DP Biology (first assessment 2025)

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Question 23M.1A.HL.TZ1.17

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Date May 2023 Marks available [Maximum mark: 1] Reference code 23M.1A.HL.TZ1.17
Level HL Paper 1A Time zone TZ1
Command term Question number 17 Adapted from N/A
17.
[Maximum mark: 1]
23M.1A.HL.TZ1.17

The table compares ribosomal RNA (rRNA) sequences of two organisms from each of the three domains by showing an association coefficient. The more similar the rRNA sequences of the organisms, the larger the coefficient.

  S. cerevisiae L. minor E. Coli B. firmus M. ruminantium M. barkeri
S. cerevisiae — 0.29 0.05 0.08 0.11 0.08
L. minor   — 0.10 0.06 0.10 0.07
E. Coli     — 0.25 0.12 0.12
B. firmus       — 0.13 0.12
M. ruminantium         — 0.24
M. barkeri           —

[Source: Woese, C.R. and Fox, G.E., 1977. Phylogenetic structure of the prokaryotic domain:
The primary kingdoms. Proc. Natl. Acad Sci. USA, 74(11), pp. 5088–5090.]

What can be concluded from the data?

A.  L. minor and E. coli are both eubacteria.

B.  S. cerevisiae and M. barkeri are in the same domain.

C.  M. ruminantium is an archaean, therefore so is B. firmus.

D.  E. coli and B. firmus are in the same domain.

[1]

Markscheme

D

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