Overall, this trial aims to compare the effectiveness of fresh, frozen, and lyophilized microbiota from a small pool of healthy donors in treating recurrent CDI when delivered by colonoscopy. The results of this study will provide valuable information on the potential benefits and drawbacks of different forms of FMT for this specific infection.
A randomised clinical trial is a type of research study that aims to compare different treatments or interventions to determine their effectiveness.
In this specific trial, the researchers are investigating the use of faecal microbiota transplantation (FMT) as a treatment for recurrent Clostridium difficile infection (CDI). FMT involves transferring healthy bacteria from a donor's stool into the intestines of a recipient.
In this trial, there are three different forms of microbiota being tested: fresh, frozen, and lyophilized. Fresh microbiota refers to using stool samples that have not been processed or preserved. Frozen microbiota involves freezing the stool samples to preserve the bacteria. Lyophilized microbiota is freeze-dried and converted into a powder form.
The researchers are also using a small pool of healthy donors, which means that they are selecting a limited number of individuals whose stool will be used for transplantation. This approach helps to standardize the characteristics of the microbiota being used.
The delivery method for FMT in this trial is colonoscopy. Colonoscopy involves inserting a flexible tube with a camera into the colon to visualize and treat any abnormalities. In this case, the microbiota would be administered during the colonoscopy procedure.
Overall, this trial aims to compare the effectiveness of fresh, frozen, and lyophilized microbiota from a small pool of healthy donors in treating recurrent CDI when delivered by colonoscopy. The results of this study will provide valuable information on the potential benefits and drawbacks of different forms of FMT for this specific infection.
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The axial skeleton can be divided into the skull, the vertebral column, and the: a. thoracic cage c. hip bones b. femur d. humerus.
The axial skeleton can be divided into the skull, the vertebral column, and the thoracic cage.
The head and boot bones of a vertebrate make up the axial skeleton, which is the portion of the skeleton. The 80 bones of the human skeleton are divided into six parts: the skull (22 bones), the ossicles of the middle ear, the hyoid bone, the rib cage, the sternum, and the vertebral column. The axial and appendicular skeletons work together to make up the entire skeleton. The bones of the vertebrae, sacrum, coccyx, cranium, ribs, and sternum are included in another definition of the axial skeleton.
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Which apply to MEIOSIS? ( check ALL 10 that apply)
produces new cells
has 8 steps and is divided into two stages
produces cells with variation
starts with diploid cell produces identical cells
copies DNA in S phase of interphase
Includes crossing over of homologous chromosomes produces haploid cells
Includes copying sister chromatids four daughter cells produces gametes/sex cells
divides once
has 4 steps (PMAT)
two daughter cells produces diploid cells produces body/somatic cells divides twice
Meiosis uses gametes and/or sex cells.
Meiosis affects which cell types?Meiosis only happens in reproductive cells because it produces haploid gametes that can be fertilized. Meiosis is necessary for sexual reproduction but is not the same as it. Due to the synthesis of gametes during meiosis, sexual reproduction requires it (sperm and eggs).
What are some instances of meiosis?Male testicles create sperm, whereas female ovaries create eggs. But the DNA must be decreased before these gametes are produced. Humans have 46 chromosomes total, which is made up of 23 unique chromosomes that exist in homologous pairings between maternal and paternal DNA.
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Which tools do meteorologists use to collect data about the weather?
Answer:
The answers are satellites, weather balloons, and automated weather stations
Tabulate two difference between growth in plants and that of animals
Which compound is produced during regeneration? PGA G3P RuBP rubisco
Answer:
RuBP is the answers
Answer:
C) RuBP is the answer on edge.
Explanation:
just made a 100%, if you ever have a question i´m glad to help!
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which of the following is a key difference between the thrift genotype hypotehsis and the thrifty phhenotype hypothesis
The key difference between thrift genotype hypothesis and the thrifty phenotype hypothesis is when specific populations have genes that were historically beneficial, that enable efficient storage of fat during times of food scarcity or in feast to compensate for times of famine.
Thrifty phenotype hypothesis states that the epidemiological associations occurs between poor fetal and infant growth and the consequent development of type 2 diabetes and many disease result from the cause of poor nutrition in early life, that produces long lasting changes in glucose-insulin metabolism.
