The expected number of Barr bodies in interphase cells of individuals with the following karyotypes: In 46 XY Barr bodies is 0, In 45 XO Barr bodies is 1, In 47 XXX Barr bodies is 2, and In 48 XXYY Barr bodies is 1.
Barr bodies are inactivated X chromosomes found in female mammals. The number of Barr bodies in interphase cells depends on the number of X chromosomes in the cell.
46 XY: This is the normal male karyotype, which has one X chromosome and one Y chromosome. Since males only have one X chromosome, they do not have Barr bodies in their interphase cells. The expected number of Barr bodies in this karyotype is 0.
45 XO (Turner's syndrome): This karyotype is a female with only one X chromosome. In this case, the single X chromosome will become the Barr body. The expected number of Barr bodies in this karyotype is 1.
47 XXX: This karyotype is a female with three X chromosomes. In this case, two of the X chromosomes will become Barr bodies, and one will remain active. The expected number of Barr bodies in this karyotype is 2.
48 XXYY: This karyotype is a male with two X chromosomes and two Y chromosomes. Since males only need one X chromosome, one of the X chromosomes will be inactivated and become a Barr body. The expected number of Barr bodies in this karyotype is 1.
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Write 5 sentences OR a list of numbered steps to explain the process of DNA replication. You’ll also complete sketch notes to supplement your written understanding.
Your writing must incorporate the following components
Direction DNA is synthesized
Description of how DNA replication is considered semiconservative
Role and location of helicase
Role and location of topoisomerase
Role and location of RNA primers
Role and location of DNA polymerase
Distinguishing between leading/lagging strand and identifying which strand replicates continuously and which replicated discontinuously
Okazaki fragments
Role and location of ligase
This results in the formation of small fragments known as Okazaki fragments. The role of the ligase is to seal the gaps between the Okazaki fragments and join them to form a continuous strand.
Here are the five sentences to explain the process of DNA replication:The process of DNA replication involves the synthesis of new DNA strands by copying the existing DNA strands.This process is considered semi-conservative since each daughter DNA molecule is composed of one parent strand and one newly synthesized strand.
During DNA replication, the enzyme helicase is responsible for unwinding the double helix by breaking hydrogen bonds between the base pairs.The enzyme topoisomerase relieves the tension that arises ahead of the replication fork.
DNA polymerase is the enzyme responsible for adding nucleotides to the growing DNA strand. It can only add nucleotides to the 3’ end of the strand.There are two strands in DNA:
the leading strand and the lagging strand. The leading strand replicates continuously, while the lagging strand replicates discontinuously. The RNA primers are located on the lagging strand and are responsible for providing a free 3’ OH group for DNA polymerase to add nucleotides.
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The shaded individuals in the pedigree diagram below are color-blind. The
only unknown is whether the mother in the first generation has normal color
vision.
What is true about the mother?
A. She must have normal color vision.
B. She must be color-blind.
C. She could be homozygous for normal color vision.
D. She could be heterozygous for color blindness.
The correct answer is D. She could be heterozygous for color blindness.
What is Color Blindness?
Color blindness is a condition in which a person's ability to perceive colors is impaired. Most cases of color blindness are genetic and are caused by a defect or absence of the color-sensitive pigment in the cones of the retina. This defect or absence can affect one or more of the three types of cones that are responsible for detecting red, green, and blue light.
In the pedigree diagram, the shaded individuals are color-blind, which is a recessive X-linked trait. This means that the individuals in the pedigree who are shaded must have inherited a color-blindness allele from both of their parents, and that the mother of the first generation could be either homozygous for normal color vision (XX) or heterozygous for color blindness (XbX).
If the mother in the first generation had normal color vision, she would have to be homozygous for normal color vision (XX), and all of her sons would also have normal color vision because they would inherit a normal X chromosome from her. However, this would mean that none of her daughters would be color-blind, and there are shaded females in the pedigree. Therefore, it is not possible for the mother to have normal color vision.
If the mother in the first generation is heterozygous for color blindness (XbX), she would have a 50% chance of passing on the color-blindness allele to each of her children. This means that all of her sons would have a 50% chance of being color-blind, as shown in the pedigree, and that her daughters would have a 50% chance of inheriting one copy of the color-blindness allele and being carriers, but they would not be color-blind themselves. This is consistent with the shaded females in the pedigree who are carriers of the color-blindness allele.
