【GR1222】 The Solution for GRE Biochemistry Practice Test [142-172]

The Solution for GRE Biochemistry Practice Test [142-172]

 

추천글 : 【GRE】 Solution for GRE Biology Practice Test

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Questions 142-145

To understand the signaling pathway leading from sperm-egg function to calcium increase, various substances were injected into sea urchin eggs. The effects of these substances on calcium increase are shown in the table below. Definitions of terms used in the table are as follows.

 

PLCγ - hydrolyzes a phospholipid to produce IP₃ and is regulated by tyrosine phosphorylation

PP2 - an inhibitor tyrosine kinase 

SHP2 - tyrosine phosphatase

Dominant-negative protein - inhibits the function of the wild-type version of the protein in the same cell

 

	Injected Substance(s)	Sperm Added	Calcium Increase
1.	Buffer control	Yes	Yes
2.	IP3	No	Yes
3.	PP2	Yes	No
4.	PP2 inactive analog	Yes	Yes
5.	PP2 followed by IP3	No	Yes
6.	Dominant-negative PLCγ protein	Yes	No
7.	Dominant-negative SHP2 protein	No	Yes

 

 

142. Calcium increase at fertilization requires which of the following?

 

 

⑴ answer: E

⑵ PP2를 첨가하면 정자가 첨가되도 칼슘 증가가 일어나지 않으므로 (3번 실험), Tyrosine kinase가 필요함

⑶ Dominant-negative PLCγ protein을 첨가하면 정자가 첨가되도 칼슘 증가가 일어나지 않으므로 (7번 실험), PLCγ가 필요함

 

 

143. A reasonable explanation for the calcium increase shown in line 7 of the table is that

 

 

⑴ answer: D

⑵ SHP2는 tyrosine phosphate이므로 PLCγ 조절에 관여

🄒 wild-type SHP2가 아니라 Dominant-negative SHP2에 대해 설명해야 함 

🄓 PLCγ의 인산화는 PLCγ의 활성화를 의미함 

 

 

144. An informative experiment to determine the position of PLCγ in the signaling pathway would be to inject eggs with which of the following?

 

 

⑴ answer: A

⑵ 2번 실험에서 IP₃가 칼슘 증가를 일으켰고, 6번 실험에서 Dominant-negative PLCγ가 칼슘 증가가 일으키지 않았음을 주목하자. Domiant-negative PLCγ + IP₃를 첨가할 경우에 칼슘 증가가 있으면 IP₃가 PLCγ Domain보다 나중에 작용하는 것이고, 칼슘 증가가 없으면 IP₃가 PLCγ Domain보다 먼저 작용하는 것이다. 

 

 

145. Based on the information provided, which of the following pairs is in the correct order to lead to calcium increase in the normal signal transduction pathway induced by fertilization?

 

 

⑴ answer: C

⑵ PLCγ가 IP₃를 만든다는 정보가 제공돼 있음 

 

 

Questions 146-148

Wild-type flies of the species Drosophila melanogaster have red eyes and wings with normal cross veins. Recessive mutant alleles of the sex-linked genes w (white) and cv (crossveinless) lead, respectively, to the formation of white eyes and crossveinless wings.

 

A virgin female fly who was heterozygous for the mutant w and cv alleles was crossed to a wild-type male. The male progeny of this cross were primarily of two types. One type had white eyes and normal wings and the other had red eyes and crossveinless wings. Smaller numbers of wild-type males and males with white eyes and crossveinless wings were also obtained.

 

 

146. The occurrence of the wild-type male progeny was most likely the result of 

 

 

⑴ answer: C

⑵ X염색체 유전 

⑶ 암컷 어미 초파리는 (w CV) (W cv) (상반)이고 수컷 초파리는 (W CV) Y라서 그 male 자손은 다음과 같음

① (w CV) Y : white eyes and normal wings. 부모형 

② (W cv) Y : red eyes and crossveinless wings. 부모형

③ (w cv) Y : white eyes and crossveinless wings. 재조합형 (교차형)

④ (W CV) Y : wild-type male progeny. 재조합형 (교차형)

 

 

147. Approximately 6 percent of the male progeny were mutant for both phenotypes. Thus, the w and cv genes are separated by approximately how many map units?

