Mike's genotype is XcY, Meg's is XX. Using Punnett square, the genotypic ratio of their offspring is 1:1 and the phenotypic ratio is 100% normal vision. If they have eight children, none are expected to be color-blind.
Explanation:In terms of genetics, color blindness is an X-linked recessive disorder. Mike, being color-blind, must have the genotype XcY. His wife, Meg, has normal color vision and is homozygous dominant, so her genotype is XX.
When we use a Punnett Square to derive the genotypic ratio of their offspring, we see that all daughters will be heterozygous carriers (XcX) and all sons will have normal vision (XY). Thus, the genotypic ratio is 1:1 (XcX: XcY).
Regarding the phenotypic ratio, all offspring are expected to have normal vision because they will inherit at least one normal vision allele. Therefore, the phenotypic ratio is 100% normal vision.
If Mike and Meg have eight children, because female offspring will be carriers and male offspring will possess the normal vision allele, we would expect none of them to be color-blind.
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what occurs during the tetrad formation?
Answer:
The tetrad occurs during the first phase of meiosis. It is the foursome of chromatids that forms when replicated homologous chromosomes align. It must be formed for crossing over to occur. It is broken apart when the homologous chromosomes separate in meiosis I.
Explanation: Hope this help (MARK BRAINLIEST)
Final answer:
Tetrad formation occurs during prophase I of meiosis, where homologous chromosomes form pairs and undergo crossing over. This process is important for producing genetically diverse gametes or spores.
Explanation:
Tetrad formation occurs during prophase I of meiosis. During this stage, homologous chromosomes form pairs and condense to become visible as X shapes, known as tetrads. The tetrads are held together at the chiasmata and undergo crossing over, which is the exchange of genetic material between non-sister chromatids. This process introduces genetic variation and is important for the production of gametes or spores.
DNA polymerases are capable of editing and error correction, whereas the capacity for error correction in RNA polymerases appears to be quite limited. Approximately one error occurs in every 104 to 105 nucleotide incorporated in RNA. Given that a single base error in either replication or transcription can lead to an error in the protein coded by the gene or mRNA. Please suggest a possible explanation for this striking difference.
Explanation:
DNA polymerase replicates the DNA supplied to all new cells produced while RNA polymerase drives DNA copy RNA synthesis. Unless corrected, error in DNA replication could result in the transmission of the error DNA to all next-generation cells.
Protein synthesis error will cause faulty copies of RNA and degraded proteins. To order to ensure the transfer of key genetic information to future generations of cells, failure to DNA replication must be corrected.
DNA polymerases, involved in DNA replication, have robust error correction mechanisms. This contrasts with RNA polymerases which do not have the same level of error correction. Additionally, DNA repair mechanisms correct errors and play a vital role in preventing mutations.
Explanation:The difference in error correction capacity between DNA polymerases and RNA polymerases can be attributed to their different roles and structures. DNA polymerases, involved in DNA replication, have a robust proofreading mechanism. As they add nucleotides, they read each newly added base. If an incorrect base has been added, they cut the phosphodiester bond and release the wrong nucleotide, replacing it with the correct one.
This is contrasted with RNA polymerases that do not have the same level of error correction. This may be due to the more 'temporary' nature of RNA in the cell; the errors in RNA do not have as long-lasting effects as those in DNA because RNA is not permanently stored in the cell.
DNA repair mechanisms are another aspect involved in maintaining the integrity of the genetic code. These mechanisms - including mismatch repair and nucleotide excision repair - excise and correct incorrectly incorporated bases, and play a significantly important role in preventing mutations and serious consequences like cancer.
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Ribosomes are the RNA-protein enzyme complexes that synthesize proteins. There are no ribosomes inside the lumen of the secretory pathway. Instead, secreted and membrane-bound proteins are synthesized in the cytosol, but have short stretches of amino acids called "signal sequences" that target the newly-forming proteins to a special channel in the membrane of the ER that enables the movement of the newly-forming proteins into the ER. This "translocation" ceases if a protein sequence is reached that will be embedded in the plasma membrane. Protein synthesis will proceed, and the membrane-spanning sequence will eventually be embedded in the ER membrane. Which of the following features would such a transmembrane amino acid sequence have?
A. The amino acids would have mostly hydrophobic side-chains.
B. The amino acids would have mostly hydrophilic side-chains.
C.There's no way to predict the relative number of hydrophobic and hydrophilic side-chains in transmembrane amino acid sequences.
