Contoh Soal Menghadapi Quiz Uji Nyali Ujian 1

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  • 1/19 Soalan

    Sintesis polinukleotida DNA dengan arah 5′→3′ terjadi melalui penambahan nukleotida baru pada karbon-3′ pada ujung polinukleotida yang telah ada dan melepaskan molekul:

    • Jika 1, 2, dan 3 benar
    • Jika 1 dan 3 benar
    • Jika 2 dan 4 benar
    • Jika hanya 4 yang benar
    • Jika 1, 2, 3, dan 4 benar
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Contoh Soal Menghadapi Quiz Uji Nyali Ujian 1 - Quiz

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  • 2. 

    Ikatan yang menghubungkan antar nukleotida dalam polinukleotida:

    • Jika 1, 2, dan 3 benar

    • Jika 1 dan 3 benar

    • Jika 2 dan 4 benar

    • Jika hanya 4 yang benar

    • Jika semua jawaban benar

    Correct Answer
    A. Jika 1, 2, dan 3 benar
    Explanation
    Jika 1, 2, dan 3 benar, maka ikatan yang menghubungkan antar nukleotida dalam polinukleotida adalah ikatan fosfodiester. Ikatan fosfodiester terbentuk antara gugus fosfat pada nukleotida dengan gugus hidroksil pada nukleotida lainnya, membentuk rantai polinukleotida.

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  • 3. 

    Pernyataan yang TIDAK BENAR tentang genom eukariotik:

    • Genom manusia lebih banyak mengandung “genome wide repeats” dibanding genom ragi

    • Genom ragi lebih banyak mengandung gen dibanding genom lalat buah

    • Genom manusia lebih banyak mengandung gen diskontinyu daripada genom ragi

    • Genom manusia lebih banyak mengandung intron dibanding genom lalat buah

    • Genom lalat buah lebih banyak mengandung intron dibanding genom manusia

    Correct Answer
    A. Genom lalat buah lebih banyak mengandung intron dibanding genom manusia
    Explanation
    The statement that is NOT TRUE about eukaryotic genomes is "Genom lalat buah lebih banyak mengandung intron dibanding genom manusia" (The genome of fruit flies contains more introns than the human genome). This statement is incorrect because the human genome contains a larger number of introns compared to the fruit fly genome.

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  • 4. 

    Rangkaian perubahan urutan dan struktur molekul RNA yang terjadi sebelum molekul RNA ditranslasi menjadi protein atau pada kasus RNA non-coding, sebelum RNA ini berfungsi di dalam sel

    • Sintesis RNA (Synthesis of RNA)

    • Pemrosesan RNA (Processing of RNA)

    • Degradasi RANA (RNA degradation)

    • Sintesis Protein (Protein synthesis)

    • Degradasi Protein (Protein degradation)

    Correct Answer
    A. Pemrosesan RNA (Processing of RNA)
    Explanation
    The correct answer is "Pemrosesan RNA (Processing of RNA)". Processing of RNA refers to the series of modifications and editing that occur to the primary transcript of RNA before it becomes a mature and functional RNA molecule. This includes processes such as splicing, where introns are removed and exons are joined together, as well as modifications like capping and polyadenylation. These processing steps are essential for the proper functioning and stability of RNA molecules.

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  • 5. 

    Selain metionin, salah satu asam amino di bawah ini juga dikode oleh kodon tunggal:

    • Histidin

    • Triptofan

    • Serin

    • Leusin

    • Isoleusin

    Correct Answer
    A. Triptofan
    Explanation
    Triptofan is one of the amino acids that is encoded by a single codon. This means that the genetic code only requires one specific sequence of three nucleotides to produce the amino acid triptofan. This is in contrast to other amino acids that may have multiple codons that can code for them. Therefore, triptofan stands out as an amino acid that has a unique and specific codon for its synthesis.

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  • 6. 

    Komponen penting penghubung transkriptom dengan proteom dalam ekspresi genom:

    • RRNA dan Kode genetik

    • Gen dan tRNA

    • MRNA dan kode genetik

    • Kode genetik dan tRNA

    • MRNA dan tRNA

    Correct Answer
    A. Kode genetik dan tRNA
    Explanation
    The correct answer is "Kode genetik dan tRNA." The code genetic refers to the sequence of nucleotides in DNA or RNA that determines the sequence of amino acids in a protein. tRNA, or transfer RNA, is a type of RNA molecule that carries the correct amino acid to the ribosome during protein synthesis, based on the codon sequence on the mRNA. Therefore, the code genetic and tRNA are both crucial components in connecting the transcriptome (mRNA) with the proteome (protein expression) in the genome.

