(1) Linkage disequilibrium refers to the non-random association of alleles at different loci, while haplotype blocks are contiguous stretches of DNA with low recombination rates that tend to be inherited together.
(2) The Hardy-Weinberg distribution helps determine the expected frequencies of alleles and genotypes in a population by predicting how alleles combine to form genotypes in a population that is at equilibrium.
The Explanation to Each AnswerThese concepts are useful in human population genetics as they can provide information on the genetic diversity, structure, and history of populations, as well as the identification of disease-causing variants and the design of efficient genetic association studies.This question should be provided as:
Explain the concepts of linkage disequilibrium and haplotype blocks and discuss why they are useful in human population genetics. Explain how the Hardy-Weinberg distributions helps determine the frequency of alleles and genotypes in a population.
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Suzie has a sore throat, fever, difficulty swallowing and a rash that covers neck and chest. A gram positive cocci bacteria, found in chains, is isolated from a swab. It has the following virulence factors: a. M proteins - b. hyaluronic acid capsule - c. Streptokinases \& streptolysins- d. Cs peptidase e. pyrogenic toxins carried ny a prophage-
f. identify this pathogen -
The pathogen that is causing Suzie's symptoms is Streptococcus pyogenes. This gram-positive cocci bacteria is known for its ability to cause a wide range of infections, including strep throat, scarlet fever, and impetigo.
The virulence factors that are associated with Streptococcus pyogenes include M proteins, which help the bacteria to adhere to host cells and avoid phagocytosis; a hyaluronic acid capsule, which also helps the bacteria to avoid phagocytosis; streptokinases, which break down blood clots and help the bacteria to spread; streptolysins, which destroy red blood cells and other host cells; C5 peptidase, which degrades complement proteins and helps the bacteria to evade the immune system; and pyrogenic toxins, which are carried by a prophage and can cause fever and other symptoms.
Based on Suzie's symptoms and the presence of these virulence factors, it is likely that she is suffering from an infection with Streptococcus pyogenes. Treatment may include antibiotics to kill the bacteria and alleviate symptoms.
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Question 14 OT 25 = View Policies Current Attempt in Progress A patient blood sample has been brought to the microbiology lab for potential pathogen identification. Identification is accomplished by (
A patient blood sample has been brought to the microbiology lab for potential pathogen identification which is accomplished by:
Performing rapid biochemical testing (A)Growing the sample on selective and/or differential media (B)Performing antibiotics sensitivity testing (D)А pаthogen brings diseаse to its host. Аnother nаme for а pаthogen is аn infectious аgent, аs they cаuse infections. Аs with аny orgаnism, pаthogens prioritize survivаl аnd reproduction. The humаn body’s immune system аcts аs а defense аgаinst pаthogens. There аre five mаin types of pаthogens: bаcteriа, viruses, fungi, protists, аnd pаrаsitic worms.
Option C can't identification because molds and yeast aren't in a patient blood sample.
Your question is incomplete, but most probably your full options were
A. Performing rapid biochemical testing
B. Growing the sample on selective and/or differential media
C. Comparing the colony characteristics with the lab reference for molds and yeast
D. Performing antibiotics sensitivity testing
Thus, the correct options are A, B, and D.
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Identify the stage of mitosis each lettered plant cell is in:
Answer:
Metaphase
Explanation:
So, the correct answer is Metaphase, chromosomes moved to spindle equator, chromosomes made up of two chromatids
Imagine that you are the lead scientist studying whether or not a change in island lizard population would be evidence for gene flow. What is the most appropriate hypothesis for this experiment in terms of the results you would expect? After a gene flow event,
Rare alleles may be lost due to genetic drift, which can also reduce the size of a gene pool.The idea that genetic drift contributes to the creation of novel species is based on the fact that it can also make a new population genetically different from its ancestral population.
How can genetic drift impact a population's genetic diversity?Due to random sampling error, small populations typically lose genetic diversity less quickly than huge populations (i.e., genetic drift).This is because small populations make it more likely for some gene variants to be lost as a result of random chance.
What makes a population with a diverse genetic make-up more able to adjust to environmental changes?By retaining a substantial genetic diversity, organisms can adapt to shifting environmental conditions and avoid inbreeding.Inbreeding occurs in small, isolated groups, which may render a breed less able to survive and reproduce.
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A cell uses active transport to move sodium ions out of the cell. Why does the cell use active transport instead of diffusion to move sodium ions out of the cell?
Answer options:
There is a lower concentration of sodium ions inside the cell than outside the cell.
The cell uses less energy when performing active transport.
Diffusion can only transport water across the cell membrane.
Diffusion requires proteins along the cell membrane to transport ions outside the cell.
The cell uses active transport instead of diffusion to move sodium ions out of the cell because there is a lower concentration of sodium ions inside the cell than outside the cell. Therefore, the correct option is A.