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A section of DNA codes for the sequence of nitrogenous bases of mRNA to initiate
protein synthesis as shown.
GATACGCT
(DNA)
What is the process involved in the coding?
O replication
O transcription
O transduction
O translation
CUAUGCGA
(mRNA)
The process that involves coding is called transcription. The correct option is b.
What is transcription?The process of transcription creates RNA from DNA using the non-coding strand of the DNA as a template. The resulting RNA is a replica of the coding strand of the DNA.
Proteins called transcription factors to help turn particular genes "on" or "off" by attaching to surrounding DNA. The transcription of a gene is increased by transcription factors known as activators, while the transcription of genes is decreased by repressors.
Therefore, the correct option is B. transcription.
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Before recommending drastic changes or additional rules for athletes, what type of evidence or experiments would you need to see linking concussions to CTE and depression? Give one example in 1-2 sentences.
Answer:
Explanation:
To link concussions to Chronic Traumatic Encephalopathy (CTE) and depression, it is important to have controlled and long-term studies that follow a large cohort of athletes who have experienced concussions, comparing their brain health and mental health outcomes to those of a similar control group of athletes who have not experienced concussions. For example, a long-term study that uses imaging techniques such as MRI or PET scans to compare the brain structure and function of athletes who have experienced concussions to those who have not, and assesses the incidence of depression and other mental health outcomes, could provide important evidence linking concussions to CTE and depression.
As light travels through the eye, it passes through several structures/chambers before reaching the retina. Which list below gives those structures in the correct order?
Option D, The correct order of structures/chambers that light passes through as it travels through the eye is: cornea, lens, iris, aqueous humor, vitreous humor
The clear, dome-shaped tissue covering the front of the eye that helps to focus light. Aqueous humor the clear, watery fluid that fills the space between the cornea and the lens. Pupil the opening in the center of the iris that controls the amount of light entering the eye. Lens the transparent structure behind the iris that changes shape to focus light on the retina. Vitreous humor the gel-like substance that fills the space between the lens and the retina. Retina the light-sensitive tissue lining the back of the eye that contains the cells responsible for vision. Therefore, the correct order of the structures/chambers that light passes through as it travels through the eye is: cornea, lens, iris, aqueous humor, vitreous humor.
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The complete Question is:
As light travels through the eye, it passes through several structures/chambers before reaching the retina. Which list below gives those structures in the correct order?
a) cornea, iris, lens, vitreous humor, aqueous humor
b) iris, cornea, lens, aqueous humor, vitreous humor
c) lens, cornea, iris, aqueous humor, vitreous humor
d) cornea, lens, iris, aqueous humor, vitreous humor
What is the most common antibiotic-resistant bacteria?
MRSA (Methicillin-resistant Staphylococcus aureus) is the most common antibiotic-resistant bacteria.
More about MRSA:
MRSA is usually spread through a wound, or when someone shares an infected object that has come in contact with an infected person, with a person who is not infected.
MRSA is also a mutation of staph.
Image found at: https://kidshealth.org/en/teens/mrsa.html
I apologize if I was wrong. I hope you have a lovely day! :)
Answer:
MRSA
Explanation:
The most common antibiotic-resistant bacteria is methicillin-resistant Staphylococcus aureus (MRSA). MRSA is a type of staph infection that is resistant to many commonly used antibiotics, including methicillin and other beta-lactam antibiotics. This makes MRSA infections difficult to treat and can increase the risk of serious health problems, such as sepsis and pneumonia. MRSA is most commonly found in healthcare settings, such as hospitals and nursing homes, but it can also occur in communities. It is important to practice good hygiene, such as washing your hands frequently and covering your nose and mouth when you cough or sneeze, to help reduce the spread of MRSA and other antibiotic-resistant bacteria.
Image : https://www.everydayhealth.com/infectious-diseases/staph-mrsa/truth-about-mrsa-infection/
Allen
What iS one reason scientists have developed a system to classify organisms?
Classification allows for organisms to interbreed and change.
A system was needed to better
track genetic changes in an organism.
Classification allows for better identification of new organisms.
