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SMPN 2 PURWOREJO Author : Roihan M. Iqbal

PENGAYAAN OSN BIOLOGI SMP 2015

PAKET 26 MCAT-based Questions

SMPN 2 PURWOREJO

13 JUNI 2014


PENGAYAAN OSN BIOLOGI SMP PART : 26

(Questions 1–7) Hemoglobin is a vital protein that transports oxygen in the red blood cells. It consists of four

polypeptide chains, and at least seven different gene codes for the subunits or chains of the

molecule. Not all these genes are equally active throughout life. All types of hemoglobin contain two

alpha chains, but the remaining two chains of the molecule vary depending on age. At different

times in the life of the individual, the proportion of hemoglobin molecules containing a particular

type of chain (pair) may vary (see y-axis). The molecule found in RBCs at a particular time is

designated by the types of chains present and how many of each (example: alpha2beta2). The figure

below reflects the timing of production of the different chains during human development.

1. According to the figure, most fetal hemoglobin contains

a. alpha and beta chains.

b. alpha and gamma chains.

c. alpha and epsilon chains.

d. gamma and beta chains.

2. By extrapolating (estimating) from the figure, what is the composition of most adult

hemoglobin molecules?

a. Alpha and beta chains

b. Alpha and gamma chains

c. Beta and delta chains

d. Alpha and delta chains

3. The hemoglobin gene that appears to stop functioning earliest encodes the

a. delta chain.

b. gamma chain.

c. epsilon chain.

d. beta chain.


4. At approximately what point does the beta chain start to appear in a higher proportion of

hemoglobin molecules than the gamma chain?

a. Three to six months before birth

b. Just before birth

c. Between birth and three months of age

d. After six months of age

5. After determining from the figure when each type of hemoglobin is present, which types would

you predict have an extremely high affinity for oxygen?

a. Alpha2gamma2 and alpha2delta2

b. Alpha2gamma2 and gamma2epsilon2

c. Alpha2epsilon2 and gamma2beta2

d. Alpha2gamma2 and alpha2epsilon2

6. Not all adult hemoglobin is of the same type. Based on information in the graph, a small

proportion of adult hemoglobin is probably

a. alpha2beta2.

b. alpha2gamma2.

c. alpha2delta2.

d. alpha2epsilon2.

7. Based on information in the figure, what is the maximum percentage of hemoglobin molecules

present at any one time that can be epsilon2delta2?

a. 0 percent

b. 55 percent–58 percent

c. 0 percent–5 percent

d. It cannot be determined.

8. Portions of the afferent arteriole, efferent arteriole, distal convoluted tubule, and renal

corpuscle make up the juxtaglomerular apparatus, which produces renin. The juxtaglomerular

apparatus is involved with regulation of

a. glucose.

b. pH.

c. wastes.

d. blood pressure.

9. The appendix is a structure on the most proximal part of the large intestine with no apparent

digestive function. Where would you find the appendix?

a. Ascending colon

b. Cecum

c. Rectum

d. Sigmoid colon


(Questions 10–15) The role of chlorophyll, a light-absorbing pigment, in photosynthesis was investigated in a pair

of experiments. Pigment was extracted from a leaf preparation and examined using paper

chromatography and spectrophotometric techniques.

Experiment 1

In order to analyze the composition of leaf extract, a sample of the pigment was applied near

the bottom of a strip of chromatographic filter paper and allowed to dry. This was repeated a

number of times to ensure that an adequate amount of pigment was present. The strip was

suspended in a test tube containing the solvent acetone. As the experiment proceeded, the

acetone soaked into the filter paper and slowly moved up across the pigment sample,

eventually reaching the top of the paper. This technique is designed to separate substances in a

mixture based on differences in their tendency to adhere to the material over which they are

passed (adsorption), as well as on their solubility in the solvent used. Substances with color can

easily be detected on the paper. The experiment revealed three separate bands of color on the

filter paper: a green band, a yellow-green band, and an orange band.

Experiment 2

Samples of each band were examined using a spectrophotometer. This instrument can send

light of various wavelengths (375 nm to 750 nm) through a sample. For each wavelength, the

amount of light passing through (percentage of transmittance) can be monitored and then

converted to absorbance. Absorbance is a measure of the amount of light absorbed as it passes

through a sample. Unless light is absorbed, the energy in the light cannot be utilized. A sample’s

absorption spectrum indicates the range of its ability to absorb different wavelengths of light.