What is active transport?Active transport is used to move molecules or ions against their concentration gradient, from an area of lower concentration to an area of higher concentration.
In this case, the cell is moving sodium ions out of the cell, where the concentration of sodium ions is higher outside the cell than inside the cell.Therefore, if the concentration of sodium ions inside the cell was already higher than outside the cell, the ions would move out of the cell via diffusion instead of active transport.Hence, the correct option is A.
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The question is incomplete, but most probably the complete question is,
A cell uses active transport to move sodium ions out of the cell. Why does the cell use active transport instead of diffusion to move sodium ions out of the cell?
Answer options:
A. There is a lower concentration of sodium ions inside the cell than outside the cell.
B. The cell uses less energy when performing active transport.
C. Diffusion can only transport water across the cell membrane.
D. Diffusion requires proteins along the cell membrane to transport ions outside the cell.
During which phase would penicillin, an antibiotic that inhibits cell-wall synthesis, be most effective?
A. death phase
B. lag phase
C. log phase
D. stationary phase
Penicillin, an antibiotic that inhibits cell-wall synthesis, would be most effective during the Log Phase. Option B.
Penicillin is an antibiotic that is used to prevent or treat bacterial infections. Penicillin, an antibiotic that inhibits cell-wall synthesis, would be most effective during the Log Phase.
In the log phase, the growth rate of the bacteria is constant, and the cells are rapidly dividing. In the presence of nutrients, bacteria continue to grow exponentially, and the population doubles at a constant rate. During this phase, bacteria are more susceptible to antibiotics and are easily killed because they are actively growing and dividing.
Because penicillin inhibits cell-wall synthesis, it is more effective in preventing bacterial cell growth and division during the Log Phase. Therefore, penicillin, an antibiotic that inhibits cell-wall synthesis, would be most effective during the Log Phase.
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food that has one of the highest contents of those vitamins. What would you choose?
Beans
Bread made with enriched flour
Fish
Beef liver
No EAR and DA have been established for biotin because the requirements are very low, in the microgram range
O True
• False
Foods that have one of the highest contents of vitamins, the correct choice would be beef liver. Beef liver is an excellent source of several vitamins, including vitamin A, vitamin B12, and vitamin B6.
Beans, enriched bread, and fish are all healthy options, but they can't compare to cow liver when it comes to vitamin richness.
Regarding the statement about biotin, it is true that no EAR (Estimated Average Requirement) and DA (Dietary Allowance) have been established for biotin because the requirements are very low, in the microgram range.
Biotin is a vitamin that is needed in small amounts to help the body convert food into energy.
Because the requirements for biotin are so low, it is typically not a concern for most people to meet their biotin needs through their diet.
Therefore, the correct choice would be beef liver regarding foods that have one of the highest contents of vitamins.
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Your friend decides to set up an experiment with four different groups with ten mice each. Each group of mice will have their respiration rate tested at a specific time of the day - Moening ( 6 AM), Noon (12 PM), Evening (6 PM) and Night (12 AM). Fach mouse was placed in the respirometer for a penod of five minutes. 5. If your friend's predietion (and hypothesis) is supported, which group of mice will have the highest respiratory rate?' (Circle ONE)' A. The Morning group B. The Day group C. The Evening group D. The Night group
Answer: A
Explanation:
A. The Morning group. Mice typically exhibit higher levels of physical activity during the morning, so it is likely that their respiratory rate will be the highest at this time.
Where would a frameshift mutation cause the most damage?
A.Near the very end of the gene
B.Near the beginning of the gene
C.Near the center of the gene
D.All frameshift mutations are equal
Down syndrome is a genetic disorder that is also called trisomy 21. A person with Down Syndrome has an extra copy of chromosome 21. What can you infer is most likely the genetic mutation that results in Down syndrome?
A.Nondisjunction during meiosis, resulting in uneven distribution of chromosomes
B.Complete duplication of chromosomes during polyploidy
C.Translocation during genetic replication, creating nonhomologous chromosomes
D.Crossing over during meiosis, leading to an exchange of genetic information
Which of the following is an example of a silent mutation?
A.A codon mutated from CCG to CCA, both coding for glycine
B.A substitution of glycine amino acid for a stop codon
C.A substitution of a serine amino acid for a glycine amino acid
D. An insertion or deletion of nucleotides that is not a multiple of three
1 - A frameshift mutation causes the most damage B. Near the beginning of the gene.
2- The most likely genetic mutation that results in Down syndrome is A. Nondisjunction during meiosis, resulting in uneven distribution of chromosomes.
3 - An example of a silent mutation is A. A codon mutated from CCG to CCA, both coding for glycine.
1 - A frameshift mutation is a type of genetic mutation that occurs when one or more nucleotides are inserted or deleted from a gene. This causes the reading frame of the gene to shift, leading to changes in the sequence of amino acids that are encoded by the gene. If a frameshift mutation occurs near the beginning of the gene, it can have a greater impact on the protein that is produced, as it will affect a larger portion of the protein's sequence. Therefore, the correct answer is B. Near the beginning of the gene.