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  • 7. 

    Strategi percobaan Hershey dan Chase dalam melabel meteri faga T2 adalah sebagai berikut:

    • DNA dilabel dengan isotop sulfur dan protein dilabel fosfor

    • DNA dan protein dilabel isotop fosfor

    • DNA dilabel isotop fosfor dan protein dilabel isotop sulfur

    • DNA dan protein dilabel isotop sulfur

    • Selubung capsid faga dilabel fosfor dan materi di dalam capsid DNA dilabel sulfur

    Correct Answer
    A. DNA dilabel isotop fosfor dan protein dilabel isotop sulfur
    Explanation
    The correct answer is that DNA is labeled with phosphorus isotopes and protein is labeled with sulfur isotopes. This is because Hershey and Chase's experiment involved labeling the DNA and protein of the bacteriophage T2 with different isotopes. By using radioactive isotopes, they were able to track the movement of these molecules and determine that DNA, not protein, is the genetic material that is passed on during viral infection.

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  • 8. 

    Dua percobaan yang mengarahkan ahli biologi molekuler menyimpulkan bahwa gen dibuat DNA dilakukan oleh dua kelompok peneliti di bawah ini:

    • Gregor Mendel dan Friedrich Miescher

    • Friedrich Miescher dan Alfred Hershey

    • Frederick Griffith dan Oswald Avery-Alfred Hershey

    • Frederick Griffith dan Alfred Hershey-Martha Chase

    • Oswald Avery dan Alfred Hershey-Martha Chase

    Correct Answer
    A. Oswald Avery dan Alfred Hershey-Martha Chase
    Explanation
    The correct answer is Oswald Avery and Alfred Hershey-Martha Chase. This conclusion is based on two experiments conducted by the molecular biologists. The experiments conducted by Oswald Avery and Alfred Hershey-Martha Chase provided evidence that DNA carries genetic information and is responsible for the production of genes. This discovery was a significant breakthrough in the field of molecular biology and laid the foundation for further research on genetics.

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  • 9. 

    Fitur penting struktur heliks ganda DNA adalah sebagai berikut, KECUALI:

    • Heliks ganda DNA adalah right-handed

    • Dua untai DNA berlawanan arah (antiparalel)

    • Heliks ganda distabilkan oleh ikatan fosfodiester dan ikatan glikosidik

    • Heliks ganda distabilkan oleh ikatan hidrogen (base pairing)

    • Heliks ganda distabilkan oleh interaksi hidrofobik (base stacking)

    Correct Answer
    A. Heliks ganda distabilkan oleh ikatan fosfodiester dan ikatan glikosidik
    Explanation
    The helical structure of DNA is stabilized by hydrogen bonding between the base pairs, specifically adenine with thymine and guanine with cytosine. This base pairing allows for the complementary and accurate replication of DNA during cell division. The antiparallel arrangement of the two DNA strands allows for efficient base pairing and ensures the stability of the helix. Additionally, the hydrophobic interactions between the stacked bases contribute to the overall stability of the DNA helix. However, the stability of the DNA helix is not primarily dependent on the phosphodiester and glycosidic bonds that connect the individual nucleotides within each DNA strand.

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  • 10. 

    Molekul penghubung sentral (central link) antra ekspresi genom dengan aktivitas biokimiawi sel: 1. DNA 2. Transkriptom 3. mRNA 4. Proteom

    • 1, 2, dan 3 benar

    • 1 dan 3 benar

    • 2 dan 4 benar

    • 4 benar

    • 1, 2, 3, dan 4 benar

    Correct Answer
    A. 4 benar
    Explanation
    The correct answer is 4, Proteom. The proteome refers to all the proteins expressed by a cell, tissue, or organism. It is the central link between the expression of the genome and the biochemical activities of the cell. Proteins are the functional molecules in cells and carry out various biochemical processes, so understanding the proteome is crucial in understanding cellular function and activity.

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  • 11. 