A system was needed to see microscopic organisms with more detail.
Answer:
Classification allows for better identification of new organisms
Fat around organs
a-chemical messenger
b-protection
c-structure of cell membranes
Fat around organs are mostly used as a form of protection as it provides a from of cushion to reduce the effects of stress.
What is Visceral body fat?This is also known as 'hidden' fat, and it is referred to as a type of fat which is stored deep inside the belly, wrapped around the organs, including the liver and intestines.
Fat around organs are mostly used as a form of protection as it provides a from of cushion to reduce the effects of stress and other unpleasant situations to ensure that the organs aren'rt damaged and are able to perform optimally.
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13. Linus Pauling proposed a DNA structure in which the prophase groups were tightly packed on the molecule's inside, leaving the nitrogenous bases sticking outward.
a) Which scientist's findings did Linus Pauling need to propose the correct 3D-structure DNA?
b) If DNA replication occurred in this structure, how would it differ from what you know is the actual process? Explain your reasoning
14. While working with a particular strain of E. coli, you discover that it transcribes the lac genes at a high level when no lactose is present in the cell's medium.
a) Explain how the lac operon functions typically.
b) List the possible causes of this abnormal activity.
c) Then design an experiment to determine which of these possible causes is the case with your bacterial colony.
The lac operon follows a negative, inducible mode of regulation. This means that the transcription of the operon is turned OFF in the absence of lactose and is only induced into activity when lactose is present in the medium.
What is lac operon?
a) If the genotype of the lac operon is normal, i.e. it does not contain any mutations, then,If lactose is present in the medium:-When lactose is present in the medium, it is translocated into the cell because the operon always produces a small amount of permease. The allolactose molecule binds to the repressor molecule and inactivates it.An inactivated repressor can no longer bind to the operator and as a result all genes are transcribed, i.e. the operon is turned on.Also, the absence of glucose leads to the production of large amounts of cAMP in the cell, which binds to the CAP protein. This complex then binds to the CAP site and increases the rate of transcription of the operon.If lactose is not present in the medium:-In the absence of lactose, the repressor remains active and binds to the operator. This prevents RNA polymerase from binding to the promoter and the operon is not transcribed, i.e. the operon is turned off.When both lactose and glucose are present:-Glucose will be used first because it is the preferred energy source and during this time there will be minimal transcription of the lac operon.b) If the lac operon is expressed at high levels even in the absence of lactose, it means that it shows constitutive expression, i.e. it is expressed regardless of the availability of lactose in the medium.This can be possible for two reasons:-1) A mutated lacI gene (lacI-) that produces an inactive or mutant repressorIn this case, the mutation alters or alters the promoter region of the lac1 gene, since this gene has a promoter separate from that of the structural genes. This means that if the promoter of that gene is altered so that it can no longer bind to the RNA polymerase enzyme, that gene will not be transcribed or expressed. As a result, the repressor protein will not be produced and the operon will therefore be constitutively active.2) Mutation operator, i.e. operator Oc In this mutation, the repressor binding site on the operator is mutated in such a way that it cannot bind to the repressor protein even in the absence of lactose. If the repressor cannot bind to the operator, then the operon will be expressed regardless of the presence or absence of lactose.C)In the case of a mutant lacI gene, we know that the promoter of this gene will be changed so that it can no longer bind to the RNA polymerase enzyme and the gene will not be transcribed. that is, this gene will not produce any mRNA or repressor protein in the absence of lactose, whereas the normal operon does,In the case of a constitutive operator, the operator will not be able to bind the repressor even in the absence of lactose. Thus, even if lactose is not available in the medium, there will be a higher concentration of free repressor protein in the cytoplasm of the cell.This means that in the case of the lacI- mutant there will be no repressor protein, while in the case of the Oc mutant there will be a high concentration of the repressor protein in the cell cytoplasm.Thus, this difference in the presence of the repressor protein can be used to distinguish between a cell with a normal and a lacI-mutated operon by Western Blot analysis.This is because this technique tests for the presence of specific proteins present in the cell sample using complementary antibodies that specifically bind to the target protein and are conjugated to a marker enzyme that stains the chromogenic substrate blue.To know more about lac operon, click the link given below:
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carbon cycle are they always parts of the same compound?