The absorption spectra of all three samples are shown in the following figure.


10. Which statement is true concerning the paper chromatography technique used in

Experiment 1?

a. All other factors being equal, the most soluble substance travels fastest and will likely be

highest on the paper.

b. All other factors being equal, the substance with the greatest adsorptive tendency will

likely be highest on the paper.

c. Both A and B

d. Neither A nor B

11. The most reasonable hypothesis that addresses the results of Experiment I is which of the

following?

a. The chlorophyll extract may contain three different pigments.

b. Chlorophyll has three different solubility/adsorption patterns.

c. The chlorophyll applied to the filter paper was not allowed to dry properly.

d. Chlorophyll can have three different conformations.

12. As seen in the figure, the three samples from Experiment 1 have different absorption

spectra. Which of the following statements is correct?

a. Each substance shows at least two different absorbance peaks of 40 percent or higher.

b. Substance A shows a low level of transmittance across a broad range of wavelengths.

c. Each pigment shows an absorbance peak at both ends of the spectrum.

d. Substance B shows its greatest absorption in the blue range of wavelengths.

13. Which colors of light are most likely utilized in photosynthesis by substance A?

a. Blue-green

b. Blue-green and red

c. Violet-blue and red

d. Violet-blue and orange

14. What do the results of Experiment 2 suggest about the question “Why are green plants

green?”

a. The light wavelengths utilized by each pigment mix to form green.

b. The pigments absorb green.

c. The pigments do not absorb green.

d. Green light is the most utilized light in photosynthesis.

15. Which hypothesis best interprets the presence of more than one pigment in green leaves?

a. The ability of different pigments to use the same wavelengths equally well makes

photosynthesis more efficient.

b. Making small amounts of more than one pigment is energetically less expensive than

making a large amount of the same pigment.

c. When light is absorbed by more than one pigment, heat can be dissipated in the plant

more easily.

d. The ability of different pigments to use different wavelengths makes photosynthesis

more efficient.


(Questions 16–21) Phenylalanine is an amino acid that is necessary in small amounts for proper human nutrition.

The normal metabolism of this substance requires the enzyme phenylalanine hydroxylase. The

ability to produce the enzyme is controlled by an autosomal dominant trait. Approximately one

in 15,000 whites in the United States is born without the capacity to produce the enzyme due to

a homozygous recessive condition called phenylketonuria (PKU). In people with PKU,

phenylalanine accumulates in the blood, and phenylpyruvic acid accumulates in the urine—

giving it a distinctive odor. High levels of phenylalanine in the brain and spinal fluid during the

first weeks after birth causes mental retardation in PKU individuals. Since PKU can be identified

by a blood test at birth, dietary treatment can be administered to prevent this damage. If

placed on a low phenylalanine diet for the first few years of life, symptoms can be avoided.

16. If a couple, both heterozygous for PKU, have a child, what is the probability that the child will

have at least one normal allele?

a. 100 percent

b. 75 percent

c. 50 percent

d. 0 percent

17. If the same couple were to have two children, what is the probability that both children will

be carriers (heterozygotes) of the disease?

a. 100 percent

b. 75 percent

c. 50 percent

d. 25 percent

18. A woman, homozygous for PKU, was successfully treated after birth to avoid symptoms of

the disorder. If she becomes pregnant, why must similar precautions be taken to control her

diet again?

a. The fetus must have the genotype for the disorder.

b. High levels of phenylalanine in her blood can cause brain damage in the fetus, regardless

of its genotype.

c. High levels of phenylalanine in her blood can stimulate her body to overproduce

phenylalanine hydroxylase, which causes damage in the fetus.

d. Phenylpyruvic acid in her urine can cause damage to the fetus.

19. Holandric traits are Y-linked. Which of the following statements is correct concerning such

traits?

A. Holandric traits will only appear in males.

B. Every holandric gene that is present will be expressed.

C. Both A and B

D. Neither A nor B


20. Amyotrophic lateral sclerosis (Lou Gehrig’s Disease) is caused by an autosomal recessive

gene. Which of the following statements is correct concerning family members of individuals

with this disorder?