2 - Down syndrome is caused by an extra copy of chromosome 21, which is typically the result of nondisjunction during meiosis. Nondisjunction is a type of genetic mutation that occurs when chromosomes do not separate properly during cell division, leading to an uneven distribution of chromosomes in the resulting cells. This can result in an extra copy of a chromosome, as is the case with Down syndrome. Therefore, the correct answer is A. Nondisjunction during meiosis, resulting in uneven distribution of chromosomes.
3 - A silent mutation is a type of genetic mutation that does not result in a change in the amino acid sequence of a protein. This can occur when a mutation changes one of the nucleotides in a codon, but the new codon still codes for the same amino acid. In the example provided, the codon CCG is mutated to CCA, but both codons code for the amino acid glycine. Therefore, this is an example of a silent mutation, and the correct answer is A. A codon mutated from CCG to CCA, both coding for glycine.
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How would the deletion of the Poly-A tail decrease protein expression?
A.) The spliceosome would incorrectly process the pre-mRNA
B.) The stability of the mRNA would be compromised in the cytoplasm
C.) Transcription factors wouldn't be able to bind to the DNA and recruit RNA polymerase for transcription
D.) Initiation factors would not be able to bind to the DNA and recruit DNA polymerase for transcription
The deletion of the Poly-A tail would decrease protein expression by (option b) The stability of the mRNA would be compromised in the cytoplasm.
The Poly-A tail is a sequence of adenine nucleotides that is added to the 3' end of the pre-mRNA molecule during the process of RNA processing.
This tail plays an important role in the stability of the mRNA molecule, as it protects the mRNA from degradation by exonucleases in the cytoplasm. If the Poly-A tail is deleted, the mRNA molecule will be more susceptible to degradation and will have a shorter half-life in the cytoplasm. This will result in decreased protein expression, as there will be less mRNA available for translation into protein.
In summary, the deletion of the Poly-A tail decreases protein expression by compromising the stability of the mRNA molecule in the cytoplasm.
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The deletion of the Poly-A tail would decrease protein expression is B.) The stability of the mRNA would be compromised in the cytoplasm.
The Poly-A tail is a long chain of adenine nucleotides that is added to the 3' end of pre-mRNA during RNA processing. The Poly-A tail serves several important functions in the cell, one of which is to protect the mRNA from degradation in the cytoplasm. Without the Poly-A tail, the mRNA would be more susceptible to degradation by ribonucleases, leading to a decrease in protein expression.
Option A is incorrect because the spliceosome is involved in the removal of introns from pre-mRNA, and the Poly-A tail does not play a role in this process. Option C is incorrect because transcription factors bind to the promoter region of the DNA, not the Poly-A tail. Option D is incorrect because initiation factors are involved in the initiation of transcription, not the stability of mRNA in the cytoplasm.
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2. Female flies with white eyes (wh) and vestigial wings (vg) were mated to males with roof wings (rf). The
F1
were all wild type. When the
F1
females were test crossed with recessive homorygous for all three traits males, the following progeny were observed: i) Determine the gene order ( 2 marks) The correct order of the loci is: ii) The genotype of
F1
was: ( 2 marks) (Chromosome 1) 4 (Chromosome 2) iii) The phenotypes "white-roof-vestigial" and "wild-Type" were formed as a result of a single crossover between (1 mark) and iv) Calculate the map distances between each pair of loci and draw the map. (Show your work 6 marks)
For 2 female flies:
The gene order is Chromosome 1: Wh-Rf and Chromosome 2: VgThe genotype of F1 is Chromosome 1: Wh-Rf/wh- rf and Chromosome 2: Vg/vgThe phenotypes are "white-roof-vestigial" and "wild-type" The map distance is 3.81 cMHow to determine gene order, phenotype and map distance?i) The gene order can be determined based on the observed progeny from the test cross. Since all the F1 females were wild type, they must have received at least one dominant allele for each of the three traits from their father. This means that the dominant alleles for roof wings (Rf) and wild type eyes (Wh) must be linked on one chromosome, while the dominant allele for vestigial wings (Vg) must be on the other chromosome. Therefore, the correct order of the loci is:
Chromosome 1: Wh-Rf
Chromosome 2: Vg
ii) Since all the F1 females were wild type, they must have received at least one dominant allele for each of the three traits from their father. Therefore, the genotype of the F1 must be:
Chromosome 1: Wh-Rf/wh- rf
Chromosome 2: Vg/vg
iii) The phenotypes "white-roof-vestigial" and "wild-type" were formed as a result of a single crossover between the loci for roof wings and wild type eyes (Rf-Wh) and the locus for vestigial wings (Vg). The crossover must have occurred between these two loci in the F1 female that was test crossed, resulting in two different gametes: one with the dominant alleles for roof wings and wild type eyes, and the recessive allele for vestigial wings (Rf-Wh vg), and the other with the recessive alleles for all three traits (rf wh vg). Therefore, the phenotypes "white-roof-vestigial" and "wild-type" were formed as a result of a single crossover between the loci for Rf-Wh and Vg.