No, the carbon cycle involves the movement of carbon in different forms and compounds, including both organic and inorganic compounds.
What happens in a carbon cycle?
In the carbon cycle, carbon moves between the atmosphere, the oceans, the land, and living organisms. Carbon can be present in different forms, such as carbon dioxide (CO2) in the atmosphere, dissolved inorganic carbon (DIC) in the oceans, and organic carbon in living organisms and in dead organic matter in soils and sediments.
Carbon also exists in various organic compounds, such as carbohydrates, proteins, and lipids, which make up the living tissues of organisms. As organisms respire, they release carbon dioxide back into the atmosphere or into the water, and when they die, their remains decompose, releasing carbon into the soil or water.
In short, the carbon cycle involves the movement of carbon in various forms and compounds, and it is not limited to a single compound or form.
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Damage to this part of the cell would most likely lead to muscle damage of the organism?
Answer:
When muscle is damaged, a protein called myoglobin is released into the bloodstream. It is then filtered out of the body by the kidneys. Myoglobin breaks down into substances that can damage kidney cells. Rhabdomyolysis may be caused by injury or any other condition that damages skeletal muscle
Three new investigational drugs were studied in an in vitro cell culture model system. The figures below summarize the data. The
figure on the left depicts the graded LDR curve of Drug X in the presence of increasing doses of Drug Y. The figure on the right
depicts the graded LDR curve of Drug X in the presence of a fixed dose of Drug Z. Based on these data, identify what type of
drug each of these agents is.
Drug X: _____
Drug Y____
Drug Z:____
Please explain how you got these answers, I am totally lost. Thank you!
Drug X appears to be an agonist, Drug Y appears to be an antagonist, Drug Z appears to be a neutral agent.
what are drugs ?
Drugs are substances that can alter the structure or function of the body or mind when introduced into the body. They can be used for medicinal purposes to treat diseases or relieve symptoms, but they can also be used recreationally for their effects on the brain and body.
Based on the information provided, we can identify the type of drug each agent is as follows:
Drug X appears to be an agonist as it is increasing the response of the system with increasing concentrations.
Drug Y appears to be an antagonist as it is reducing the response of the system when co-administered with increasing concentrations of Drug X.
Drug Z appears to be a neutral agent as it does not affect the response of the system in the presence of increasing concentrations of Drug X.
It is important to note that these classifications are based solely on the data provided in the figures and additional experiments may be required to confirm these classifications.
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A disease kills the mice. Explain how the maize harvest will be affected.
maize ->mice->snakes->
Answer:
Two hshshehdhehwjwkwkwooeidudd
1. What are the necessary components required to start the transcription of a
gene? What needs to be added in order to create a protein?
2. How does the length of the transcribed region impact the length of the gene
(DNA)?
3. You accidentally produce a gene (strand of DNA) after you have filled up all
five of your protein shapes? How would you eliminate this DNA?
4. Where does the negative transcription factor work?
5. Why are there more than one positive transcription factors on each of the
three genes?
Answer:
Explanation:
1) The necessary components required to start the transcription of a gene include the DNA template strand, RNA polymerase, and a promoter region. The promoter region is a specific sequence of DNA that signals the start site of transcription and determines where RNA polymerase will bind to the DNA. In addition, transcription factors (both positive and negative) can also be involved in the initiation of transcription by binding to the promoter region and influencing the activity of RNA polymerase.
To create a protein, the transcribed RNA molecule must undergo post-transcriptional modifications such as splicing, capping, and polyadenylation. This processed RNA is called messenger RNA (mRNA) and can be translated into a protein by ribosomes in a process called translation.
1) The length of the transcribed region of a gene (the region that is transcribed into RNA) can impact the length of the gene, as the length of the transcribed region determines the length of the resulting mRNA molecule. If a portion of the gene is not transcribed, then it will not be included in the mRNA and will not contribute to the final protein. However, the length of the gene itself does not directly impact the length of the transcribed region or the resulting mRNA.
2) If you accidentally produce a gene (strand of DNA) after filling up all five of your protein shapes, you could eliminate this DNA through degradation. Enzymes called nucleases can degrade the excess DNA and remove it from the cell. In some cases, cells may also use processes such as programmed cell death (apoptosis) to remove cells that contain abnormal or excess DNA.