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which group of hormones is responsible for the development of the sex organs?
The group of hormones responsible for the development of sex organs is known as sex hormones.
Sex hormones are primarily produced by the gonads, which are the testes in males and the ovaries in females.
In males, the sex hormones are primarily androgens, with testosterone being the most prominent hormone. Testosterone plays a crucial role in the development of male sex organs, such as the testes, , and prostate gland.
In females, the main sex hormones are estrogen and progesterone. Estrogen is responsible for the development of female sex organs, including the ovaries, uterus, fallopian tubes, and breasts.
These sex hormones play essential roles in the development and functioning of the reproductive system, secondary sexual characteristics, and the regulation of reproductive processes throughout life.
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In 1953, Watson and Crick discovered the structure of DNA by examining data from many different experiments. How did using Chargaff’s data help Watson and Crick determine the structure of DNA ?
Answer: Chargaff's observation was an important clue to the A-T and G-C pairings that Watson and Crick eventually proposed. ... The structure of DNA suggested that the order of bases contains information. Because A is always paired with T and G with C, the order of bases on one strand determines the order on the other strand.
Explanation:
Chargaff’s data helped Watson and Crick determine the structure of DNA
through determining that it has a double helical structure.
Chargaff's rule states that a DNA consists of nitrogenous base pairs in which
the pairs have equal units and specific binding mate.
Adenine binds with Thymine Cytosine binds with Guanine.The base pairs present in the DNA makes its shape to be double helical.
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the product of the mouse xist gene is a 15 kb inactive x-specific transcript containing no conserved orf and located in the nucleus.
The mouse Xist gene produces a 15 kb non-coding RNA transcript specific to the inactive X chromosome. It plays a crucial role in X chromosome inactivation, coating the nucleus and contributing to gene silencing without encoding a protein.
The product of the mouse Xist gene is a 15 kb long transcript that is specific to the inactive X chromosome (Xi). This transcript plays a critical role in X chromosome inactivation (XCI), a process that occurs during early development to silence one of the two X chromosomes in female mammals.
The Xist transcript is a non-coding RNA, meaning it does not encode a protein. It lacks a conserved open reading frame (ORF), suggesting that it does not produce a functional protein product. Instead, Xist functions as an RNA molecule involved in the regulation of gene expression.
Upon transcription, the Xist RNA localizes to the nucleus, where it coats the future inactive X chromosome. This coating leads to the recruitment of various protein complexes and epigenetic modifications, ultimately resulting in the silencing of genes on the Xi. Xist is thought to contribute to the establishment and maintenance of the inactive state by mediating long-range chromatin interactions and influencing the organization of the Xi.
Understanding the molecular mechanisms of Xist and XCI is important for unraveling the complexities of gene regulation and the development of sex-specific traits in mammals.
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describe the study organism and why it was ideal for this kind of study? johnson et al (2013)
The study organism used in the Johnson et al. (2013) study was the eastern oyster (Crassostrea virginica), which was deemed ideal due to its ecological and economic significance, as well as its sensitivity to environmental stressors.
The oyster's filter-feeding behavior makes it a valuable indicator of water quality, and its economic importance as a food source highlights the need for understanding the impact of environmental stressors on the species. Additionally, oysters are sensitive to changes in temperature, salinity, and dissolved oxygen, making them ideal for studying the effects of climate change and other environmental stressors on marine ecosystems.
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--The complete question is, What was the study organism used in the Johnson et al. (2013) study, and why was it deemed ideal for the research?--
What does an enzyme do?
A. Binds to a substrate
B. Lowers activation energy
C. Speeds up chemical reactions
D. All of the above
E. None of the above
Which of the following statements about the effects of climate change on phenology is true?
A.Phenological changes from climate warming have been documented, and more are expected.
B.Phenological events are occurring later in the spring, because winters are colder and longer.
C.Phenological changes do not affect reproduction and therefore do not affect fitness.
D.Phenological changes are more likely when events are cued by day length, not temperature.
The true statement about the effects of climate change on phenology is A.
Phenological changes from climate warming have been documented, and more are expected. Climate change has caused shifts in the timing of phenological events such as the start of spring, the blooming of flowers, and the arrival of migratory birds. These changes can have significant impacts on the fitness of organisms, as they may cause mismatches between the timing of key life cycle events and the availability of resources or suitable conditions.