 

 

⑴ answer: C

⑵ 교차율 = #재조합형 / (#부모형 + #재조합형) = 2 × #열성동형접합 / (#부모형 + #재조합형) = 2 × 6% = 12% 

 

 

148. The percent of phenotypically wild-type females from this cross is predicted to be approximately

 

 

⑴ answer: A

⑵ 암컷 어미 초파리는 (w CV) (W cv) (상반)이고 수컷 초파리는 (W CV) Y라서 그 female 자손은 다음과 같음

① (w CV) (W CV) : wild-type

② (W cv) (W CV) : wild-type

③ (w cv) (W CV) : wild-type

④ (W CV) (W CV) : wild-type 

 

 

Questions 149-150

Lactate dehydrogenase (LDH) can exist as various tetrameric isozymes. Extracts from different rat tissues were run on a nondenaturing polyacrylamide gel and stained for LDH activity, as shown below. Standard tetramers of M subunits (M₄) and H subunits (H₄) were run alongside the tissue extracts.

 

 

 

149. Which of the following is the subunit composition of the isozyme running at position 2?

 

 

⑴ answer: B

⑵ 젖산탈수소효소 

⑶ (1, 2, 3, 4, 5) = (M₄, M₃H, M₂H₂, MH₃, H₄) 

 

 

150. Which of the following conclusions can be drawn from the experiment regarding the heart isozymes?

 

 

⑴ answer: D

 

 

Questions 151-153

Cholesterol is an essential component of some membrane microdomains. FRET (fluorescence resonance energy transfer) is a process by which the energy absorbed by a fluorescent molecule can be transferred to an adjacent fluorescent molecule. When FRET occurs, the neighboring molecule, rather than the excited molecule, fluoresces. To determine if two membrane proteins, X and Y, are near neighbors present in a microdomain of the plasma membrane, the proteins were tagged with CFP or YFP (CFP = cyan fluorescent protein; YFP = yellow fluorescent protein). CFP is excited by 435 nm light and emits fluorescent light at 480 nm. YFP is excited by 488 nm light and emits light at 535 nm. The following data were obtained.

 

 

 

151. The reason that the curve in Figure 2 is flat is 

 

 

⑴ answer: D

 

 

152. The reason that there are two peaks of emitted fluorescent light in Figure 4 is 

 

 

⑴ answer: A

⑵ FRET : 특히 CFP emission wavelength 480 nm가 YFP의 excitation wavelength 488 nm와 매우 유사하므로 가능

 

 

153. These data suggest which of the following?

 

 

⑴ answer: C

⑵ 해석 : membrane microdomain의 중요한 요소인 콜레스테롤이 없어지면 X-CFP, Y-YFP가 더이상 인접하지 않음

 

 

Questions 154-157

An E. coli culture is grown on a mixture of glucose and lactose as carbon sources. The growth curve for the culture is shown in the figure below.

 

 

 

154. Which region(s) of the curve correspond(s) to exponential growth of the culture?

 

 

⑴ answer: E

⑵ 회분 배양의 생장 곡선 

 

 

155. Which of the following statements is true regarding region Y of the growth curve?

 

 

⑴ answer: A

⑵ 통성혐기성(facultative) : 호기성과 혐기성 조건에서 모두 살 수 있게 대사경로를 바꿀 수 있는 미생물

 

 

156. The growth experiment described above is repeated using an E. coli strain with a deletion of the lacZ gene. Which of the following statements is correct about the strain?

 

 

⑴ answer: B

⑵ 베타갈락토시다아제(β-galactosidase, lac Z) : 젖산 분해효소, 젖당을 알로젖당으로 전환

 

 

157. The growth curve is the result of 

 

 

⑴ answer: B

⑵ lac 오페론의 양성적 조절 : lac operon 작동보다 효율적인 포도당을 우선적으로 사용하여 lac operon 상의 불필요한 효소를 합성하지 않음

 

 

Questions 158-160

In a series of experiments designed to test the effect of glycosylation on proteins, the hen egg white enzyme lysozyme (MW ~ 14 kD) was genetically modified so as to introduce a site for N-glycosylation. When the protein was expressed in yeast, two forms of the enzyme were isolated: one with a short oligomannose chain, and one with a long polymannose chain. These proteins were examined for enzymatic activity with two different substrates: a soluble substrate, glycol chain, and an insoluble substrate, a cell wall suspension from the bacterium M. luteus, as shown in the table below. Further experiments were carried out in order to determine the thermal stability of the proteins, as shown in the figure below. Thermal stability was estimated by measuring the developed turbidity (precipitation) and residual enzymatic activity when samples were heated from 30 ℃ to 95 ℃. Turbidity was measured as optical density at 500 nm. Residual enzyme activity was measured by hydrolysis of glycol chitin.