3) Negative transcription factors work by inhibiting the activity of RNA polymerase and preventing the transcription of a gene. They can bind to the promoter region or to specific sequences within the gene itself, blocking the binding of RNA polymerase and suppressing transcription.
4) There are more than one positive transcription factors on each of the three genes because different genes may be regulated by different sets of transcription factors. Positive transcription factors can increase the rate of transcription by binding to specific sequences within the promoter region and enhancing the binding and activity of RNA polymerase. Having multiple positive transcription factors on a gene allows for more complex and nuanced regulation of gene expression, as different factors may have different binding specificity and responsiveness to different stimuli.
How does Alzheimer’s affect homeostasis
Answer:
Central to the neurodegenerative process is the inability of neurons to properly regulate intracellular calcium levels. Increased levels of amyloid beta-peptide (Abeta) induce oxidative stress, which impairs cellular ion homeostasis and energy metabolism and renders neurons vulnerable to apoptosis and excitotoxicity.
Part A The evolution of cuticle presented land plants with a challenge that threatened their ability to live on land. What is this challenge, and why do stomata represent a solution? Match the words in the left column to the appropriate blanks in the sentences on the right. Reset Help prevent(s) Cuticle allow(s) water loss from leaves and prevent(s) entry of CO2 required for allow(s) photosynthesis. Stomata allow(s) CO2 to diffuse and can close to allow(s) water loss. Submit Previous Answers Request Answer X Incorrect; Try Again; One attempt remaining You filled in 2 of 4 blanks incorrectly.
The evolution of cuticles presented land plants with the challenge of preventing water loss from leaves and preventing the entry of CO2 required for photosynthesis, while stomata allow CO2 to diffuse and can close to prevent water loss.
1. Cuticle prevents water loss from leaves and prevents entry of CO2 required for photosynthesis.
2. Stomata allow CO2 to diffuse and can close to prevent water loss.
The waxy cuticle on the surface of leaves helps prevent water loss, but it also creates a barrier that prevents gases from entering and leaving the plant. This presents a problem for photosynthesis, which requires the uptake of carbon dioxide (CO2) and the release of oxygen (O2).
Stomata, have tiny pores on the leaf surface, allowing CO2 to diffuse into the leaf and O2 to diffuse out. Additionally, they can close to prevent excessive water loss through transpiration. The evolution of stomata allowed plants to thrive on land by overcoming the challenge of gas exchange while minimizing water loss.
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Complete Question:
The evolution of cuticles presented land plants with a challenge that threatened their ability to live on land. What is this challenge, and why do stomata represent a solution?
1. Cuticle ___ water loss from leaves and ___ entry of CO2 required for photosynthesis.
2. Stomata ____ CO2 to diffuse and can close to ______ water loss.
Fill the blanks with allow or prevent appropriately.
Male patterned baldness is a sex-linked recessive disorder that
causes people to lose their hair at a younger age. It occurs more
often in males than females. A father who is not bald and a mother
who is a carrier for this trait have offspring. Select all of the
possible genotypes for their kids.
Choose all that apply:
X B x b
X b X b
X B Y
YY
The possible genotypes of the offspring are given below: XY, XXb, XY, XXb
What are sex-linked recessive disorders?Genetic disorders connected to mutations in genes on the X chromosome are referred to as having X-linked or sex-linked recessive inheritance. Because he contains just one X chromosome, a male who carries this mutation will be affected. A female who carries a gene mutation in one X chromosome but has a normal gene on the other X chromosome usually has no symptoms.
The possible genotypes of the offspring of a father who is not bald and a mother who is a carrier for this trait is given below:
XY x XXb ---> XY, XXb, XY, XXb
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Aude cych in Humai s 16 chrom ne conden At this po "heir size, nad hai that shown in part Ill of the figure above, if the zygote's chromosome 41) In a life cycle such as that shown in part III of the figur number is ten, which of the following statements will be true? utele chromosome number per cell is ten and the gametophytes is five A) The sporophyte's chromosome number per cell is ten and B) The sporophyte's chromosome number per cell is five and the ome number per cell is five and the gametophytes is ten. C) The sporophyte and gametophyte each have ten chromosomes per cell D) The sporophyte and gametophyte each have five chromosomes per cell
Based on the given information, the correct answer to the question is (B) - The sporophyte's chromosome number per cell is five, and the gametophyte's chromosome number per cell is five.