Answer:
mostly hydrophobic
Explanation:
A membrane is made of amphipathic molecules with the hydrophobic part pointing to the lumen and cytosol and the hydrophobic part (much bigger makes up most of the membrane) in the middle, the protein chains that are embedded in the membrane are thus mostly hydrophobic because the membrane is mostly hydrophobic and like likes like
The authors of a cross-sectional study hypothesized that lack of regular exercise is associated with obesity in children. Their study of 12 children in Michigan, however, failed to show a statistically significant association between exercise habits and obesity (OR = 1.9, p = 0.11). Is this A) Selection Bias, B) Confounding, C) Information Bias, D) Random Error
Answer:
A) Selection Bias
Explanation:as it depends on sampling method of data collection
Coat color is determined by two loci in large cats. Two pink panthers fall in love and produce a large litter of baby panthers with the following phenotypic ratios: 12/16 pink; 3/16 black; and 1/16 white. What is the genotype of the white progeny?A) A_ B_B) A_ bbC) aa B_D) aa bb *Correct answerE) A_ B_ and A_ bb
Answer:
D) aa bb
Explanation:
Whenever there are two loci involved in a certain trait and after a cross the phenotypic ratio of the offspring is 12:3:1 (instead of the expected 9:3:3:1) there is dominant epistasis, where the presence of a dominant allele in one locus masks the expression of the other locus. In this case, for loci A and B, we could say that the offspring have the following genotypes:
9 A_B_ : pink3 A_bb : pink3 aaB_: black1 aabb: whiteThe presence of the dominant allele A in the A/a locus gives pink coat color, regardless of the alleles in the B/b locus.
Final answer:
The genotype of the white progeny in large cats given the phenotypic ratio is aa bb, indicating complete dominance when dominant alleles are present, and an autosomal epistatic interaction determines the white coat color.
Explanation:
When considering the phenotypic ratio of the large cats' offspring, which are 12/16 pink, 3/16 black, and 1/16 white, and given that both parents are pink panthers, we can deduce the inheritance pattern. The progeny that are white must exhibit the recessive phenotype at both loci, indicating a genotype of aa bb. This genotype suggests that the A and B genes are showing complete dominance when present in the dominant form. For the white progeny to be white, they must lack both dominant alleles responsible for the pigmentation.
In terms of coat color inheritance in cats, we know that it is usually related to genes located on the X chromosome. However, the white coloring mentioned in the phenotypic ratio suggests an autosomal epistatic interaction where a second gene can override the expression of the first gene, determining the coat color.
The question on the phenotypes of offspring from a cross between a yellow and a black cat needs further information, particularly the specific genotype of the parents, as multiple outcomes could occur based on X-linked inheritance and potential for codominance or incomplete dominance.
how do you think the declaration of independence changed life in the colonies
Answer:
indepent change every body live because it set slaves free
Explanation:
read tx history book up under geroge Washington
The declaration of independence gave birth to what is known today as USA . The document is symbolic of American democracy and one of the charecters of freedom . The words stated in the declaration railed support from colonists at home and colonists abroad .
Here's your answer...
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Which of the following statements is/are true of corals?
Choose all that apply.
A. Corals are animals.
B. Corals are benthic organisms.
C. Corals secrete silica to build the coral reef.
D. Corals are solitary.
E. Corals live in tropical water.
Answer:
The correct answers are option A. "Corals are animals". B. "Corals are benthic organisms"., and E. "Corals live in tropical water".
Explanation:
Corals are animals, what we know as a coral, is in fact a group of small animals called polyps that need food to survive. Corals are benthic organisms because they live at the bottom of the sea. The subclass of benthic organisms that corals belong is called Macrobenthos, for being large enough to be seen at the naked eye. Corals live mostly in tropical waters, because they do not tolerate waters with a temperature below 18 Celsius.
Answer:
a
Explanation:
What is the energy requirement in kJ/mol to transport a proton across the mitochondrial inner membrane in plant cells at night when the outside temperature is 15ºC, the pH differential across this membrane is 1.4 pH units, and the membrane potential of 160 mV. Choose the ONE correct answer.
