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【单选题】
SI13包括两个bit位CCN_ACTIVSI_STATUS_IND来指示（）功能是否激活。
A.
DYMA
B.
DTM
C.
EDA
D.
NACC
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举一反三
【单选题】Directions: There are 20 blanks in the following passage, and for each blank there are 4 choices marked A, B, C and D at the end of the passage. You should choose ONE answer that best fit into the pas...
A.
giving
B.
receiving
C.
accepting
D.
learning
【单选题】有如下事件过程：Private Sub Form_Load() Show Dim Arr() As Variant Arr = Array(1, 3, 5, 7, 9, 11, 13, 15) Call converse(Arr) For i = 0 To 7 Print arr(i); Next iEnd Sub 以上程序的功能是，通过调用Converse过程，将数组Arr中的元素逆序存放并输出...
A.
Private Sub converse(dim a() as Variant ) Dim T For i = LBound(A) To UBound(A) T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
B.
Private Sub converse(dim a() as Variant) Dim T Dim J As Integer J = (LBound(A) + UBound(A)) / 2 For i = LBound(A) To J T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
C.
Private Sub converse(a() As Variant) Dim T For i = LBound(A) To UBound(A) T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
D.
Private Sub converse(a() As Variant) Dim T Dim J As Integer J = Int((LBound(A) + UBound(A)) / 2) For i = LBound(A) To J T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
【单选题】酶原的激活是由于
A.
激活剂将结合在酶原分子上的抑制剂除去
B.
激活剂使酶原的空间构象发生变化
C.
激活剂携带底物进入酶原的活性中心
D.
激活剂使酶原分子的一段肽水解脱落从而形成活性中心
E.
激活剂使酶含量增加
相关题目：
【单选题】Directions: There are 20 blanks in the following passage, and for each blank there are 4 choices marked A, B, C and D at the end of the passage. You should choose ONE answer that best fit into the pas...
A.
giving
B.
receiving
C.
accepting
D.
learning
【单选题】有如下事件过程：Private Sub Form_Load() Show Dim Arr() As Variant Arr = Array(1, 3, 5, 7, 9, 11, 13, 15) Call converse(Arr) For i = 0 To 7 Print arr(i); Next iEnd Sub 以上程序的功能是，通过调用Converse过程，将数组Arr中的元素逆序存放并输出...
A.
Private Sub converse(dim a() as Variant ) Dim T For i = LBound(A) To UBound(A) T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
B.
Private Sub converse(dim a() as Variant) Dim T Dim J As Integer J = (LBound(A) + UBound(A)) / 2 For i = LBound(A) To J T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
C.
Private Sub converse(a() As Variant) Dim T For i = LBound(A) To UBound(A) T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
D.
Private Sub converse(a() As Variant) Dim T Dim J As Integer J = Int((LBound(A) + UBound(A)) / 2) For i = LBound(A) To J T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
【单选题】酶原的激活是由于
A.
激活剂将结合在酶原分子上的抑制剂除去
B.
激活剂使酶原的空间构象发生变化
C.
激活剂携带底物进入酶原的活性中心
D.
激活剂使酶原分子的一段肽水解脱落从而形成活性中心
E.
激活剂使酶含量增加
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