ヒトゲノムのセンセーショナルなタイトル

このQ&Aのポイント
  • ヒトゲノムの遺伝子数とDNA識別標の関係について解説します
  • DNA配列の長さとヒトゲノムの特異性について詳しく説明します
  • ヒトゲノムのDNA配列における識別標の最小長さについて考えます
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  • ベストアンサー

ヒトゲノムの質問です。答えが解りません。

There are about 25,000 genes in the human genome. If you wanted to use a stretch of the CAN of each gane as a unique identification tag , roughly what minimum length of DNA sequence would you need? To be unique , the length of DNA in nucleotides would have to have a diversity ( the number of different possible sequences ) equivalent to at least 25,000 and would have to be present once in the haploid human genome ( 3.2 × 109 nucleotides)? (Assume that A,T,C and G are present in equal amounts in the human genome.) 日本語訳 ヒトゲノムには、約25,000の遺伝子がある もしあなたが独特の識別標としてそれぞれのDNAのひと配列を使いたいならば、あなたはおよそどのようなDNA配列の最小の長さを必要とするでしょうか? 独特であるために、ヌクレオチドのDNAの長さが多様性(異なる可能な配列の数)を少なくとも25,000に等しくしなければならない。そして、一倍体のヒトゲノム(3.2 × 109ヌクレオチド)でいったん存在しなければならない。 (A,T,C,Gは、ヒトゲノムで等しい量で存在すると仮定してください。) 訳は自分で訳しました。この問題の答えがさっぱり解りません。 明日までなのでだれか助けてください。

質問者が選んだベストアンサー

  • ベストアンサー
noname#91223
noname#91223
回答No.3

もう宿題に間に合わないかもしれませんが、、 > なんで16乗が最小の16ヌクレオチドと結びつくのですか? 1ヌクレオチドだったら4種類のパターンですから 4の1乗(G,A,T,C) 2ヌクレオチドだったら1塩基目が4種類、2塩基目も4種類で、それらを組み合わせると 4の1乗x4の1乗、つまり4の2乗種類のパターン(GG, GA, GT, GC, AG, AA, AT, AC, TG, TA, TT, TC, CG, CA, CT, CC) 3ヌクレオチドだったら1塩基目が4種類、2塩基目も4種類、3塩基目も4種類で、それらを組み合わせると4の1乗x4の1乗x4の1乗、つまり4の3乗種類のパターン(例は省略) 同様な論理で、16ヌクレオチドだったら4の16乗種類のパターンが存在します。このような計算の詳細は確率統計(?)の教科書をご参照ください。

hekarosu
質問者

お礼

なるほど。確率ですね。 最小で16ヌクレオチドあれば同じ組み合わせができないとゆことですか!すごいわかりやすいです。 宿題は急に休講になったので間に合いました。ほんとうにありがとうございました。

その他の回答 (2)

noname#91223
noname#91223
回答No.2

#1です。「連続する8ヌクレオチドは」は「連続する特定の長さのヌクレオチドは」と読み替えてください。間違えました。

hekarosu
質問者

お礼

4の何乗が3.2 × 10^9を超えるか考える必要があります(GATC4種類で配列を作るため)。これは4の16乗になります。つまり最小で16ヌクレオチド必要ということです。 やっぱりこのぶぶんがわかりません。 なんで16乗が最小の16ヌクレオチドと結びつくのですか?

noname#91223
noname#91223
回答No.1

塩基組成以外にも様々な仮定が必要ですが、そういったことを無視して計算してみます。 > have to be present once in the haploid human genome ということは半数体ゲノム中にユニークな配列でなければなりません。3.2 × 10^9ヌクレオチド中に存在する連続する8ヌクレオチドは、互いの重複も全て含めると全部で約3.2 × 10^9個です。その中でユニーク(present once)であるためには、4の何乗が3.2 × 10^9を超えるか考える必要があります(GATC4種類で配列を作るため)。これは4の16乗になります。つまり最小で16ヌクレオチド必要ということです。 これで半数体ゲノム中の任意の特定部位が識別できるということになりますから、必然的に25000個であろうと何個であろうと任意の特定遺伝子が識別できます。 25000個と敢えて強調しているのは罠でしょうか、それとも私の大きな勘違いでしょうか・・。間違っていたらすみません。

hekarosu
質問者

お礼

回答ありがとうございます。 わかりやすい回答でした。 あと英文DNA→CANなど間違っていました。すいませんでした。 これでおこられずにすみます。

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