A. +21.8 kJ/mol
B. +75.4 kJ/mol
C. +23.1 kJ/mol
D. +46.2 kJ/mol
E. -21.8 kJ/mol
F. -23.1 kJ/mol
G. +19.0 kJ/mol
H. +44.6 kJ/mol
Answer:
C. +23.1 kJ/mol
Explanation:
the formula to use to calculate the energy requirement in kJ/mol to transport a proton across the mitochondrial inner membrane in plant cells is:
ΔGt = RTIn [tex]\frac{C_{2} }{C_1}[/tex] + ZFΔV
let's list the values of the data we are being given in the question to make it easier when solving it.
Z= 1
F= 96500C (faraday's constant)
ΔV= 160mV = 0.160V
R= 8.314( constant)
T= 15ºC ( converting our degree Celsius into kelvin, we will have 273.15k+ 15 = 288.15K)
∴ T= 288.15K
Putting it all together in the formula, we have:
ΔGt = 8.314 × 288.15 × 2.303 log [tex]\frac{C_{2} }{C_1}[/tex] + 1 × 96500 × 0.160
ΔGt = 5517.25 [tex][ -(log(H^{+}_{out} ) + log(H^{+}_{in} )_{in}[/tex] +15440
ΔGt = 5517.25 [tex](pH_{out} - pH_{in}[/tex] +15440
Given that the pH differential gradient across the membrane is 1.4pH units. It implies that;
ΔGt = 5517.25 × 1.4 + 15440
= 7724.15 +15440
= 23164.15 Joules/moles
= +23.1 KJ/mole
The correlation between the area of an island and the number of species found there is referred to as:
A. species-area relationship.
B. latitudinal gradient.
C.theory of island biogeography.
D. immigration and extinction.
Answer: Option A.
It is called specie- area relationship.
Explanation:
Specie- area relationship in ecology describe the relationship between the area or habitat and the total number of species present in the area. It is also called species-area curve.
A larger area will contain a larger number of species.
A lot of factors determine the slope and elevation of specie-area relationship and they are, immigration and extinction rate, predator- prey dynamics, rate of disturbances between large and small area e.t.c.
Which of the following statements about interphase is FALSE?
A. Most cells in adult human bodies are in interphase (including G0)
B. DNA is replicated in S phase
C. Once a cell enters interphase, it always moves on to M phase again
D. Cells grow in G1 and G2
E. Cells can stop the cell cycle by going to G0
Answer:
C. Once a cell enters interphase, it always moves on to M phase again
Explanation:
Interphase is the first stage of cell cycle and is followed by M phase. Interphase includes two growth phases called G1 and G2. The S phase of interphase is marked by DNA replication. The G2 phase of interphase is followed by M phase. However, not all the cells that have entered the interphase ends up in M phase.
A cell can be withdrawn from the cell cycle if the checkpoints are not passed through. Many cells halt the process of cell division for some time or even longer and may or may not resume the process of cell division again. For instance, if replication fork is stalled or there is error in DNA replication, the cell will be withdrawn from cell cycle at the G2 checkpoint.
In an organism's DNA, what makes one nucleotide different from another?
A. The sugar
B. The phosphate
C. The base
D. Only some nucleotides have carbon atoms
E. Only some nucleotides have nitrogen atoms
Answer:
C. The base
Explanation:
The nitrogenous bases, Adenine, Guanine, Thymine and Cytosine in DNA are the variable elements in the nucleotides. Nucleotides are constituted by three elements: The sugar (deoxyribose) and the phosphate group constitutes the backbone of the double helix and are constant in all nucleotides. Also, all nucleotides have carbon and nitrogen atoms.
C) The base is what makes one nucleotide different from another in an organism's DNA.
Explanation:In an organism's DNA, what makes one nucleotide different from another is the base. Nucleotides are composed of three components: a sugar molecule, a phosphate group, and a nitrogenous base. The base is the variable part of the nucleotide and can be adenine (A), thymine (T), cytosine (C), or guanine (G). The sequence of bases in a DNA molecule is what determines the genetic code or instructions for building and functioning of an organism.
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What causes ozone depletion?
A. Flying airplanes especially at high altitudes.
B. Combustion of a sulfur containing fuel like coal.
C. CFCs leaking from old refrigerators and air conditioners.
D. Deforestation especially in mountainous regions.
Final answer:
Ozone depletion is primarily caused by emissions of chlorofluorocarbons (CFCs) and other similar chemicals into the atmosphere, where they release chlorine atoms that destroy ozone molecules. This process allows more harmful ultraviolet radiation to reach the Earth's surface.
