1 ) When the kinase ATM is activated after DNA damage occurs in the cell

a. it can directly phosphorylate p53, which is then stabilized
b. it can directly phosphorylate p53, which is then targeted for degradation by MDM2
c. it can phosphorylate the p53 gene, leading to an increase in its transcription
d. Two of the above are correct
e. All of the above are correct

2) Which of these is responsible for the lac operon of E. coli not being expressed when both glucose and lactose are present?
a. low cyclic AMP levels
b. high cyclic AMP levels
c. binding of Lac Repressor
d. binding of CAP ( my answer but not sure)
e. permease expression

3) In corn Zea mays, purple kernels is the wild type condition. The appearance of purple-spotted kernels in the colorless mutant strains of corn that Barbara McClintock studied was due to

a. movement of a transposable element into the pigment gene in some cells of the kernel ( my answer)
b. movement of a transposable element out of the pigment gene in some cells of the kernel
c. a chromosomal translocation
d. breakage of the chromosome at the site of the pigment gene
e. unequal crossing over between highly related pigment genes

4 ) What effect would a mutation in the gene encoding CAP have on expression of the lac operon in E. coli?
a. it could no longer be repressed and would be constitutively expressed
b. it could no longer be strongly activated in the presence of inducer
c. it could no longer be strongly activated in the absence of inducer
d. expression would increase only when glucose was present
e. none of the above

---->>> Use this information to answer the next two questions: A new sugar, sugarose, induces synthesis of two enzymes from the sug operon of E. coli. Four genes are know to be involved in sugarose utilization. The table below shows the induction behavior of the two enzymes in strains carrying deletion mutations (loss-of-function) affecting each one of the four genes.
Here, ‘+’ means the enzyme is induced normally, ‘C’ means the enzyme was synthesized constitutively, ‘0’ means the enzyme is never detected.
mutation of / enzyme 1 / enzyme 2
gene A / + / 0
gene B / 0 / +
gene C / 0 / 0
gene D / C / C

5 ) Knowing what you do about operon regulation in E. coli, which gene is likely to encode a repressor protein that regulates the genes for enzymes 1 and 2 in wild type cells?
a. gene A
b. gene B
c. gene C
d. gene D
e. none of the above

6 )Knowing what you do about operon regulation in E. coli, which gene is likely to be the structural gene for enzyme 1?
a. gene A
b. gene B
c. gene C
d. gene D
e. none of the above

1) The correct answer is d. Two of the above are correct. When the kinase ATM is activated after DNA damage occurs in the cell, it can directly phosphorylate p53, which is then stabilized (option a). It can also phosphorylate the p53 gene, leading to an increase in its transcription (option c).

2) The correct answer is c. The lac operon of E. coli is not expressed when both glucose and lactose are present due to the binding of Lac Repressor. When glucose is present, it inhibits the production of cyclic AMP (cAMP), which is required for the binding of CAP (Catabolite Activator Protein) to the lac operon. Additionally, Lac Repressor binds to the operator region of the lac operon in the presence of glucose, preventing transcription of the operon.

3) The correct answer is a. In corn Zea mays, the appearance of purple-spotted kernels in the colorless mutant strains studied by Barbara McClintock was due to the movement of a transposable element into the pigment gene in some cells of the kernel. This resulted in a disruption of the normal pigment production, leading to the purple-spotted phenotype.

4) The correct answer is b. A mutation in the gene encoding CAP would result in the lac operon not being strongly activated in the presence of an inducer. CAP (Catabolite Activator Protein) is required for the full activation of the lac operon in the absence of glucose. It helps to recruit RNA polymerase to the lac promoter, enhancing transcription. Thus, a mutation in the gene encoding CAP would weaken the activation of the lac operon.

5) The likely gene to encode a repressor protein that regulates the genes for enzymes 1 and 2 in wild-type cells is gene C. Based on the given table, gene C mutation results in enzymes 1 and 2 never being detected (indicated by 0). This suggests that gene C likely encodes a repressor protein that normally regulates the expression of these enzymes.

6) The likely gene to be the structural gene for enzyme 1 is gene B. Based on the given table, gene B mutation results in enzyme 1 being constitutively synthesized (indicated by C). This suggests that gene B is likely the structural gene for enzyme 1, as its mutation leads to constant production of the enzyme regardless of inducer presence.