This is because the chromosome number in the life cycle is ten, and the sporophyte and gametophyte are haploid, meaning they have half the chromosome number of the diploid zygote. Therefore, each sporophyte and gametophyte would contain five chromosomes per cell.
The given scenario describes the life cycle of a hypothetical plant species named "Aude cych in Humai". The figure shows the different stages of the life cycle and the changes in chromosome numbers during the process of meiosis and fertilization.
At the point where the zygote is formed, it contains 16 uncondensed chromosomes. If the chromosome number in the life cycle is ten, the resulting chromosome numbers in different stages can be determined based on the process of meiosis and fertilization.
In this life cycle, the sporophyte is the diploid stage, and the gametophyte is the haploid stage. During meiosis, the sporophyte undergoes two rounds of cell division, resulting in four haploid spores. These spores then develop into haploid gametophytes.
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A student runs an experiment to figure out if the size of a wire changes how well i
the resistance in the wires.
s electricity. She uses 5 different widths of copper wire and attaches each to a multimeter to measure the resistance in the wires. The data collected is shown below:
Wire 1- 12 Ohms
Wire 2- 17 Ohms
Wire 3- 9 Ohms
Wire 4- 14 Ohms
Wire 5- 6 Ohms
What is the independent variable in this experiment?
A The multimetet.
B. The resistance values.
C. The material of the wires, copper.
D. The widths of the wires.
The widths of the wires was the independent variable in the experiment.
define resistance ?Resistance is a measure of the degree to which an object opposes the flow of electric current through it. It is the ratio of the voltage applied to a conductor or circuit to the resulting current that flows through it, and is typically measured in units of ohms (Ω). The higher the resistance of a material, the more difficult it is for electric current to flow through it. Factors that can affect resistance include the material of the conductor, its temperature, and its dimensions.
Resistance is an important concept in understanding and designing electrical circuits, as it can impact the overall performance and efficiency of a system. The widths of the wires was the independent variable in the experiment.
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Which middle eastern plateau is surrounded by mountains and contains two uninhabitable deserts?a. Saudi arabia's central plateaub. Iran's interior plateauc. The anatolian plateaud. The gaza strip please select the best answer from the choices provided. Abcd.
The Middle Eastern plateau that is ringed by mountains and has two inhospitable deserts is known as Iran's Interior Plateau, thus the correct choice is B.
Numerous plateaus are formed as magma from deep within the Earth pushes towards the surface but is unable to break through the crust. Instead, the magma raises the enormous, flat, impervious rock above it. Geologists believe that a magma cushion may have provided the Colorado Plateau with its last lift beginning around ten million years ago. The Iranian plateau is bordered to the west by the Zagros Mountains, to the north by the Caspian Sea, to the south by the Persian Gulf, and to the east by the Indo-Gangetic Plains.
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The complete question is:
Which Middle Eastern plateau is surrounded by mountains and contains two uninhabitable deserts?
A. Saudi Arabia's Central Plateau
B. Iran's Interior Plateau
C. The Anatolian Plateau
D. The Gaza Strip
Answer:B.
Iran's Interior Plateau
took the test. B is correct
Explanation:correct on edg
3. In Jimsonwed, purple flower (P) is dominant to.white (p), and spiny pods
(S) are dominant to smooth(s). In a cross between a Jimsonweed
homozygous for white flowers and spiny pods and one homozygous for
purple flowers and smooth pods, what are all possible phenotypes and
genotypes of the offspring?
The possible offspring are:
PP, Ss - This offspring would have purple flowers and spiny pods. Its genotype would be heterozygous for both traits.
Pp, Ss - This offspring would also have purple flowers and spiny pods, but it would be heterozygous for only the flower color trait.
Pp, ss - This offspring would have purple flowers and smooth pods. Its genotype would be heterozygous for only the pod trait.
pp, ss - This offspring would have white flowers and smooth pods. Its genotype would be homozygous for both traits.
What is a Jimsonweed?Jimson weed (Datura stramonium) is a plant species in the nightshade family (Solanaceae). It is also commonly known as thorn apple, devil's trumpet, and locoweed.