Explanation:
The primary cause of ozone depletion is chlorofluorocarbons (CFCs) and other ozone-depleting substances (ODS) that are emitted into the atmosphere. These chemicals were commonly used as refrigerants, solvents, and propellants in various industries. Once released, these stable molecules do not dissolve in rain and eventually ascend to the stratosphere, approximately 10 kilometers above the Earth's surface. In the presence of ultraviolet radiation, CFCs release chlorine atoms that react with ozone (O3), destroying it and leading to a thinner ozone layer. This reduction in ozone allows more harmful ultraviolet radiation, specifically UVB, to reach the Earth's surface, increasing the risk of skin cancer and other health issues.
The Montreal Protocol is a global agreement that has greatly contributed to the mitigation of ozone layer depletion by phasing out the production of CFCs and other harmful chemicals.
What effect does histone acetylation have on transcription? Group of answer choices Extremely decrease expression Moderately decrease expression No change in expression Measurably increase expression Either increase or decrease expression
Answer:
Measurably increase expression
Explanation:
Acetylation of histones, or more precisely their amino acids lysine, causes their positive charge to be neutralized, and weakening of the interaction of the histone tail with negatively charged local DNA induces local opening of chromatin structures. Thus, local DNA is exposed, increasing access to transcription factors and promoting significant increases in gene transcription.
In other words, histone acetylation results in the decompression of chromatin, which measurably increases expression.
Which of the following does not influence the responsiveness of a target cell to a hormonea) the number of receptors on the target cellb) the temperature of the tissuec) the presence and effects of other hormonesd) the concentration of the hormonee) previous exposure to a hormone
Answer:
The temperature of the tissue.
Explanation:
Hormones are the chemical messenger of the body that regulates, maintains the proper functioning of the body. Two main types of hormones are exocrine hormones and endocrine hormones.
The hormone binds to its particular receptor and initiates the signaling cascade mechanism of the body. The binding of the hormone to its receptor is independent of the temperature conditions. The temperature do not have any effect on the response of target cell to the hormone.
Thus, the correct answer is option (b).
The temperature of the tissue does NOT influence the responsiveness of a target cell to a hormone.
A hormone is a chemical messenger that sends information to different parts of the body.The responsiveness of a hormone depends on the number of receptors present in target cells.The temperature of a given tissue is a physic property that may or may not be associated with hormone activity.In conclusion, the temperature of the tissue does NOT influence the responsiveness of a target cell to a hormone.
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What is the area of this graph called where the population reaches its maximum level? A) S-curve B) demographics C) carrying capacity D) exponential growth
Answer:the answer is D
Explanation:
Answer: C
Explanation: HINT WORD (capacity)
Choose true or false for the following statements.
1. Alleles of every gene in a population make up the population’s gene pool.
2. A population in genetics always includes all members of a particular species.
3. Once a population is established, its genetic composition does not change over time.
4. Population genetics looks at changes in genetic variation that occur over time.
Answer No 1: True
Gene pool can be described as the number of genes which are present in a population of organisms which can interbreed. All the genetic information that is present in a population is the gene pool. As genes are made up of alleles, all the alleles which are present in a population will make up the gene pool.
Answer No 2: False
In genetics, a population can be described as the members of a species which can interbreed and produce fertile offsprings that are living in a certain area during a certain time period.
The option 2 is false because it does not specify area and time period in the definition.
Answer No 3: False
Over time, the genetic composition of a population will change. The organisms which will better adapted to survive to the environmental changes will survive and reproduce. The organisms which are not properly adapted might not be able to survive at all or they might become lesser in number. Hence, the genetic composition keeps changing depending on environmental changes.
Answer No 4: True
Population genetics can be described as a study of the genetic variations which occur in a certain population. It evaluates the gene frequency and how the gene frequency changes for a certain gene changes over time in a population.
Population genetics helps us to recognize the alleles which are better suited for an environment and which are present in a certain environment in abundance.
The statements about population genetics are evaluated for their accuracy: the gene pool is made up of all alleles in a population, a genetic population does not necessarily include all species members, genetic compositions of populations do change over time due to various factors, and population genetics does study changes in genetic variation over time.
Explanation:The student presents four statements about population genetics and asks whether each is true or false. Let's evaluate each statement:
Alleles of every gene in a population make up the population's gene pool. - True
A population in genetics always includes all members of a particular species. - False (A genetic population consists of members of a species that interbreed and share a gene pool.)
Once a population is established, its genetic composition does not change over time. - False (Allele frequencies within a population can change due to natural selection, genetic drift, gene flow, and mutations.)
Population genetics looks at changes in genetic variation that occur over time. - True
Population genetics is the study of how selective forces change the allele frequencies and genetic variation within a population over time, affecting its gene pool. Factors such as natural selection, genetic drift, gene flow, and mutations contribute to these changes, contrasting with the notion that a population's genetic makeup remains static.
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n a certain plant, blue (B) flower color is dominant to white (b). You have a blue flowered plant and a white flowered plant. What do you know about the genotype and phenotype of each plant? Now you cross the plants and all the offspring are blue. What have you learned about the genotypes of the original plants?
Answer:
If blue flower colour is dominant to white then there will be two conditions for the dominant plant. It will either have homozygous dominant (BB) alleles or it will have heterozygous alleles (Bb). The phenotype will be blue flower colour.
As the white flower is recessive,it will have the alleles bb. The phenotype will be white flower colour.
If all the offsprings produced by the cross are blue, then it means that the dominant blue plant was homozygous for the trait. All the offsprings produced will have blue colour and a heterozygous genotype.
In the plant species concerned, the phenotype of the plants is determined by their genotype. The blue-flowered plant (which is phenotypically and genotypically dominant) must be homozygous dominant (BB) while the white-flowered plant is homozygous recessive (bb). The fact that all offspring are blue after crossing the plants confirms this.
Explanation:In the context of this question, the blue flower color (B) is dominant and the white (b) is recessive in a certain plant species. From this, we can assert that the genotype of the blue-flowered plant can be BB (homozygous dominant) or Bb (heterozygous), while the genotype of the white-flowered plant has to be bb (homozygous recessive), because white is a recessive trait. The phenotype is the color we see, hence, blue for the first plant and white for the second.
When you cross these plants and all the offspring are blue, this discloses that the blue-flowered parent plant must have been homozygous dominant (BB) because if it were heterozygous (Bb), there would have been a 75% chance of blue offspring and a 25% chance of white offspring, in accordance with Mendel's hybridization experiments. The absence of any white offspring indicates that there were no recessive alleles (b) in the blue-flowered parent plant.
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Two nutrient solutions are maintained at the same pH. Actively respiring mitochondria are isolated and placed into each of the two solutions. Oxygen gas is bubbled into one solution. The other solution is depleted of available oxygen. Which of the following best explains why ATP production is greater in the tube with oxygen than in the tube without oxygen?
A. The rate of proton pumping across the inner mitochondrial membrane is lower in the sample without oxygen. B. Electron transport is reduced in the absence of a plasma membrane. C. In the absence of oxygen, oxidative phosphorylation produces more ATP than does fermentation. D. In the presence of oxygen, glycolysis produces more ATP than in the absence of oxygen.
Was told the answer is A. but i need help justifying that and why it is not D
Answer:
Glycolysis is not an oxygen dependent process and the rate of glycolysis is not affected by the amount of oxygen present. The amount of oxygen present affects the rate of oxidative phosphorylation in the mitochondria. Hence with more oxygen present, there is increased oxidative phosphorylation in the presence of high amounts of final electron acceptor O2 which pumps more protons across the membrane and results in a steeper proton gradient across the mitochondrial membrane thus increasing the rate of ATP production. Thus the answer is A as the rate of proton pumping is increased with higher rate of oxidative phosphorylation due to the abundance of final electron acceptor O2.
Oxygen doesn't affect glycolysis' rate. Oxygen affects oxidative phosphorylation. With more oxygen, oxidative phosphorylation rises, pumping more protons across the mitochondrial membrane and increasing ATP synthesis.
What is oxidative phosphorylation?During the process of oxidative phosphorylation, electrons that were obtained from NADH and FADH2 mix with oxygen (O2). The energy that is released as a result of these oxidation and reduction events is then used to fuel the synthesis of ATP from ADP.
Glycolysis does not require oxygen, and the presence of oxygen has no effect on the rate at which it occurs. Oxygen has an effect on the oxidative phosphorylation that occurs in mitochondria. In the presence of high amounts of final electron acceptor O2, the process of oxidative phosphorylation speeds up when there is more oxygen present. This results in the transfer of more protons across the mitochondrial membrane, which boosts the production of ATP.
Therefore, option (A) is correct and (D) can't be true.
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Hemoglobin consists of four globular proteins, each formed from a polypeptide chain. Each globular protein contains a nonprotein group with an iron ion. This iron ion binds to an oxygen molecule. Thus, every hemoglobin protein can bind four oxygen molecules. Select the correct statement about oxygen transport below.
A. Myoglobin transports oxygen through the bloodstream to tissue.
B. To carry oxygen, iron must be in its nonheme form.
C. Erythrocytes contain abundant myoglobin for transporting oxygen.
D. Hemoglobin is found in red blood cells, which transport oxygen to tissues
Answer:
d
Explanation:
Hemoglobin is found in red blood cells, which transport oxygen to tissues. Therefore, option (D) is correct.
Red blood cells carry oxygen via hemoglobin. Haemoglobin transports oxygen from the lungs to bodily tissues and organs via the circulation. It is the main circulatory protein that transports oxygen. Four globular proteins containing iron ions can bind four oxygen molecules to each haemoglobin molecule.
Myoglobin, a related protein present in muscle tissue, stores oxygen in muscle cells, not in the circulation. Therefore, option (D) is correct.
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Cancer is characterized by uncontrolled cell division. Cell division is preceded by DNA replication. Several proteins of the DNA-replication machinery can be targets for chemotherapeutic agents. One such protein is topoisomerase. Select the effects that result from the inhibition of topoisomerase. Supercoils accumulate, resulting in chromosomal instability.
Inhibition of topoisomerase can result in the accumulation of supercoils, leading to chromosomal instability.
Explanation:
Cancer is characterized by uncontrolled cell division, which is preceded by DNA replication. Inhibiting the protein topoisomerase, which plays a role in DNA replication, can have effects such as the accumulation of supercoils. Supercoils are twists or turns in the DNA structure, and their accumulation could lead to chromosomal instability.
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_____ arise from a single fertilized egg and are thus genetically almost identical.
Answer:
Monozygotic twins
Explanation:
Monozygotic or identical twins are those that originate from a single ovary_ and a single sperm and therefore share the same genetic load.
What happens in these cases is that the embryo splits in two, and depending on the moment in which this happens, different configurations can be expected in the development of the placenta.
When the embryo is cleaved within the first four days after fertilization it results in a bicorial-biamnotic pregnancy, that means, two sacs and two independent chorions will be visible (exactly the same as in the case of dizygotic twins from two different embryos).
Answer:
Zygote
Explanation:As it only organism after fertilization formed
After growing a pure culture of this bacteria, what strategy would be most effective for sequencing its genome?
A. Transcriptome sequencing of the RNA.
B. Metagenomic sequencing of the water from a Galveston beach.
C. Whole-genome shotgun sequencing from the bacterial culture.
D. None of the above.
Answer:
b
Explanation:
The best strategy for sequencing the genome of a grown pure bacterial culture is whole-genome shotgun sequencing, as it provides a complete view of the organism's genomic sequence.
After growing a pure culture of a bacteria, the most effective strategy for sequencing its genome would be C. Whole-genome shotgun sequencing from the bacterial culture. This method is designed for analyzing the DNA from a single organism and provides a comprehensive view of the entire genome. In contrast, transcriptome sequencing of the RNA would be more relevant for understanding gene expression profiles, and metagenomic sequencing of water would be best for the study of microbial communities from environmental samples, not pure cultures.
- Proper aseptic technique is crucial to ensuring growth of pure bacterial colonies. What is one experimental way you can test your practices to confirm that you are using proper aseptic technique? (5 points)- In a laboratory setting, what are three ways you can properly sterilize culturing equipment?
- For each inoculation tool, give one scenario in which use of that tool would be appropriate.
- Do some research and describe two or three scenarios in which it would be preferable to use a stab tube vs. a growth plate.
(Hint: What do bacteria use to help them move? Can motility be used to help identify many medically important pathogenic bacteria such as the Enterobacteriaceae?)
Answer:
1. Aseptic techniques can be tested for their suitability by flaming the loop or by using agar slants.
2. Culturing equipment can be sterilized by three methods and these include wet heat, dry heat and filtration.
3. Inoculating needles are used for transferring bacterial culture to a soft agar medium. Inoculating loops are used to transfer culture into liquid medium or plate.
4. Growth plates or agar plates, Liquid media and Stab-tube media are used for growth of bacterial colonies.
Explanation:
1. A-septic techniques can be tested for their suitability by flaming the loop or by using agar slants. The loop is flamed to an extent that it becomes red-hot and is then cooled before picking up the organisms from the bacterial culture to be transferred. The hot loop is kept into the tube for a few seconds before removing the inoculum so that a bacterial aerosol is not produced.
2. Culturing equipment can be sterilized by three methods and these include wet heat, dry heat and filtration. Wet heat method is the most preferred method for sterilizing culture material using an autoclave. The material is heated using pressurized steam in an autoclave; which is an effective procedure for killing microbes, spores and viruses at 100◦C. However, it is interesting to know that pressurized steam has 7 times more heat than water at 100◦C which allows rapid delivery of heat and good penetration for sterilizing dense materials. High pressured steam hydrolyzes bacterial protein and removes any chances of contamination.
Dry heat method kill microbes by oxidation of cells thus requires more energy and is conducted at a higher temperature of 121◦C for efficient sterilization.
Filtration is another method of sterilizing liquids without heating and is performed by passing the solution through a filter with a pore diameter of 0.2 µm. The filter material used for sterilization can be heat-fused glass funnels or cellulose membranes. However, viruses can pass through these filters and filtration is not a preferred sterilization method.
3. Inoculating needles are used for transferring bacterial culture to a soft agar medium because needles supports appropriate spreading of culture and produces growth along the stab line also.
Inoculating loops are used to transfer culture into liquid medium or plate because it is free following and adequate spreading of culture is not required.
4.Growth plates or agar plates are the best bet when morphology of the species is to be identified or a bacterial colony needs to be isolated. In this scenario, growth plates allow the separation of a single clone and aids microscopic analysis.
In order to expand bacterial culture for the purpose of isolating DNA plasmid, liquid media is used. These liquid cultures allow easy inoculation of growth parameters such as antibiotics, and control of temperature and air pressure.
Stab tube media is prepared by filling test tube with soft agar and is specifically designed for growth of bacteria in low oxygen conditions. The stab tube allows immersion of inoculation loop into the agar depth where oxygen is extremely low and bacterial colonies can be produced without any shaking of media.
Final answer:
Testing aseptic technique can be done using control cultures, and equipment sterilization can be achieved through autoclaving, dry heat, or chemical methods. Inoculation tools like loops, needles, and pipettes are used for specific purposes. Stab tubes are preferable for studying anaerobic bacteria and bacterial motility, crucial for identifying certain pathogens.
Explanation:
Proper aseptic technique is crucial in microbiology to prevent contamination and ensure the growth of pure bacterial cultures. An experimental way to test your aseptic technique is by performing a control culture without introducing bacteria, allowing it to grow in the same conditions as your experimental cultures. If no growth is observed, it suggests that the aseptic technique was successful.
There are several methods for sterilizing culturing equipment, including:
Autoclaving, which uses pressurized steam to sterilize.Dry heat sterilization, where equipment is heated in an oven.Chemical sterilization, using disinfectants or antiseptics for items that cannot be exposed to high temperatures.Each inoculation tool, such as loops, needles, and pipettes, has its specific use. Loops are typically used for streaking bacteria on agar plates to isolate single colonies. Needles are for transferring bacteria to slants or stab cultures. Pipettes are used to transfer liquids, essential for creating dilutions or adding bacteria to broth cultures.
Stab tube cultures are used in scenarios where it's important to study anaerobic growth or bacterial motility. A stab tube provides an environment with reduced oxygen exposure, ideal for growing anaerobic bacteria. Moreover, the stabbing action can demonstrate motility, which is important for identifying pathogenic bacteria such as those in the Enterobacteriaceae family.
n mice, black fur (B) is dominant to white (b) . At different locus, a dominant allele (A) produces a band of yellow just below the tip of each hair in mice that have black fur, which gives a frosted appearance that is termed agouti. Expression of the recessive (a) allele results in a solid coat color. If double heterozygous agouti parents are crossed, what color will most of the offspring be ?
Answer:
Black fur with solid coat
Explanation:
Given -
Black fur allele is represented by "B"
and white fur allele sis represented by "b"
Yellow band allele is represented by "A" and
solid coat color allele is represented by "a"
Genotype of double heterozygous agouti parents will be
"BbAa"
Crossing BbAa we get -
BA Ba bA ba
BA BBAA BBAa BbAA BbAa
Ba BBAa BBaa BbAa Bbaa
bA BbAA BbAa bbAA bbAa
ba BbAa Bbaa aaAa bbaa
Most of the offsrping will be Black fur with solid coat
If alleles of the two genes assort independently, what is the predicted ratio of parental:recombinant types from cross (iii)?
A. 3:1
B. 1:3
C. 4:0
D. 0:4
E. 2:2
Answer:
Option E , 2:2
Explanation:
As per Mendel's theory of independent assortment, alleles for two or more different traits are expressed independently which means expression of one allele is not affected by presence of unaffected allele.
Recombination types of offspring are those which have genetic structure different from their parents.
In the cross (iii), the number of parental genetic combination is equal to the number of recombination types.
Hence, the ration of parental and offspring genetic combination is 2:2
Option E is correct
The genetic code is said to be "degenerate" becauseA. A single tRNA may couple a single mRNA codon to multiple amino acids. B. A single codon in an mRNA can couple to the anticodons of many different types of tRNAs. C. A single tRNA specific for a particular amino acid may respond to multiple codons in an mRNA D. More than one type of tRNA can be charged with a particular amino acid.
Answer:
The correct answer is option C. "A single tRNA specific for a particular amino acid may respond to multiple codons in an mRNA".
Explanation:
Genetic code is said to be "degenerate" because there are more codons than amino acids, which is possible because a single tRNA specific for a particular amino acid may respond to multiple codons in an mRNA. tRNA recognize the codons of the mRNA by a specific nucleotide sequence called anticodon, however the nucleotide sequence of the anticodon can recognize more than one codon sequence. This phenomenon takes place at the third position of the codon and is known as "wobble".
Broadleaf deciduous trees of temperate forests A) Have small leaves to conserve energy. B) reabsorb chlorophyll, resulting in the colorful autumn pigments becoming visible. C) grow slower than narrow-leaved trees during summer months. D) are the only type of plant life found in this biome.
Answer:
Broadleaf deciduous trees of temperate forests reabsorb chlorophyll, resulting in the colorful autumn pigments becoming visible.
Explanation:
During autumn and winter, deciduous trees from temperate forests need to store different nutrients that will be used for the plant growth during the following spring. Storaging nutrients in leaves require too much energy and constant photosynthetic activity, which might be very difficult for the plant to support during these colder seasons. To confront this situation, these species have developed some strategies such as leaves senescence. The tree stops supplying water and nutrients to the leaves, so these last ones stop producing chlorophyll. When this molecule is completely lost, other pigments that were masked by the chlorophyll, show up. Before senescence occurs, pigments such as carotenoid, anthocyanin, or pheophytin reveal yellow, orange, red, purple and brown tones, which are the characteristic autumn colors.
On the planet Seltaeb, there is a rare recessive condition known as Hofner Bassis (BB) in the Blue Meanies, a race whose genomes are identical to that of humans. Individuals with this condition can be clearly identified, as they are very musically gifted. The two alleles at this locus are B (the trait allele) and b.If both parents are known to be carriers for the recessive gene, what is the probability that they will have an unaffected child?A. 0B. 0.75C. 0.10D. 0.50E. 0.25
Answer:
The correct option is B. 0.75
Explanation:
The punnet square for this cross can be shown as follows:
B b
B BB Bb
b Bb bb
As the trait is a recessive disorder, both the alleles of the gene should be BB for the trait to occur. According to the punnet square, there is 25% chance (0.25) that the organism will have the recessive condition.As the punnet square shows, there is a 75% (0.75) chance that the offsprings will be unaffected. However, 50% will be carriers.
What is the most common type of fossils? Group of answer choices:
O Trilobite (class Trilobita)
O Snail (class Gastropoda)
O Bryozoan (phylum Bryozoa)
O Foram (order Foraminiferida)
O Cephalopod (class Cephalopoda)
Answer:
Snail (class Gastropoda)
Explanation:
The most common body fossils (here we exclude ichnofossils, that are fossils of footprints, trails and other activities of an organism) in number of occurrences in the collections are of snails of the genus Turitella, which are present in the fossil record from the Cretaceous to the recent periods.
_________ account for the hereditary influence on the phenotype in a cumulative way.
Answer:
Additive alleles
Explanation:
Additive alleles account for the hereditary influence on the phenotype in a cumulative way.
Alleles are nothing but variations of a gene. that simply gives variation to the traits.
Whereas, Additive alleles are alleles that contribute to most observable traits, such as height, weight, hair color, eye color,and complexion. Therefore, they account for the hereditary influence on the phenotype.