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Based on this data which method should department use it as the most economic effective and time-saving solution for the problem of soil erosion
By planting more and more trees, as the roots of the trees hold the soil in place for the problem of soil erosion.
Soil erosion is a sluggish process that takes place when the effect of water or wind detaches and removes soil debris, inflicting the soil to go to pot. Soil erosion and occasional water excellent due to erosion and floor runoff have come to be intense issues international.
Soil erosion is a obviously taking place system that influences all landforms. In agriculture, soil erosion refers to the wearing away of a discipline's topsoil via the natural physical forces of water and wind or thru forces associated with farming sports including tillage.
Soil erosion with the aid of Water Liquid water is the most important agent of erosion on the earth. Rain, rivers, floods, lakes, and the ocean deliver away bits of soil and sand and slowly wash away the sediment. Rainfall produces 4 styles of soil erosion: splash erosion, sheet erosion, rill erosion, and gully erosion.
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What is the lowest level of organization to include both biotic and abiotic factors? a. organism b. ecosystem c. population d. biome please select the best answer from the choices provided a b c d
Ecosystem is the name given to the lowest level of structure that includes both biotic and abiotic elements. B is the right answer, thus.
The ecosystem is typically thought of as the lowest level of structure that includes nonliving (abiotic) elements. That would imply that abiotic influences are present in ecosystems, biomes, and the biosphere.
An ecosystem is a region where a bubble of life is created by plants, animals, and other species interacting with the weather, environment, and other factors. In ecosystems, biotic and abiotic factors—or nonliving components—coexist. Biotic factors include plants, animals, and other species.
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give atleast 5 name of seed
Answer:
Sunflower Seeds
Chia Seeds
Flax
Pumpkin Seeds
Sesame
Pleaeeee help me I love u so much if I do :) your the besttttttt
you need vitamins to use the carbohydrates, fats, and proteins you consume and to sustain numerous chemical reactions in your body.
True, B vitamins help human bodies utilize minerals, produce new red blood cells, use carbohydrates, protein, and fat, and maintain a healthy nervous system. The three most well-known B vitamins are thiamin, riboflavin, and niacin. Everybody has to take B-Complex vitamins every day.
Thiamin- Vitamin B1
Riboflavin- Vitamin B2
Niacin- Vitamin B3
Red Blood Cells (RBC)- Red blood cells are the cells responsible for the oxygen transport to all the cells of the living organism. RBC have hemoglobin which is responsible for red color in blood.
B-Complex vitamins contains 12 different types of Vitamin B which also includes, B1, B2, and B3.
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one of the 20 codis core loci strs used in examining noncoding dna sequences is d1s1656 and has the sequence (gata)n. from a statistical approach, how often would you expect this dna sequence to appear in a noncoding region of human dna that has relatively equal distributions of the four nucleotides (i.e., it is not g/c or a/t rich)?
To determine how often we would expect the DNA sequence (GATA)n to appear in a noncoding region of human DNA with relatively equal distributions of the four nucleotides, we can consider the probability of each nucleotide occurring at each position in the sequence. The probability of the sequence (GATA) occurring at a specific position is approximately 0.00390625, or about 0.39%.
The sequence (GATA)n consists of four nucleotides: G, A, T, and A. Since we are assuming relatively equal distributions of the four nucleotides in noncoding regions, each nucleotide has a 25% chance of occurring at each position.
Therefore, the probability of the sequence (GATA) occurring at a specific position is calculated as follows:
Probability = (Probability of G) * (Probability of A) * (Probability of T) * (Probability of A)
= 0.25 * 0.25 * 0.25 * 0.25
= 0.00390625
This means that the probability of the sequence (GATA) occurring at a specific position is approximately 0.00390625, or about 0.39%.
The number of repeats, represented by 'n', will affect the overall probability of the complete sequence appearing. If the number of repeats is known, we can multiply the probability calculated above by the number of repeats to estimate the overall frequency of the sequence (GATA) in noncoding regions of human DNA.
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What happens to the air pressure of an air parcel as it flows up the side of a mountain?
Answer:
it reduces,
Explanation:
bcos pressure decreases with altitude or heights.
Please help, appreciated if you do :) Which of these accurately describes the change in the muscle cells of a person running a marathon? Decrease in the number of mitochondria Increase in myoglobin Increase in the number of skeletal muscles Decrease in myoglobin
Answer:
Hey there!
The correct answer would be a Decrease in myoglobin. Myoglobin contains a large amount of oxygen, and desaturates when muscles move. When you run a marathon, your muscles move and thus you have decreased myoglobin.
The change in muscle cells of person a Decrease in myoglobin.
What is Myoglobin?Myoglobin contains a large amount of oxygen, and desaturates when muscles move. When you run a marathon, your muscles move and thus you have decreased myoglobin.
striated muscles, which include your cardiac muscles and skeletal muscles (the muscles linked to your bones and tendons), contain the protein known as myoglobin. Its primary job is to give your muscles' cells oxygen (myocytes).
body's cells require oxygen to function. They transform the stored energy using oxygen. Due to their regular use, your cardiac muscles and skeletal muscles need a lot of oxygen and energy.
Therefore, The change in muscle cells of person a Decrease in myoglobin.
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You are studying a population of Dlugosch’s Small Octopus that has been suffering from a skin disease. In your work, you find that some individuals have more resistance (immunity) to the disease than others. You remember learning about life history trade-offs in your Ecology class, and you hypothesize that immunity trades off with investments in other major life history traits in the octopuses. You decide to design a study to test whether higher immunity leads to a lower number of eggs produced by females.
9. State a prediction for what the results of your study would look like if they support the hypothesis given in the Mission.
Make a figure of results that would be consistent with that prediction:
Label y axis with the variable that should be on the
Label x axis with the variable that should be on the
Draw results bars or lines consistent with your prediction
If the infection reduces the population size of the octopus, would you predict an r or a K strategy would be more likely to evolve (based on what we discussed in class)? Explain why that life history strategy would be favored in a small population, and give at least two traits associated with the strategy that you predict.
If higher immunity leads to a lower number of eggs produced by females in Dlugosch's Small Octopus, then there will be a negative correlation between immunity level and egg production.
A K approach would be more likely to develop if the illness causes the octopus population to decline. Resources are few in a tiny community, therefore long-term stability and survival are prioritized over rapid population expansion. The K approach emphasizes having fewer kids while allocating more resources to ensuring their survival and welfare. In a small population, the following two characteristics connected to a K strategy might be more prevalent:
Increased parental care: Parents invest more time and resources in raising a smaller number of offspring to increase their chances of survival and reproductive success.
Delayed maturation: Individuals take longer to reach reproductive maturity, allowing them to invest more time in growth, development, and building up resources before reproduction.
These traits help ensure the survival and success of the limited number of offspring in a small population, aligning with the K-selected life history strategy.
In this figure, the y-axis represents the number of eggs produced by females, and the x-axis represents the immunity level of the octopuses. The bars or lines would show a decreasing trend as the immunity level increases, indicating a negative relationship between immunity and egg production.
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Plenty of food, water and suitable temperature are the conditions required for the growth of fungi. Is it true or false?
Explanation:
Some species of fungi do grow better at warm temperatures (70-90°F), but there are some that thrive in very high temperatures of 130-150°F and some that will thrive in very low temperatures below 32°F (below freezing). ... Water: Fungi need lots of water to grow. For most fungi this is true. :)
Suppose you wanted to investigate the sustainability of agriculture in California using the triple bottom line. What types of questions would you ask in each domain? What data could you collect to answer those questions?
A sustainable agriculture is one that improves the environmental quality and resource base on which agriculture is based over the long term;" accommodates essential human food and fiber needs; is financially attainable; and improves farmers' and society's overall quality of life."
The "three legs" of the sustainable agriculture—environmental health, economic viability, and social and economic equity—are the fundamental objectives of sustainable agriculture.
Permaculture, agroforestry, mixed farming, multiple cropping, and crop rotation are all examples of sustainable agricultural practices.
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answer multiple choice: 1- If you water your palms, do so in the:
a. night b. morning 0. afternoon d. any time
2- Like animals, fungi are a. autotrophic b. heterotrophic c. mesotrophic d. all the answers
is true 3- The Oomycota are now members of the kingdom Chromista rather than Fungi but continue to be treated as fungi because
of: a. their many other similarities to them b. not similarities to them 0. have cell walls composed of chitin d. have cell wells composed
of cellulose. 4- The pathogen attacks other host (barberry) to complete its
life cycle, this is: a. fusarium poxydoms b. Puccini graminet tritica c. Ustilago nuda tritici d. Helminrhosporium maydis 5- Downy mildew effected the leaves surface of:
a. wheat b. palm c. maize d. barely 6- the symptoms of leaf blight its; a. the effect at young stage b. small yellowish round to oval spots
on leaves 0. a and b correct d. dark and brown color 7- Net blotch that symptoms, can be extensive in winter but affected leaves die back and new leaves in spring can be
symptom-free caused in: a. barley b. maize c. palm d. wheat 8- Sesame grown since antiquity for its: a. leaves b. seeds 0. roots (1. all the answers its true
9-The Sesame seeds are also high in: a. protein b. Vitamin B6 c. thiamin d. all the answers its true ... Show more"
1. Watering palms can be done at any time.
2. Fungi are heterotrophic organisms.
3. Oomycota, although now classified under the kingdom Chromista, are still treated as fungi due to their many similarities.
4. The pathogen that completes its life cycle by attacking another host (barberry) is Helminrhosporium maydis.
5. Downy mildew affects the leaves of maize.
6. The symptoms of leaf blight include small yellowish spots on leaves.
7. Net blotch, which can cause extensive symptoms in winter but symptom-free new leaves in spring, is found in barley.
8. Sesame is primarily grown for its seeds.
9. Sesame seeds are high in protein, Vitamin B6, and thiamin.
1. When it comes to watering palms, it can be done at any time without a specific preference for a particular period of the day.
2. Fungi, unlike autotrophic organisms, are heterotrophic, meaning they obtain nutrients by consuming organic matter.
3. Oomycota, previously classified as fungi but now considered part of the kingdom Chromista, are still treated as fungi due to their numerous similarities with true fungi.
4. The pathogen Helminrhosporium maydis attacks barberry plants as part of its life cycle.
5. Downy mildew affects the surface of maize leaves, causing damage and potential crop loss.
6. Leaf blight is characterized by small, round to oval, yellowish spots on leaves, typically occurring at a young stage of plant development.
7. Net blotch, a fungal disease, can cause extensive symptoms in winter but does not affect new leaves in spring, particularly in barley plants.
8. Sesame is primarily cultivated for its seeds, which are rich in oil and have various culinary uses.
9. Sesame seeds offer nutritional benefits, as they are high in protein, Vitamin B6, and thiamin, providing essential nutrients for a healthy diet.
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The ABO blood group is a classification of blood depending on the presence of how many antigens?
Answer:
a. two
Explanation:
a). Explain the meaning of the term gas exchange surface in human lungs.
b). List three features of gas exchange surfaces.
c). Explain how each feature in your list helps gas exchange to happen efficiently.
Answer:
a)Gas exchange is the process by which respiratory gases pass across respiratory surfaces
b) 1. They are moist to prevent cells to prevent cells from drying and dissolve gases
2.They have a large surface area so that a lot of gas can diffuse across at the at the same time
3.They have a high concentration gradient maintained by the movement of air and blood
Explanation:
Which of the following would you expect to see in a positive feedback system?
1. a reduction in output
2. a magnification of output
3. a stabilization of output
4. a movement toward equilibrium
Answer: a magnification of output
Explanation: positive feed back means that you should be able to do the same amount of work but have a greater output
You are performing a recombinant DNA experiment using a restriction enzyme that makes staggered cuts and using the GFP gene as a reporter. The plasmid DNA gets cut properly, but the different pieces do not rejoin properly. Most likely, you have a defective __________________.
The defective component in the recombinant DNA experiment with staggered cuts and the GFP gene as a reporter is most likely the ligase enzyme responsible for joining the DNA fragments.
In a recombinant DNA experiment, the use of restriction enzymes helps in making precise cuts in the plasmid DNA. These enzymes create staggered cuts that produce single-stranded overhangs or sticky ends. The next step involves the use of a ligase enzyme, which catalyzes the joining of these DNA fragments by creating phosphodiester bonds. However, in this scenario, the different pieces of DNA fail to rejoin properly, indicating a defect in the ligase enzyme.