A. If not affected themselves, both parents of an affected individual have to be carriers.

B. If not affected themselves, all siblings of an affected individual must be carriers.

C. Both A and B

D. Neither A nor B

21. SOAL ALTERNATIF Polydactyly (extra toes or fingers) is an autosomal dominant trait. Identify

the correct statement concerning this trait.

1. All children of homozygotes must have the trait.

2. Affected children must have at least one affected parent.

3. Twenty-five percent of the children of two heterozygote parents must have the trait.

22. A countercurrent mechanism in the kidney helps maintain a high concentration of salt in the

interstitial fluid of the medulla, the site adjacent to where the hypotonic tiltrate passes

before its elimination out of the body. This permits efficient

A. secretion of excess salts.

B. adjustment of filtrate pH.

C. filtration of venous blood returning to the heart.

D. reabsorption of water.

23. Point mutations are changes in a single DNA nucleotide base pair. An addition results in an

extra base pair, while a deletion results in an omitted base pair along the DNA molecule. In

contrast, a substitution results in an incorrect base pair replacing the correct one. Which

point mutations have the highest probability of causing drastic results?

A. Substitutions

B. Deletions

C. Additions and deletions

D. All three have equal chances of causing drastic results

24. Water and small proteins leak out of capillaries at their arterial ends because hydrostatic

pressure (exerted mainly by blood pressure pushing outward against the capillary walls) is

greater than colloid osmotic pressure (a fluid-retaining force caused by large solutes in the

blood). Most of the fluid returns at the venule end because blood pressure

A. increases and large solutes decrease.

B. decreases and large solutes decrease.

C. increases and large solutes stay the same.

D. decreases and large solutes stay the same.


25. Under the microscope, numerous structures observed in the renal pyramids have a relatively

“straight-line” appearance. What parts of the nephron are probably present in the renal

pyramids? Cross the column provided for each TRUE options.

No Options

A Loops of Henle

B Proximal convoluted tubules

C Bowman’s capsules

D Collecting ducts

The evolutionary origin of viruses remains a mystery.

Hypothesis A

One hypothesis is that the ancestors of modern viruses were free-living heterotrophs that

evolved in the early seas and fed on the organic molecules around them. By the time these

nutrients became depleted, other heterotrophic and autotrophic forms may have begun to

appear, and viruses adapted to a parasitic lifestyle.

Hypothesis B

A second hypothesis favors the idea that viruses evolved from a cellular form and subsequently

became highly specialized parasites. So specialized were they, that they “lost” almost all of their

cellular components except the nuclear (genetic) material.

Hypothesis C

A third hypothesis is that viruses arose from bits of nucleic acid that escaped from cellular

organisms. They may have survived at first without any form of outer covering before evolving

their specialized coats. Various viruses may have arisen this way from different kinds of

organisms.

26. Which hypothesis views viruses as having evolved independently from prokaryotes and

eukaryotes?

A. Hypothesis A

B. Hypothesis B

C. Hypothesis C

D. None of the above

27. Which of the following questions would be especially difficult to answer for a proponent of

the view that viruses predate cells?

A. Why are viruses so much smaller than cellular forms?

B. How could viruses reproduce before cells evolved?

C. From what precursor molecules could viral DNA and viral proteins have formed?

D. From what source could viruses have obtained nourishment?


28. Which piece of evidence would strongly support only hypothesis C?

A. Viral genetic material is usually made of the same type of nucleic acid as that of the

viruses’ hosts.

B. Sometimes, viral genes are virtually identical to host-cell genes.

C. Viruses are well adapted as parasites to their host cells.

D. Viruses are all very similar in their genetic makeup.

29. Which hypothesis appears strongest in view of the frequent observation today that there is

greater genetic similarity between viruses and their hosts than between types of viruses?

A. Hypothesis A

B. Hypothesis B

C. Hypothesis C

D. Hypotheses B and C

30. Which of the following is not implied by hypothesis A?

A. Photosynthetic organisms evolved after viruses.

B. Viral ancestors could absorb.

C. Viral ancestors required host cells to reproduce.

D. Viral ancestors could metabolize food on their own.

End of papers

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