Tính số liên kết hóa trị biết L = 3978Å và thời gian sao mã 1 phân tử mARN hết 6 phút 30 giây

Chiều dài của gen là 3978 A^{o. Thời gian sao mã 1 phân tử mARN hết 6 phút 30 giây. Thời gian sao mã đối với mỗi loại nu trên mạch mang mã gốc của gen đó theo thứ tự X:A:T:G tương ứng với tỉ lệ về thời gian sao mã là 1:2:4:6.}

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- Vận tốc trượt của riboxom trên phân tử mARN.

- Khoảng cách về độ dài từ riboxom thứ nhất đến riboxom thứ sáu khi chúng còn đang trượt trên phân tử mARN.

Theo dõi Vi phạm

Sinh học 9 Bài 16 Trắc nghiệm Sinh học 9 Bài 16 Giải bài tập Sinh học 9 Bài 16

Trả lời (1)

a. Xác định số liên kết hóa trị nối giữa các đơn phân của gen

Số nu một mạch đơn của gen là: 3978 / 3,4 = 1170 nu

Vận tốc phiên mã là: 1170 / 390s = 3 nu/s

Số nu từng loại trên mạch gốc của gen:

X = 3*390/13 = 90 nu;

A = 90 * 2 = 180 nu;

T = 90 * 4 = 360 nu;

G = 90 * 6 = 540 nu.

Số nu từng loại của gen là:

A = T = 180 + 360 = 540 nu;

G = X = 90 + 540 = 630 nu

Số lk hóa trị nối giữa các đơn phân của gen: (1170 * 2) – 2 = 2338 lk.

b. Số lượng ribonu từng loại trên mARN là: mạch mã gốc → mARN

A₁ = 180 nu → U_m = 90 ribonu;

T₁ = 360 nu → A_m = 360 ribonu;

G₁ = 540 nu → X_m = 540 ribonu;

X₁ = 90 nu → G_m = 90 ribonu;

c. *Vận tốc trượt của riboxom trên mARN là: 61,2 / 1,2 = 51 A^o/s

* Khoảng cách về độ dài từ riboxom thứ nhất đến thứ 6 khi chúng còn đang trượt trên mARN là: 61,2 * 5 = 306 A^o.

bởi Nguyen Long 19/11/2018

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