 

 

 

158. What is the primary effect of glycosylation on enzymatic activity assayed with the soluble or insoluble substrates?

 

 

⑴ answer: D

⑵ 주어진 표를 보면 enzymatic activity를 줄이는 대신 선택성을 높이는 것으로 보임 

 

 

159. Which of the following most accurately describes the observed effects of glycosylation on thermal stability of the enzyme?

 

 

⑴ answer: E

⑵ ● > ■ > ◆ 순으로 열적 안정성이 증가하는 것으로 보아 E가 맞는 설명

 

 

160. A second glycosylation site is engineered into the lysozyme gene, and protein expression in yeast results in additional polymannosylation at this second site. Which of the following is the most likely result of this additional glycosylation relative to polymannosylation at the single site?

 

 

⑴ answer: B

⑵ 158번과 비슷한 결론 

 

 

Questions 161-163

The pedigree below follows the inheritance pattern of a late-onset (after age thirty) genetic disease that is 100 percent penetrant; affected individuals are represented as a shaded circle (female) or square (male). A restriction fragment length polymorphism (RFLP) analysis of each individual's DNA is shown below the pedigree. Individuals IIIa through IIIi are under the age of thirty.

 

 

 

161. Which grandchildren (IIIb-IIIi) will eventually be affected by the disease?

 

 

⑴ answer: E

⑵ Fragment A는 모두가 공유하므로 중요하지 않음. 질환자는 Ib, Ic, IIb인데 Fragment D가 IIb에서 관찰되지 않으므로 중요하지 않음. 따라서 Fragment C가 있는지로 질환자 여부를 가릴 수 있음 

⑶ 물론 그 Fragment의 존재성이 바로 질환자가 되는지(즉, 우성인지)는 별도로 확인해야 하지만, 다른 사실관계(예를 들어, IIIa는 정상)를 비교·대조하면 E로 답이 확정될 수 있을 것으로 보임

 

 

162. Individual IIIb and the indicated woman (IIIa) whose family has no history of the disease, have a child. What is the probability that the child will be affected?

 

 

⑴ answer: D

⑵ 결국 자손이 Fragment D를 보유할지에 대한 확률이므로 50% 

 

 

163. What is the most likely explanation for the inheritance of fragment A?

 

 

⑴ answer: B

 

 

Questions 164-166

Chlamydomonas is a unicellular green alga containing two flagella per haploid cell. Under certain conditions, two haploid cells can be induced to mate, producing a single cell containing a mixture of the cytoplasm from both haploid cells. This state is called a dikaryon, and the single cell now contains four flagella instead of two.

 

In addition to dikaryon formation, two other characteristics of this research organism allow investigators to study flagellar assembly: First, the flagella on individual cells can be easily removed, and the cell will regenerate new flagella as shown by the kinetics in Figure 1 below. Second, new protein synthesis is required to produce full-length flagella, as shown in Figure 2, by using the drug cycloheximide (whose effects can be rapidly reversed simply by removal of the drug from the medium).

 

 

In an experiment to observe the response of cells to dikaryon formation (Figure 3), cell A has been deflagellated and allowed to reassemble its flagella in the presence of cycloheximide. Cell B, which carries a myc-tagged tubulin gene, had intact flagella. Cell A and cell B are then induced to form a dikaryon in medium lacking cycloheximide. After time sufficient to allow full-length flagellar regeneration, fluorescently tagged antibodies to tubulin were used to identify all four flagella of the dikaryon (Figure 4a), while anti-myc antibodies were used to identify tubulin carrying the myc epitope tag (Figure 4b).

