.ointment
c.poultice
d.solution
Which argument is BEST supported by the
information given?
A. The earthworm and lizard evolved at the same
time because of the absence of legs and hair.
B. The lizard and humans share a common
ancestor because of the presence of a
backbone and legs.
C. The trout and humans share a common
ancestor because of the presence of a
backbone.
D. Earthworms were the most recent to evolve
because of the absence of the derived
characteristics.
Answer:
B
Explanation:
It tells us a lot about why the argument is correct.
The _____ on the out side of the virus are "sticky" and attracted to specific molecules on the outside of animal cells. A. Protein Spikes B. Carbohydrate Molecules C. Cholesterol Markers
Answer:
A. Protein Spikes
Explanation:
A viruses are defined as the quintessential parasites in the living world. They are a chain of the nucleic acids that lives in a host cell. Without the host cell, all the viruses cannot carry out their life sustaining activities or functions. They are always housed in a protein layer or a protein coat. All the viruses contains the RNA or the DNA. The protein spikes which is on the outside of a virus are "sticky" and they are attracted to the specific molecules of the outside of an animal cells.
Using this illustration as a guide, indicate the level one would need to drill to gain a sample containing the highest ratio of parent rock in comparison with organic material.
horizon C
horizon A
horizon B
horizon O
Answer:
horizon C
Explanation:
The horizon C has the highest ratio of parent rock as compared to organic matter. This horizon starts after horizon B and present above the bed rock. This horizon consists of weathered and partially decomposed rocks. It is the actual layer of soil of that place whereas the other horizon A and B comes with the passage of time when the wind bring the soil particles and these particles deposits on this layer of the soil making horizon A and B.
it helps the plant to pollinate the flowers
Answer:
Bees or similar inscets.
Environmental scientists calculating pollutant half-lives often define a transport rate constant that is analogous to a reaction rate constant and describes how a pollutant moves out of an ecosystem. In a study of the gasoline additive MTBE in Donner Lake, California, scientists from the University of California, Davis, found that in the summer the half-life of MTBE in the lake was 26.01 days. Assume that the transport process is first order. What was the transport rate constant of MTBE out of Donner Lake during the study
Answer:
k ≈ [tex]2.7 *10^{-2} d^{-1}[/tex]
Explanation:
For a first order reaction
Half life ( [tex]t_{1/2}[/tex] ) = 0.693 / k
k = rate constant
[tex]t_{1/2}[/tex] = 26.01
hence k = 0.693 / ( 26.01)
k ≈ 2.7 * 10^-2 d^-1
This is a detailed simple solution as required
A cleft chin in humans is coded by the dominant allele C. A chin without a cleft is coded by the recessive allele c. A man with a cc allele combination for the trait produces a zygote with a woman with a Cc allele combination for the trait. Which allele combinations could occur in the zygote?
Answer:
The allele combinations are: Cc, cc, Cc, cc
Phenotypically there is a 50% chance that the child will be born with or without the cleft
The allele combinations that could occur in the zygote is Cc or cc.
What is an allele combination?An identification of the specific alleles found at each of the relevant mouse loci on the two homologous chromosomes (i.e., all allele pairs of a genotype). One or more allele pairs can make up an allele combination.
Genetic linkage is the capacity of a chromosome to transmit alleles that are near to one another as a whole unit through meiosis. Inheritance of traits is also understood by this principle. It is useful in locating the genes that are in charge of characteristics and diseases.
The person is homozygous for the allele if the two alleles are the same. The person is heterozygous if the alleles are different.
Therefore, the zygote may include the allele combinations Cc or cc.
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what is the scientific name of frog
Answer: Anura
Explanation:
Answer:
The scientific name for a frog is Anura.
In Drosophila, the genes for eye color, wing shape, and wing length are located on chromosome II. Purple eyes (pr), arc bent wings (a), and vestigial wings (vg) are the mutant forms of the wild type traits red eyes, straight wings, and long wings, respectively. You've discovered some data in your genetics laboratory which indicates that the distance between vg and pr is 12.5 m.u., the distance between a and pr is 44.7 m.u., and the distance between a and vg is 32.2 m.u. what is the chance of a double crossover between these three genes
Answer:
The expected double recombinant frequency = 0.0402 = 4.02%
Explanation:
Available data:
genes for eye color, wing shape, and wing length are located on the same chromosomePurple eyes (pr) is the mutant form of the wild type traits red eyes (pr+)Arc bent wings (a) is the mutant form of the straight wings (a+)Vestigial wings (vg) are the mutant forms of the long wings (vg+)the distance between vg and pr is 12.5 m.uthe distance between a and pr is 44.7 m.uthe distance between a and vg is 32.2 m.u→ Two genes that are very close will have a few recombination events and are strongly bounded.
→ The map unit is the distance between the pair of genes for which every 100 meiotic products, one of them results in a recombinant one.
1) First, we need to know their order in the chromosome, and to do so, we need to compare the distances between them.
The most distant pair is a-pr
----a---------------------------------------------pr---
║-------------------44,7 MU-------------║
From gene a to vg there are 32.2 MU
----a-----------------------------vg--------------pr---
║-------------------44,7 MU-------------║
║----------32.2 MU------║
Finally, de distance from vg to pr is 12.5 MU
----a-----------------------------vg--------------pr---
║-------------------44,7 MU---------------║
║----------32.2 MU------║--12.5 MU--║
We can name region I to the distance between a-vg, and region II to the distance between vg-pr.
2) To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cM. And that the maximum recombination frequency is always 50%.
1 MU -----------1% of recombination
32.2 MU -----X = 32.2 % of recombination
12.5 MU-------X = 12.5 % of recombination
3) The expected double recombinant frequency = recombination frequency in region I x recombination frequency in region II.
The expected d. recombinant frequency = 32.2% x 12.5% = 0.322 x 0.125
The expected double recombinant frequency = 0.0402 = 4.02%
6. New experimental data does not support a currently accepted hypothesis. Which
course of action should the researcher take?
A) Do the experiment again.
B) Change the data to fit the hypothesis.
C) Form a new hypothesis and plan a new experiment
D) Change the procedure to obtain the desired outcome.
what is a false scorpion
Answer:
huh im not prcicley undertsa
Answer:
A pseudoscorpion, also known as a false scorpion or book scorpion, is an arachnid belonging to the order Pseudoscorpiones, also known as Pseudoscorpionida or Chelonethida. Despite their appearance to true scorpions and ticks, they are NOT harmful to people. Physical removal is the only necessary control.
Explanation:
In a flower, the anther is attached to filament. What advantage does the filament provide to the anther?
A. The filament elevates the anther to allow for easier transport of pollen by wind or insects.
B. The filament provides the anther with nutrients to produce pollen grains.
C. The filament protects the anther from harsh sunlight and insects.
D. The filament produces a sticky surface to allow the anther to capture pollen.
The correct answer is A. The filament elevates the anther to allow for easier transport of pollen by wind or insects.
Explanation
The filament of a flower is the sterile part of the stamen (male organ of the flower) that is located just below the anther. Its function is structural to hold the sling so that it is accessible to insects. The filament is usually cylindrical and fine. Therefore, the correct answer is A. The filament elevates the anther to allow for easier transport of pollen by wind or insects.
What is a prokaryote
Answer:
Prokaryote is a microscopic single-celled organism which has neither a distinct nucleus with a membrane nor other specialized organelles, including the bacteria and cyanobacteria.and also blue green algae.