iv) To calculate the map distances between each pair of loci, use the formula:
Map distance = (Number of recombinant offspring / Total number of offspring) x 100
Use the following data to calculate the map distances:
Recombinant progeny:
wh rf vg: 16
Wh Rf vg: 9
Non-recombinant progeny:
wh Rf vg: 312
Wh rf vg: 318
Total number of offspring = 655
Map distance between Wh and Rf:
Map distance = (9/655) x 100 = 1.37 cM
Map distance between Rf and Vg:
Map distance = (16/655) x 100 = 2.44 cM
Map distance between Wh and Vg:
The map distance between Wh and Vg can be calculated by adding the map distances between Wh and Rf and between Rf and Vg:
Map distance = 1.37 cM + 2.44 cM = 3.81 cM
Therefore, the map for the three loci would look like:
Chromosome 1: ----Wh(1.37 cM)---Rf(1.37 cM)----
Chromosome 2: ----Vg(3.81 cM)----
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in the RSS separating sequences palindromes are found? a. only in 1 turn spacer sequences b. only i 2 turn spacer sequences c. in both 1 and 2 turn spacer sequences d. flanking the spacers specially i heptamer regions e. in the heptamers and 1 turn spacer sequence regions of 1 turn RSS
in the RSS separating sequences palindromes are found is c. in both 1 and 2 turn spacer sequences.
Palindromes are sequences of DNA that read the same forwards and backwards. These sequences are found in both 1 and 2 turn spacer sequences in the RSS (recombination signal sequence) of the DNA. The RSS is a specific sequence of DNA that is recognized by the RAG (recombination activating gene) proteins and is important for the process of V(D)J recombination in the immune system.
The presence of palindromes in the RSS allows for the formation of hairpin structures, which are important for the recombination process. Therefore, palindromes are found in both 1 and 2 turn spacer sequences of the RSS.
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During chromosome duplication, a copy of a chromosome is produced.
These "two"
copies are referred to as ____.
a. chromatin
b. homologous chromosomes
c. sister chromatids
d. gametes
During chromosome duplication, a copy of a chromosome is produced. These "two" copies are referred to as sister chromatids. The correct choice c, which refers to sister chromatids.
During the S phase of the cell cycle, as the DNA is being copied to make two identical copies of each chromosome, chromosomes duplicate themselves. This process is known as chromosomal duplication.
These two copies of the genetic material are referred to as sister chromatids, and they remain together at the centromere until the process of cell division causes them to become divided.
Sister chromatids are genetically identical to the original chromosome and to each other. They also have the same appearance.
Consequently, the response that is appropriate to this question is choice c, which refers to sister chromatids.
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While we primarily learned about how fertilization occurs in mammals, many people want to prevent fertilization from occurring unexpectedly. In fact, preventing fertilization, in the form of birth control, is a large and competitive industry. Contraception currently focuses on physical barriers to fertilization, which can be used by either sex, or on medications or devices that regulate ovulation.
Describe a process necessary for fertilization that might be a target for a sperm/male-specific contraceptive medication?
What aspect(s) of fertilization in humans make it difficult to produce sperm/male-specific contraceptive medication?
One process necessary for fertilization that might be a target for a sperm/male-specific contraceptive medication is sperm motility.
Sperm motility is the ability of sperm to move and swim through the female reproductive tract in order to reach the egg for fertilization. If a medication can be developed to reduce or inhibit sperm motility, it could potentially prevent fertilization from occurring. One aspect of fertilization in humans that makes it difficult to produce sperm/male-specific contraceptive medication is the fact that there are millions of sperm released during ejaculation. This means that even if a medication is able to reduce the motility of a large percentage of the sperm, there may still be enough sperm that are able to reach the egg and fertilize it. Additionally, the process of fertilization is complex and involves multiple steps, so targeting just one aspect of the process may not be enough to effectively prevent fertilization.
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The rate of migration of a protein in SDS-PAGE is NOT influenced by
Size of the protein
Strength of the electric field
Charge of the protein
Pore size of the gel
The rate of migration of a protein in SDS-PAGE is not influenced by the charge of the protein.
Thus, the correct answer is the charge of the protein (C).
SDS polyаcrylаmide gel (SDS-PAGE) is а type of аnаlyticаl technique thаt is used to sepаrаte proteins. SDS pаge is а type of electrophoresis technique thаt sepаrаtes substаnces on the bаsis of size. The rаte of migrаtion through in SDS polyаcrylаmide gel will not be influenced by chаrge in the protein. SDS, thаt's аdded in the gel is responsible for negаting the chаrge of аll proteins.