    RNA non-coding baik pada prokariot maupun eukariot: 1. rRNA 2. mRNA 3. tRNA 4. snRNA

    • 1, 2, dan 3 benar

    • 1 dan 3 benar

    • 2 dan 4 benar

    • 4 benar

    • 1, 2, 3, dan 4 benar

    Correct Answer
    A. 1 dan 3 benar
    Explanation
    The correct answer is 1 and 3 are correct. This is because rRNA (ribosomal RNA) is a non-coding RNA found in both prokaryotes and eukaryotes, and tRNA (transfer RNA) is also a non-coding RNA found in both types of organisms. mRNA (messenger RNA) is not a non-coding RNA, as it carries the genetic information from DNA to the ribosomes for protein synthesis. snRNA (small nuclear RNA) is a non-coding RNA, but it is only found in eukaryotes, not prokaryotes. Therefore, options 2 and 4 are incorrect.

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  • 12. 

    Ikatan β‐N‐glikosidik dibentuk antara gula dengan basa nitrogennya sebagai berikut: 1. Karbon-1 gula dengan nitrogen-1 basa adenin 2. Karbon-1 gula dengan nitrogen-1 basa timin 3. Karbon-1 gula dengan nitrogen-9 basa sitosin 4. Karbon-1 gula dengan nitrogen-9 basa guanin

    • 1, 2, dan 3 benar

    • 1 dan 3 benar

    • 2 dan 4 benar

    • 4 benar

    • 1, 2, 3, dan 4 benar

    Correct Answer
    A. 2 dan 4 benar
    Explanation
    The correct answer is 2 and 4 are correct. This is because in the formation of β‐N‐glycosidic bond, carbon-1 of the sugar molecule forms a bond with nitrogen-1 of adenine (base) and carbon-1 of the sugar molecule also forms a bond with nitrogen-9 of guanine (base).

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  • 13. 

    Analisis transkriptom pada tahapan sporulasi Saccharomyces cerevisiae memberikan gambaran aplikasi studi transkriptom untuk mengidentifikasi: 1. Interaksi antara genom dan sinyal lingkungan 2. protein yang urutannya belum diketahui 3. gene yang fungsinya belum ditentukan 4. urutan genom yang urutannya belum ditentukan

    • 1, 2, dan 3 benar

    • 1 dan 3 benar

    • 2 dan 4 benar

    • 4 benar

    • 1, 2, 3 dan 4 benar

    Correct Answer
    A. 1 dan 3 benar
    Explanation
    The correct answer is 1 and 3 are correct. This is because transcriptome analysis during the sporulation stage of Saccharomyces cerevisiae can be used to identify the interaction between the genome and environmental signals (option 1) as well as to determine the function of genes that have not been characterized yet (option 3).

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  • 14. 

    Secara umum fitur genom prokariot dan eukariot berbeda dalam hal: 1. Ukuran gennya 2. keberadaan intron 3. tipe urutan berulang 4. monomer penyusunnya

    • 1, 2, dan 3 benar

    • 1 dan 3 benar

    • 2 dan 4 benar

    • 4 benar

    • 1, 2, 3 dan 4 benar

    Correct Answer(s)
    A. 1, 2, dan 3 benar
    A. 2 dan 4 benar
    Explanation
    The correct answer is 1, 2, and 3 are correct because prokaryotic and eukaryotic genomes differ in terms of their gene size, presence of introns, and types of repetitive sequences. Additionally, 2 and 4 are also correct because eukaryotic genomes have introns and are composed of monomers.

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  • 15. 

    Fitur penting molekul DNA adalah pada urutan nukleotidanya SEBAB Jika molekul DNA telah diketahui urutannya maka gen yang terkandung di dalamnya dapat diidentifikasi dan aktivitas gennya dapat diperlajari secara rinci

    • Pernyataan dan alasan benar, keduanya ada hubungan sebab akibat

    • Pernyataan dan alasan benar, keduanya tidak ada hubungan sebab akibat

    • Pernyataan benar dan alasan salah

    • Pernyataan salah dan alasan benar

    • Pernyataan dan alasan salah

    Correct Answer
    A. Pernyataan dan alasan benar, keduanya ada hubungan sebab akibat
    Explanation
    The explanation for the given answer is that the statement and the reason both state that the important feature of DNA molecules is their nucleotide sequence. The reason supports the statement by explaining that if the sequence of a DNA molecule is known, the genes contained within it can be identified and their activities can be studied in detail. This cause-and-effect relationship between knowing the nucleotide sequence and being able to identify and study genes supports the conclusion that both the statement and reason are correct and are related to each other.