 

 

 

164. The most likely explanation for the effect of cycloheximide on flagellar assembly (Figure 2) in the cells is that it

 

 

⑴ answer: C

 

 

165. Which of the following statements best summarizes the results of the experiment as represented in Figures 3 and 4?

 

 

⑴ answer: B

⑵ Figure 4b를 보면 왼쪽 2개는 기존에 Cell B에 있던 거고, 오른쪽 2개는 새로 합성된 것. cycloheximide 때문에 Cell A에 튜불린이 없으므로 Cell B에서 온 튜불린이 오른쪽 2개 합성에 관여한 것

 

 

166. The data shown in Figures 3 and 4 indicate that during flagellar assembly, new tubulin subunits are incorporated

 

 

⑴ answer: D

⑵ Fibure 4b에서 오른쪽 2개가 새로 합성된 건데, cycloheximide가 영향을 주는 건 proximal half가 아니라 distal half

 

 

Questions 167-169

Cell cycle-specific protein kinases are believed to be involved in the activation of the phosphatase Cdc25. To monitor phosphorylation of Cdc25, different kinases were mixed with Cdc25 and ³²P-labeled ATP. Following incubation under the experimental conditions described below, the reaction mixtures were subjected to SDS-PAGE and autoradiography (Figure 1). The appearance of a band on the autoradiogram represents newly phosphorylated Cdc25. In parallel, Cdc25 activity was measured and the results graphed directly below the corresponding gel lane (Figure 2).

 

 

 

167. According to the data, which of the following is correct?

 

 

⑴ answer: B

⑵ SDS-PAGE가 아미노산마다 붙으면 음하전 간의 반발력으로 모든 단백질이 동일한 전하밀도를 가지는 2차 구조가 됨

⑶ Lane 4를 보면 p38이 먼저 작용하고 Polo kinase가 작용해야 이중 활성화가 가능하고, Polo kinase가 먼저 작용하면 p38이 추가로 인산화하지 못함을 알 수 있음

 

 

168. Which of the following is supported by the data?

 

 

⑴ answer: C

⑵ Lane 4만 활성이 있고, 이는 Cdc25의 활성에는 p38, Polo kinase가 모두 작용해야만 함을 알 수 있음

 

 

169. The data support which of the following statements concerning the phosphorylation and activation of Cdc25?

 

 

⑴ answer: D

⑵ Lane 4 및 167번 설명과 동일함 

 

 

Questions 170-172

The effects of an RNA polymerase inhibitor were determined using HeLa cells in culture. Before the addition of inhibitor, ³H-uridine and ³⁵S-methionine were added to separate cultures, and the accumulation of labeled RNA (Figure 1) and labeled protein (Figure 2) was determined over time. After 10 hours of labeling, the cultures were split and the RNA polymerase inhibitor was added to one culture of each type, as indicated by the arrows in the graphs below. In control cultures, incubation was continued in the absence of the inhibitor as shown. 

 

 

To further characterize the effects of the inhibitor on HeLa cells, a nuclear extract was made and fractionated by anion exchange chromatography using a DEAE Sephadex column. After the column was washed, bound proteins were eluted using a gradient of increasing (NH₄)₂SO₄ concentration. The individual fractions were examined for RNA polymerase activity in the absence (solid lines) or presence (dotted lines) of the RNA polymerase inhibitor (Figure 3).

 

 

 

170. In the cell culture experiments, the incorporation of ³⁵S into newly synthesized protein continued for several hours in the presence of the inhibitor. In these experiments, continuation of protein synthesis occurred mostly because of the time it takes for

 

 

⑴ answer: A

⑵ ³⁵S는 단백질을 정량하는 것이고, 억제제 처리 후 단백질이 계속 합성되는 것은 이미 전사된 mRNA 때문이라는 결론 

 

 

171. As determined by the in vivo cell labeling experiments in Figure 1, the percent of the total RNA synthesis that is sensitive to the inhibitor is closest to

 

 

⑴ answer: B

⑵ %sensitive = 1 - #insensitive / (#sensitive + #insensitive) = 1 - (3000 - 2000) / (6000 - 2000) = 0.75 

 

 

172. The material eluting from the column in peak 1 (Figure 3) localized to the nucleolus. This RNA polymerase is most likely directly involved in the production of 

 

 

⑴ answer: A

⑵ 인(nucleolus) : rRNA 합성, telomerase 합성, 리보솜 소단위체의 조립 장소

 

입력: 2026.05.03 00:52