Even though the ionic detergents SDS hydrolyzes аnd binds to molecules to mаke them evenly negаtively chаrged, SDS-PАGE predominаntly sepаrаtes proteins by mаss. Аll SDS-bound molecules in а sаmple will therefore move аcross the gel аnd towаrds the positive chаrged electrode whenever а current is delivered.
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PLEASE HELP
b) State one scientific question. (2 points)
c) State one nonscientific question. (3 points)
a.) An example of scientific question is, does the effect of smoking increase the rate of lung cancer in a population?
b.) An example of non-scientific question is, why does the earth exist?
What are scientific and non scientific questions?A scientific question is the type of question that contains dependent and independent variables which can be proven by experimental processes and methods.
Example of scientific question is does the effect of smoking increase the rate of lung cancer in a population?
A non scientific question is the type of question that cannot be proven by any scientific means.
An example of a non scientific question is, Why does the earth exist?
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In humans, the "smeller"-tasting locus has two alleles, S and s, where S is dominant and is the "smeller" allele; s is recessive and is the "non-smeller" allele. Suppose in a population of 500 people, 180 have the "smeller" phenotype. The population is in Hardy-Weinberg equilibrium with respect to this locus (6 points) a. What is the frequency of the S allele? b. What is the frequency of the s allele? c. How many people have the SS genotype? d. How many people have the Ss genotype? e. How many people have the ss genotype? f. Suppose that the next generation has 600 people. How many are predicted to be "smellers"?
a. The frequency of the S allele can be calculated by taking the square root of the proportion of "smeller" individuals in the population. Therefore, the frequency of the S allele is √(180/500) = 0.6.
b. The frequency of the s allele can be calculated by subtracting the frequency of the S allele from 1. Therefore, the frequency of the s allele is 1 - 0.6 = 0.4.
c. The frequency of the SS genotype can be calculated by squaring the frequency of the S allele. Therefore, the frequency of the SS genotype is (0.6)² × 500 = 180.
d. The frequency of the Ss genotype can be calculated by multiplying the frequencies of the S and s alleles and then doubling the result (since there are two possible ways to obtain the Ss genotype). Therefore, the frequency of the Ss genotype is 2 × 0.6 × 0.4 × 500 = 240.
e. The frequency of the ss genotype can be calculated by squaring the frequency of the s allele. Therefore, the frequency of the ss genotype is (0.4)² × 500 = 80.
f. If the population is still in Hardy-Weinberg equilibrium, the frequency of the S allele will remain the same. Therefore, the predicted number of "smellers" in the next generation can be calculated by multiplying the total population size by the frequency of the SS and Ss genotypes, which are 180/600 and 240/600, respectively.
Thus, the predicted number of "smellers" in the next generation is (180/600 + 240/600) × 600 = 420.
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A graduate student sets out to study the mode of inheritance of widow's peak in humans. To do so, they screened individuals from 1st, 2nd, and 3rd year students in biology. (Assume that each student was screened once, in other words each student is only taking one of the courses). The results are listed here:
- 1339 students had a widow's peak
- 460 students without a widow's peak
a. Are the results of the screen consistent with mendelian ratios expected from of monohybrid (F1 X F1) cross if widow's peak is the dominant phenotype and not having a widow's peak is the recessive phenotype? Perform a chi2 analysis and provide your conclusion.
b. Is the design of the experiment appropriate to make such a conclusion? If yes, describe how the experiment is appropriate to make such a conclusion. If no, explain the reason and describe how the experiment could be conducted to provide conclusive evidence for the genetic basis of the widow's peak phenotype.
The results from a monohybrid (F1 X F1) cross are not consistent if the difference between observed and expected values is significant. The experimental design is also inappropriate to draw such a conclusion.
a. In order to answer this question, a Chi-squared analysis should be conducted. This involves calculating the observed number of widow's peak phenotypes (1339) and the expected number of widow's peak phenotypes from a monohybrid cross (F1 X F1) with a dominant and recessive phenotype (900).
If the difference between the observed and expected values is significant, then the results of the screen are not consistent with the Mendelian ratios expected from a monohybrid (F1 X F1) cross.
b. The design of this experiment is not appropriate to make such a conclusion, as the sample size of individuals screened (1899) is too small to draw meaningful conclusions. To make a conclusion, a larger sample size should be screened to ensure that the results are not influenced by any outliers.
Additionally, the experiment should also include a control group to compare the widow's peak phenotype with a known genetic basis. By doing so, the experiment can provide conclusive evidence for the genetic basis of the widow's peak phenotype.
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In a particular species of mice, a single gene (B or b) determines tail length. The short-tail allele is sex-linked and dominant. Shading in the pedigree below indicates that an individual has a short tail. The pedigree shows the pattern of inheritance of the short-tail allele over three generations of mice.
Individual 6 is female. What is the genotype of individual 6?