The possible phenotypes are purple flowers with spiny pods, purple flowers with smooth pods, white flowers with spiny pods, and white flowers with smooth pods. The possible genotypes are heterozygous for both traits, heterozygous for only the flower color trait, heterozygous for only the pod trait, and homozygous for both traits.
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What are the approximate coordinates of the star Rigel?
5 hours 10 minutes right ascension and -8⁰ declination
5 hours 40 minutes right ascension and -2⁰ declination
5 hours 40 minutes right ascension and 10⁰ declination
6 hours 50 minutes right ascension and 6⁰ declination
Answer:
5 hours 10 minutes right ascension and -8⁰ declination
Explanation:
The approximate coordinates of the star Rigel are 5 hours 14 minutes right ascension and -8⁰ declination.
Put the following in order from smallest to largest:
Giardia lamblia protozoa (7µm)
Bacterial Plasmids (20nm)
Mastadenovirus (30nm)
Cionorchis sinensis trematode (5mm)
Gram-pos bacteria (6µm)
All together from littlest to biggest: Bacterial Plasmids (20nm)<Mastadenovirus (30nm) < Gram-pos microscopic organisms (6µm) < Giardia lamblia protozoa (7µm) < Cionorchis sinensis trematode (5mm)
Anything more modest than 200 nm can't communicate with noticeable light. Losing on the stage is excessively simple. Noticeable light is just great at frequencies under 390 nm.
Objective Focal point or Mathematical Opening This piece of a compound magnifying instrument is the focal point nearest to the example.
The level stage is where you place your slides. Stage cuts hold the slides set up. Spinning Nosepiece or Turret: This is the part that holds at least two objective focal points and can be pivoted to change power without any problem.
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hemophilia is caused by sex linked recessive allele. Diane's brother has hemophilia there's no history of hemophilia and her husband Craigs family is it possible for Diane and Craig to have a child with hemophilia
PLS HELP
It not possible for Diane and Craig to have a child with hemophilia even though the gene is recessive in the wife
Can genes be recessive?A recessive gene is a gene that will only express its characteristic (phenotype) when two copies of the gene are present, one from each parent. If an individual has only one copy of the gene (they are heterozygous), the dominant gene will be expressed instead.
In general, dominant genes mask the expression of recessive genes, so the phenotype of an individual is determined by the dominant gene. However, in certain cases (such as homozygosity for a recessive gene), the recessive gene can be expressed.
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Plant and animal cells release energy stored in the bonds of glucose molecules when they perform
OA. meiosis.
OB. metamorphosis.
OC. photosynthesis.
OD. cellular respiration.
1. Which of the following diseases is the first to be vaccinated against?
1. smallpox
2. pertussis
3. influenza
4. polio
Helppppp it’s due today
Stablizing selection is represented in graph A, and distructive selection is represented in graph B. Thus, option B is correct.
What will be the equation?
Given inequality is,
x + 2y ≤ 47
This inequality has a sign of less than equal to ( ≤ ). Properties of this inequality when graphed will be sign of equal to in the given inequality shows that the graph of the line will be a solid line.
Sign of less than represents the shaded region below the line.The graph lying in first quadrant and graph of horizontal line lying in the fourth quadrant are representing functions.
Thus, Stablizing selection is represented in graph A, and distructive selection is represented in graph B. Thus, option B is correct.
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Which lipid is responsible for stabilizing a plasma membrane?
Cholesterol
Triglyceride
Phospholipid
Saturated fatty acid
Answer: phospholipid Bilayer
Explanation:
What is a hypertonic, hypotonic, and isotonic solution?
Answer:
Hypotonic has a lower concentration of fluid, sugars and salt than blood. Hypertonic has a higher concentration of fluid, sugars and salt than blood. Isotonic has similar concentration of fluid, sugars and salt to blood.
When does the egg begin the 2nd round of meiosis?
A) when the egg was first made
B) before it is released from the ovary
C) after a sperm enters the egg cell
Answer: After ovulation the oocyte is arrested in metaphase of meiosis II until fertilization. At fertilization, the secondary oocyte completes meiosis II to form a mature oocyte (23,1N) and a second polar body.
Explanation:
