Any guidance or explanation for the answer (cell biology) to this question would be extremely helpful. Thank you so much!
1.) You've sequenced the genes for two newly discovered GLUT1 alleles, GLUT1A and GLUT1B. The gene encoding GLUT1A has a single missense mutation that changes GLN161 to E. The gene encoding GLUT1B has a single missense mutation that changes Thr310 to S. Otherwise, the amino acid sequences of GLUT, GLUT1A and GLUT1B are identical. Based solely on this information, what can you most likely predict about how, or if the Kms of GLUT1A and GLUT1B to bind glucose would differ from the Km for GLUT1?
A.) Both GLUT1A and GLUT1B would most likely have significantly higher Kms than GLUT1.
B.) Both GLUT1A and GLUT1B would most likely have significantly lower Kms than GLUT1.
C.) The Km for GLUT1A is most likely to be significantly higher than the Km for GLUT1. The Km for GLUT1B is most likely to be about the same as the Km for GLUT1 or perhaps only slightly higher.
D.) The Km for GLUT1B is most likely to be significantly higher than the Km for GLUT1. The Km for GLUT1A is most likely to be about the same as the Km for GLUT1 or perhaps only slightly higher.
E.) GLUT1, GLUT1A and GLUT1B would most likely all have the same Km.
Any guidance or explanation for the answer (cell biology) to this question would be extremely helpful. Thank you so much! 1.) You've sequenced the genes for two newly discovered GLUT1 alleles, GLUT1A and GLUT1B. The gene encoding GLUT1A has a single missense mutation that changes GLN161 to E. The gene encoding GLUT1B has a single missense mutation that changes Thr310 to S. Otherwise, the amino acid sequences of GLUT, GLUT1A and GLUT1B are identical. Based solely on this information, what can you most likely predict about how, or if the Kms of GLUT1A and GLUT1B to bind glucose would differ from the Km for GLUT1? A.) Both GLUT1A and GLUT1B would most likely have significantly higher Kms than GLUT1. B.) Both GLUT1A and GLUT1B would most likely have significantly lower Kms than GLUT1. C.) The Km for GLUT1A is most likely to be significantly higher than the Km for GLUT1. The Km for GLUT1B is most likely to be about the same as the Km for GLUT1 or perhaps only slightly higher. D.) The Km for GLUT1B is most likely to be significantly higher than the Km for GLUT1. The Km for GLUT1A is most likely to be about the same as the Km for GLUT1 or perhaps only slightly higher. E.) GLUT1, GLUT1A and GLUT1B would most likely all have the same Km. |
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1. Humans with mutations in the AQP1 (aquaporin-1) gene do not produce the AQP1 protein. The absence of AQP1 causes an inability to ____ in these individuals.â
a. | produce concentrated urineâ | |
b. | ârecognize thirst | |
c. | âmake large volumes of dilute urine | |
d. | âcontrol sweating | |
e. | âSalivate |
2. Two solutions of differing glucose concentration are placed in a container separated by a selectively permeable membrane that restricts large molecules like glucose, but allows the free diffusion of water. What will be the glucose concentration in the container after dynamic equilibrium has been reached?â
a. | âThe water molecules will be evenly distributed in the container, with more glucose molecules on one side than the other. | |
b. | âThe volumes of the solution will be different, with a higher volume on the side that originally contained the higher concentration of glucose. | |
c. | âThe volumes of the solution will be different, with a lower volume on the side that originally contained the higher concentration of glucose. | |
d. | âThe glucose concentration of the two solutions will be identical on both sides of the membrane, but the volumes will remain unchanged. | |
e. | âThe volume and concentrations will be unchanged on both sides of the membrane. |
3. Which mutation is an example of an adaptation?â
a. | A mutation results in decreased sperm count in humans.â | |
b. | âA mutation is found to be the cause of Alzheimer's Disease. | |
c. | âA mutation results in hairless cats, reducing allergies in humans. | |
d. | âA mutation increases the size of tomato plants. | |
e. | âA mutation renders an individual immune to HIV infection. |
4. The prokaryotic chromosome is comprised of a ____.
a. | âhighly folded mass of a single, linear molecule of DNA | |
b. | âloosely folded mass of a single, linear molecule of DNA | |
c. | âhighly folded mass of a double, circular molecule of DNA | |
d. | âhighly folded mass of a single, circular molecule of DNA | |
e. | âloosely folded mass of a single, circular molecule of DNA |
5. Which methodology was used by researchers to identify the nuclear localization sequence?â
a. | infection of cells in culture with viruses containing sequential deletions of amino acids in a protein known to localize to the nucleus, followed by microscopic determination of cellular localizationâ | |
b. | âtreatment of cells in culture with pharmacological agents that inhibit the nuclear pore complex, followed by microscopic determination of cellular localization | |
c. | âtransfection of cells in culture with mutant human proteins containing sequences believed to be required for nuclear localization followed by microscopic determination of cellular localization | |
d. | âfluorescently labeling cells with known nuclear proteins and tracking changes in localization in the presence or absence of nuclear pore complex proteins | |
e. | âtreatment of cells in culture with pharmacological agents that binds to the nuclear localization sequence and prevents nuclear transport, followed by microscopic determination of cellular localization27. Cells actively secreting enzymes would likely carry out more ____ than other cells.â |
6. Cells actively secreting enzymes would likely carry out more ____ than other cells.â
a. | exocytosisâ | |
b. | âosmosis | |
c. | âendocytosis | |
d. | âconjugation | |
e. | âFractionation |
7. What is a function of prokaryotic common pili?â
a. | âbiofilm formation | |
b. | âtransfer of DNA | |
c. | âprotein polymerization | |
d. | âenergy production | |
e. | âlipid synthesis |
Homozygous recessive loss of function mutations in any one of 4 different genes in C. elegans worms (genes A, B, C, or D) results in female worms that don't form proper egg-laying structures, called vulvas, as shown in Lines 2-5 of the data table below. In vulvaless females, the fertilized eggs hatch and baby worms develop inside the mother, killing her in the process.
genotype | phenotype | |
1 | AA BB CC DD (wild type) | normal vulva |
2 | aa BB CC DD | No vulva |
3 | AA bb CC DD | no vulva |
4 | AA BB cc DD | no vulva |
5 | AA BB CC dd | no vulva |
You would like to figure out the order in which these genes act during the creation of the vulva (you cannot assume it will be A--->B--->C--->D as the gene names are arbitrary). Your friend tells you to perform epistasis analysis by making double mutants between the different homozygous recessive mutants and analyzing the phenotype of the double mutant. For example, she asks you to examine the phenotype of aabbCCDD (double mutant of genes A and B) and compare this to the single mutants to figure out whether gene A acts earlier than gene B, or vice versa. You know this won't work. Why not? Pick the ONE BEST choice:
a. The phenotype of the single mutant is already pretty severe. The double mutant will most likely be dead, therefore epistasis analysis won't be possible.
b. Each of the single mutants (aaBBCCDD) and (AAbbCCDD) has the same mutant phenotype i.e., no vulva. Epistasis analysis between any 2 genes is only possible when the mutant phenotypes for each gene is different.
c. Epistasis analysis involves making triple mutants (such as aabbccDD) in order to learn something about how the genes are ordered.
d. Epistasis analysis can never be carried out with null loss of function mutations. The mutations being analyzed all have to be dominant gain of function alleles.
To address your concern, you first generate overactive alleles of either gene A (denoted as A*) or gene B (denoted as B*). As shown in lines 6-7 of the table below, C. elegans that have one of these overactive alleles produce multiple vulvas.
Genotype | Phenotype | |
6 | A*A BB CC DD | multiple vulvas |
7 | AA B*B CC DD | multiple vulvas |
To find out the order in which these genes act, you combine the overactive alleles with different loss of function alleles and observe the phenotype in double mutants (see Lines 8-11).
Genotype | Phenotype | |
8 | A*A bb CC DD | multiple vulvas |
9 | A*A BB cc DD | multiple vulvas |
10 | A*A BB CC dd | no vulva |
11 | AA B*B cc DD | multiple vulvas |
Based on the data in the table, what is the order in which these 4 genes normally act in wild-type C. elegans in order to produce a wild-type/normal vulva?
a. A----> B----->C----->D ---> Vulva
b. D----> B----->C----->A----> Vulva
c. B----> C----->A----->D----> Vulva
d. C----> B----->A----->D----> Vulva
e. C----> B----->D----->A----> Vulva
f. don't have enough data to make any conclusions
Based on the vulva formation phenotype of the double mutants, which of the following statement(s) accurately describes the genetic interactions between the A* allele and alleles of other genes affecting vulva formation? Pick ALL that apply:
a. The A* allele is epistatic to homozygous recessive loss of function mutations in gene B
b. A homozygous recessive loss of function mutation in gene B is epistatic to the A* allele
c. Homozygous recessive loss of function mutation in gene D is epistatic to the A* allele
d. The A* allele is epistatic to homozygous recessive loss of function mutation in gene D
e. The A* allele enhances the homozygous recessive loss of function mutation in gene D