XBXb
XbXb
XBY
XbY
In a particular species of mice, individual 6, who is female, has the genotype XBXb. Individual 6 is female. Therefore, the correct answer is the first option.
The pedigree given above shows that a single gene B or b decides the tail length of a particular species of mice. The short-tail allele is dominant and sex-linked. In the pedigree, shaded individuals indicate that the mice have short tail.
The following conclusions can be drawn from the pedigree:
For males, the X chromosome and the Y chromosome determine their sex. On the other hand, females have two X chromosomes. For males, the presence of the B allele on their X chromosome will decide whether they have a short tail or a long tail. In females, the presence of the B allele on one or both X chromosomes will determine whether they have a short tail or a long tail. If there is no B allele on any of their X chromosomes, they will have a long tail.In the pedigree, Individual 6 is female, and she is shaded, which means she has a short tail. Individual 6 is female, so she has two X chromosomes. She has a short tail, which means that she must have the dominant short-tail allele. If she had B alleles on both X chromosomes, she would have been BB, which is not possible because there are no BB females in the pedigree. If she had no B alleles on both X chromosomes, she would have been bb, which is also not possible because she has a short tail. Therefore, the genotype of individual 6 is XBXb, which means she has a short tail because of the B allele on one of her X chromosomes.
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Part A- Testing for Mono and Disaccharides
1. Turn an electric plate on high and place a 500 mL beaker half full of water, to make a hot water bath (about 80 degrees Celsius). 2. Measure 3 mL of water and of each of the provided solutions (Not #5,#6 or #7) using a graduated cylinder. Place in clean test tubes and label each tube.
3. Add 15-20 drops of Benedict’s Solution to each test tube (this is about 1mL).
Part A- Testing for Mono and Disaccharides:
1. Turn an electric plate on high and place a 500 mL beaker half full of water, to make a hot water bath (about 80 degrees Celsius).
2. Measure 3 mL of water and of each of the provided solutions (Not #5,#6 or #7) using a graduated cylinder. Place in clean test tubes and label each tube.
3. Add 15-20 drops of Benedict's Solution to each test tube (this is about 1mL).
4. Place the test tubes into the hot water bath and leave for a few minutes. Observe the color of the solution in the test tubes and compare it to the color chart.
5. If the solution changes color (other than to a yellow color), then a monosaccharide or disaccharide is present. The more intense the color, the greater the amount of monosaccharide or disaccharide.
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By law of --- ---, the relationship between free (unoccupied) and bound receptors can be described as follows: _____or re-arranged:_____If we let Ro = total number of receptors, then ____Replace [R] by [Ro] - [LR] and re-arrange above equations:____ where [LR]/Ro is the --- of --- --- --- by ---.
By the law of mass action, the relationship between free (unoccupied) and bound receptors can be described as follows: [L][R] = Kd[LR] or re-arranged: [LR] = ([L]/Kd)[R]. If we let Ro = total number of receptors, then [R] = Ro - [LR]. Replace [R] by [Ro] - [LR] and re-arrange above equations: [LR] = ([L]/Kd)(Ro - [LR]) or [LR] = (Ro[L])/(Kd + [L]) where [LR]/Ro is the fraction of total receptors occupied by ligand.
The law of mass action explains the equilibrium between a ligand (L) and its receptor (R), where [L] is the concentration of the ligand, [R] is the concentration of free (unoccupied) receptors, and [LR] is the concentration of ligand-bound receptors. The amount of ligand at which half of the receptors are bound is indicated by the dissociation constant, or Kd.
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How have the different shades of skin color (from pinkish white to dark brown) evolved throughout human history?
Since they boosted fitness for early human people residing throughout equatorial Africa, darker skin tones evolved.Darker skin prevents the breakdown of circulating folate.
How have the various skin tones changed over the course of human history?Due to societal standards, environmental variances, and restrictions on the biochemical consequences of ultraviolet radiation reaching the skin, distinctions between groups have evolved through biological evolution or sexual selection.
What helps explain how humans acquired a variety of skin pigmentation colors?To balance UV radiation levels and constitutive pigmentation levels, natural selection produced two clines in the color of human skin (UVR).One cline formed by high UVR near the equator has an impact on the development of black, neuroprotective, eumelanin-rich pigmentation.
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Nutrition, nursing, biology
1. Intestinal malabsorption syndromes or pancreatic insufficiency may result in deficiency of any of the following vitamins except
a. D
b. A
c. C
d. E
e. K
2. Chronic alcoholism results in deficiency of all of these vitamins except
a. B1
b. B2
c. B3
d. Folic Acid
e. None of the above
3. Diarrhea, dermatitis, and dementia are characteristics of deficiency of
a. B1
b. B2
c. B3
d. B6
e. Folic Acid
4. High cardiac output failure suggests deficiency of which vitamins?
a. Riboflavin
b. Pyridoxine
c. Niacin
d. Thiamine
e. Ascorbic acid
1. C. Vitamin C is not typically affected by these conditions.
2. e. None of the above. Alcoholism can lead to deficiencies in all of the listed vitamins, including B1 (thiamine), B2 (riboflavin), B3 (niacin), and folic acid.