Defective ligase enzymes can be caused by various factors such as incorrect folding, mutations in the gene encoding the ligase, or the absence of essential cofactors required for its activity. The malfunctioning ligase enzyme prevents the proper sealing of the DNA fragments, leading to an inability to generate recombinant DNA molecules.
To troubleshoot this issue, it is recommended to verify the integrity and functionality of the ligase enzyme used. Additionally, ensuring the presence of all necessary cofactors and optimal reaction conditions may help in resolving the problem.
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Ocho vw from the main vide
Data
Activity 1
1. Ddel produces sticky ends. For the wild-type beta-globin sequence, how many DNA fragments are present in the following digestion by Dder
2. For the mutant beta-globin sequence, how many DNA fragments are present in the following digestion by Ddel? If the number of fragments is different than that of the wild- type beta-globin sequence following Ddel digestion, explain why.
3. Count the number of base pairs for each fragment. Remember that a base pair includes two bound nucleotides. If a base is not paired because of a sticky end, do not count it. This count determines the approximate size of the fragment.
a. For the Ddel-digested, wild-type beta-globin sequence, how many nucleotides are in each fragment?
b. For the Ddel-digested, mutant beta-globin sequence, how many nucleotides are in each fragment?
4. On the basis of fragment size, how can the difference between the wild-type sequence and the homozygous mutant sequence be recognized?
5. What fragments would be present following Ddel digestion of a sample from someone with a heterozygous beta-globin genotype?
6. On the gel diagram, indicate where the DNA fragment(s) in the Ddel-digested beta-globin samples would be expected to run. Draw a line for each band. Use the DNA marker with known base pair sizes to orient the bands. A Ddel-digested sample from a newborn with an unknown beta-globin genotype is drawn in the last lane.
1. DdeI is a restriction enzyme that recognizes the sequence 5′-CTNAG-3′. It cleaves the double-stranded DNA within its recognition sequence, leaving a single-stranded overhang (sticky end) at each end. In the given wild-type beta-globin sequence (5' CTG ACT CCT GAG 3'), there is only one recognition sequence of DdeI. It is CTGA CTC, which is present between the seventh and eighth base pairs. The recognition sequence has five nucleotides. DdeI will cut the DNA at this site, producing two fragments. Therefore, the Ddel digestion of the wild-type beta-globin sequence produces two fragments: 5'-CTGACTC-3' and 5'-CTGGAG-3'.
2. In the mutant beta-globin sequence (5' CTG ACT CCT GTG 3'), the recognition site for DdeI is changed by one nucleotide from 5'-CTGA CTC-3' to 5'-CTG ACC-3'. The mutated sequence does not contain a recognition sequence for DdeI, so there will be no fragments produced by digestion.
3. To calculate the number of nucleotides in each fragment, you need to count the number of base pairs in each fragment. A base pair consists of two nucleotides. For the DdeI-digested wild-type beta-globin sequence, there are two fragments. The first fragment is 8 base pairs long, and the second fragment is 1 base pair long. So, for the digested fragment there are 19 nucleotides in one fragment and 5 in the other fragment.
5' CTG ACT CC T GAG 3'
3' GAC TGA GGA CT C 5'
The mutant fragment has no restriction site so there are 12 base pair and 24 nucleotides.
4. The difference between the wild-type and mutant beta-globin sequences can be recognized by the number of fragments produced by DdeI digestion. The wild-type sequence produces two fragments, while the mutant sequence does not produce any fragments Each fragment migrates at a different speed, placing the fragments at different distances.
5. If someone has a heterozygous beta-globin genotype, one of the chromosomes will be wild-type, and the other will be mutant. DdeI digestion of this sample will produce three fragments: two fragments from the wild-type chromosome and no fragments from the mutant chromosome. The sizes of the fragments will be the same as those produced from the wild-type sequence.
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How is cell differentiation different than mitosis?
Answer:Cell differentiation is the process by which cells specialize to achieve their required functions. In order for a cell to differentiate, it will express specific genes. mitosis produces cells that are identical to each other (clones).
Explanation:
Answer:
Though the options aren't presented to the question, I believe the correct answer is: Cell differentiation is different than mitosis, for during cell differentiation, it produces cells with different gene expressions. Meanwhile, during mitosis, it produces clones of the same genes.
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