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  • 16. 

    “Tandemly repeated DNA” disebut juga DNA satelit SEBAB fragmen DNA yang mengandung urutan berulang tersebut membentuk pita-pita satelit ketika DNA genomik difraksinasi dengan teknik elektroforesis gen agarosa.

    • Pernyataan dan alasan benar, keduanya ada hubungan sebab akibat

    • Pernyataan dan alasan benar, keduanya tidak ada hubungan sebab akibat

    • Pernyataan benar dan alasan salah

    • Pernyataan salah dan alasan benar

    • Pernyataan dan alasan salah

    Correct Answer
    A. Pernyataan benar dan alasan salah
    Explanation
    The statement is correct because "tandemly repeated DNA" refers to DNA fragments that contain repeated sequences, which can form satellite bands when genomic DNA is fragmented using agarose gel electrophoresis. However, the reason is incorrect because it does not provide a cause-and-effect relationship between the two statements.

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  • 17. 

    Organisasi genetik genom ragi jauh lebih ekonomis dibanding genom manusia SEBABgenom ragi mempunyai densitas gen lebih rendah dibanding genom manusia.

    • Pernyataan dan alasan benar, keduanya ada hubungan sebab akibat

    • Pernyataan dan alasan benar, keduanya tidak ada hubungan sebab akibat

    • Pernyataan benar dan alasan salah

    • Pernyataan salah dan alasan benar

    • Pernyataan dan alasan salah

    Correct Answer
    A. Pernyataan benar dan alasan salah
    Explanation
    The statement is true because the genetic organization of yeast is indeed more economical compared to the human genome. This is because the yeast genome has a lower gene density compared to the human genome. However, the reason provided in the second statement is incorrect, as it states the opposite. Therefore, the statement is true, but the reason is false.

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  • 18. 

    Pasangan basa yang paling mungkin membentuk ikatan hidrogen adalah basa purin-pirimidin SEBAB pasangan basa purin-purin terlalu terlalu besar dan pasangan pirimidin-pirimidin terlalu kecil untuk membentuk ikatan hidrogen.

    • Pernyataan dan alasan benar, keduanya ada hubungan sebab akibat

    • Pernyataan dan alasan benar, keduanya tidak ada hubungan sebab akibat

    • Pernyataan benar dan alasan salah

    • Pernyataan salah dan alasan benar

    • Pernyataan dan alasan salah

    Correct Answer
    A. Pernyataan dan alasan benar, keduanya ada hubungan sebab akibat
    Explanation
    Pasangan basa purin-purin terlalu besar untuk membentuk ikatan hidrogen, karena memiliki struktur yang kompleks dan banyak gugus fungsional. Di sisi lain, pasangan basa pirimidin-pirimidin terlalu kecil untuk membentuk ikatan hidrogen, karena memiliki sedikit gugus fungsional yang dapat berinteraksi dengan basa lainnya. Oleh karena itu, pasangan basa yang paling mungkin membentuk ikatan hidrogen adalah basa purin-pirimidin.

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  • 19. 

    Kode genetik tidak bersifat universal SEBAB beberapa asam amino dikode oleh lebih dari satu kodon

    • Pernyataan dan alasan benar, keduanya ada hubungan sebab akibat

    • Pernyataan dan alasan benar, keduanya tidak ada hubungan sebab akibat

    • Pernyataan benar dan alasan salah

    • Pernyataan salah dan alasan benar

    • Pernyataan dan alasan salah

    Correct Answer
    A. Pernyataan dan alasan benar, keduanya tidak ada hubungan sebab akibat
    Explanation
    The correct answer is "Pernyataan dan alasan benar, keduanya tidak ada hubungan sebab akibat." This means that the statement and the reason are both correct, but they do not have a cause-effect relationship. The statement is true because the genetic code is not universal, meaning that different organisms may have different codons for the same amino acid. The reason is also true because some amino acids are coded by more than one codon. However, the reason does not explain why the genetic code is not universal, so there is no cause-effect relationship between the statement and the reason.

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  • Current Version
  • Mar 21, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Oct 10, 2008
    Quiz Created by
    Hendropramono

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