3. c. B3. These are common symptoms of niacin deficiency, also known as pellagra.
4. d. Thiamine. Thiamine (vitamin B1) deficiency can lead to a condition called beriberi, which is characterized by high cardiac output failure, among other symptoms.
Nutrition is an important aspect of maintaining good health. Vitamins play a crucial role in this regard, and a deficiency or insufficiency of certain vitamins can lead to various health problems, including cardiac issues. Here are the answers to the questions:
Intestinal malabsorption syndromes or pancreatic insufficiency may result in deficiency of any of the following vitamins except c. C. Vitamin C is not typically affected by these conditions.
Chronic alcoholism results in deficiency of all of these vitamins except e. None of the above. Alcoholism can lead to deficiencies in all of the listed vitamins, including B1 (thiamine), B2 (riboflavin), B3 (niacin), and folic acid.
Diarrhea, dermatitis, and dementia are characteristics of deficiency of c. B3. These are common symptoms of niacin deficiency, also known as pellagra.
High cardiac output failure suggests deficiency of which vitamins? d. Thiamine. Thiamine (vitamin B1) deficiency can lead to a condition called beriberi, which is characterized by high cardiac output failure, among other symptoms.
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How do you get floppy ears in Phet simulation?
To get floppy ears in the Phet simulation, click on the Human tab, select the Ears section, and use the draggable slider to adjust the size of the ears.
To get floppy ears in Phet simulation, follow the steps below:Open the Phet Simulation software that is installed on your computer.
Find the "Rabbit Population by Natural Selection" simulation and launch it on your system.
After the simulation is launched, you'll see that it has many different customizable options. These include different color options, shapes, sizes, and other features like floppy ears, long ears, short ears, and more.
To get floppy ears in the Phet Simulation software, you'll need to select the right combination of features that will enable you to customize your rabbit and give it the floppy ears you want.
There are two different options to select from when customizing your rabbit's ears: the long ear and the floppy ear. If you want floppy ears, choose this option from the menu.
After choosing the floppy ear option, you can further customize the ears by adjusting their size, shape, and other features. This will allow you to create a custom rabbit with floppy ears that is unique to your preferences and specifications.
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Occurs from a combination of immaturity of the hematopoietic system combined with the destruction of RBC because of low levels of Vit. E
This condition you are describing is called Hemolytic Anemia, and is caused by a combination of the immaturity of the hematopoietic (blood) system, coupled with the destruction of red blood cells (RBCs) due to low levels of Vitamin E in the body.
Hemolytic anemia occurs when the red blood cells (RBC) are destroyed faster than they can be produced by the hematopoietic system, leading to a decrease in the number of RBC in the body. One of the causes of hemolytic anemia is a deficiency in vitamin E, which is necessary for the proper functioning of the hematopoietic system. Without sufficient levels of vitamin E, the hematopoietic system cannot produce enough RBC to replace those that are destroyed, leading to anemia.
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the sequence of events from firing an action potential on the
axon hillock of a somatic efferent neuron to the resulting change
in membrane potential in skeletal muscle
Action potential firing in the axon hillock of a somatic efferent neuron results in the release of acetylcholine from the synaptic terminal into the synaptic cleft.
Acetylcholine binds to receptors on the motor end plate of skeletal muscle, initiating depolarization. This depolarization opens voltage-gated calcium channels in the muscle cell, leading to the release of calcium ions from the sarcoplasmic reticulum.
Calcium ions bind to troponin, causing a conformational change in tropomyosin, which exposes the binding sites on actin for myosin heads. Myosin heads attach to actin, forming cross-bridges that generate force and cause the sliding of actin filaments past myosin filaments.
The resulting change in membrane potential in skeletal muscle causes contraction.
In summary, the firing of an action potential in the somatic efferent neuron leads to the release of acetylcholine, which initiates depolarization of the muscle cell.
The depolarization triggers the release of calcium ions from the sarcoplasmic reticulum, leading to muscle contraction through the interaction of actin and myosin filaments.
This process is known as excitation-contraction coupling and is essential for the movement and functioning of the skeletal muscle.
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Question 1: What serious health consequence of prolonged anorexia nervosa in a teen-aged girl will be likely in the later years of her life? Explain why this is the case.
Question 3: Compare and contrast bulimia nervosa with binge-eating disorder. Identify behavioral characteristics that are shared between these two disorders, and behavioral characteristics that are unique to each.
Answer 1: One serious health consequence of prolonged anorexia nervosa in a teenage girl can be infertility later in life.
Answer 3: Both bulimia nervosa and binge-eating disorder are characterized by episodes of consuming large amounts of food in a short period of time, followed by feelings of guilt and shame.
Prolonged anorexia can lead to infertility because anorexia nervosa can disrupt hormonal balance, leading to amenorrhea (absence of menstrual periods) and potentially causing damage to the reproductive system.
Both bulimia nervosa and binge-eating disorder are characterized by episodes of consuming large amounts of food in a short period of time, followed by feelings of guilt and shame. However, individuals with bulimia nervosa will also engage in compensatory behaviors, such as self-induced vomiting or excessive exercise, in an attempt to prevent weight gain.
On the other hand, individuals with binge-eating disorders do not engage in compensatory behaviors and may experience weight gain as a result. Additionally, individuals with bulimia nervosa may have a distorted body image and place a greater emphasis on their weight and appearance, while individuals with binge-eating disorder may not have these concerns.
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Metabolic Pathways • multistep series or pathway, with each step catalyzed by _____
Enzyme Structure • Simple enzymes consist of ____ alone • Conjugated enzymes contain protein and nonprotein molecules - referred to as a holoenzyme - Apoenzyme: _____ portion - Cofactors: - either organic molecules called coenzymes e.g. ____ - or inorganic elements e.g. ____ Metabolism and the Role of Enzymes • Metabolism: - Pertains to ____ chemical reactions of cell • Anabolism: A building and bond-making process that forms _____ macromolecules from ones. - Requires _____ of energy (= endergonic) • Catabolism: - Breaks bonds of ____ molecules into _____ molecules - Releases energy (=exergonic) Enzymes • Biological _____ : - ____ rate of chemical reactions - Do not become part of product(s) - Are not consumed - Do not create a reaction - ____ activation energy - May need coenzyme or inorganic cofactor
Metabolic Pathways • Multistep series or pathway, with each step catalyzed by enzymes.
Enzyme Structure • Simple enzymes consist of proteins alone • Conjugated enzymes contain protein and nonprotein molecules - referred to as a holoenzyme - Apoenzyme: protein portion - Cofactors: - either organic molecules called coenzymes e.g. vitamins - or inorganic elements e.g. metal ions Metabolism and the Role of Enzymes • Metabolism: - Pertains to all chemical reactions of cell • Anabolism: A building and bond-making process that forms large macromolecules from small ones. - Requires input of energy (= endergonic) • Catabolism: - Breaks bonds of large molecules into small molecules - Releases energy (=exergonic) Enzymes • Biological catalysts: - Increase rate of chemical reactions - Do not become part of product(s) - Are not consumed - Do not create a reaction - Lower activation energy - May need coenzyme or inorganic cofactor
Metabolic pathways are a series of multistep reactions that are catalyzed by enzymes. Enzymes are biological catalysts that speed up the rate of chemical reactions without becoming part of the product or being consumed. Enzymes are composed of a protein portion called the apoenzyme and a nonprotein portion called a cofactor. Cofactors can be either organic molecules called coenzymes, such as vitamins, or inorganic elements, such as metal ions.
Metabolism is the sum of all the chemical reactions that occur within a cell. There are two types of metabolic reactions: anabolism and catabolism. Anabolism is a building and bond-making process that forms larger macromolecules from smaller ones and requires an input of energy, making it an endergonic reaction. Catabolism, on the other hand, breaks bonds of larger molecules into smaller ones and releases energy, making it an exergonic reaction.
Enzymes play a crucial role in metabolism by lowering the activation energy required for a reaction to occur. This allows reactions to proceed at a faster rate and makes them more efficient. Enzymes may also require the assistance of a coenzyme or inorganic cofactor in order to function properly.
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Mus musculus may have black, brown or golden coloured fur. House mice with black fur of identical genotypes were mated and produced the following offspring:
25 mice with black fur, 15 mice with golden fur, and 2 mice with brown fur
What epistatic ratio is approximated by these offspring?
Explain how you determined this ratio (verify).
What type of epistasis is seen in the offspring?
The epistatic ratio approximated by these offspring is 25:15:2. This ratio was determined by counting the number of mice with each fur color and comparing them to each other.
For example, there are 25 mice with black fur, 15 mice with golden fur, and 2 mice with brown fur. These numbers can be simplified to a ratio of 25:15:2.
The type of epistasis seen in the offspring is recessive epistasis. This is because the recessive allele of one gene masks the expression of the other gene. In this case, the recessive allele for black fur masks the expression of the other genes that determine fur color, resulting in the majority of the offspring having black fur. This is evident in the fact that 25 out of the 42 offspring have black fur, while only 15 have golden fur and 2 have brown fur.
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why is it necessary for externally developing organisms to lay many eggs and sperm
Answer:
External fertilization in an aquatic environment protects the eggs from drying out. Broadcast spawning can result in a greater mixture of the genes within a group, leading to higher genetic diversity and a greater chance of species survival in a hostile environment.