Molecular Biology of the Cell by Bruce Alberts, Alexander Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter Walter by by Bruce Alberts, Alexander Johnson, Julian Lewis, David Morg

1088 Chapter 19: Cell Junctions and the Extracellular Matrix
transepithelial electrical
resistance (ohms cm 2 )
10,000
8000
6000
4000
2000
0
(A) BASOLATERAL
– toxin
+ toxin
wash
(B) APICAL
wash
0 8 16 24 32 40 48 0 8 16 24 32 40 48
hours
hours
Figure Q19–2 Effects of Clostridium toxin on the barrier function
of MDCK cells (Problem 19–7). (A) Addition of toxin from the
basolateral side of the epithelial sheet. (B) Addition of toxin from
the apical side of the epithelial sheet. For a given voltage, a higher
resistance (ohms cm 2 ) gives less paracellular current.
Problems p19.08/19.12
the absence of claudin-4, the cells remain healthy and
the epithelial sheet appears intact. The mean number
of strands in the tight junctions that link the cells also
decreases over 24 hours from about four to about two, and
they are less highly branched. A functional assay for the
integrity of the tight junctions shows that transepithelial
resistance decreases dramatically in the presence of the
toxin, but the resistance can be restored by washing out
the toxin (Figure Q19–2A). Curiously, the toxin produces
these effects only when it is added to the basolateral side
of the sheet; it has no effect when added to the apical surface
(Figure Q19–2B).
A. How can it be that two tight-junction strands
remain, even though all of the claudin-4 has disappeared?
B. Why do you suppose the toxin works when it is
added to the basolateral side of the epithelial sheet, but
not when added to the apical side?
19–8 It is not an easy matter to assign particular functions
to specific components of the basal lamina, since
the overall structure is a complicated composite material
with both mechanical and signaling properties. Nidogen,
for example, cross-links two central components of the
basal lamina by binding to the laminin γ-1 chain and to
type IV collagen. Given such a key role, it was surprising
that mice with a homozygous knockout of the gene for
nidogen-1 were entirely healthy, with no abnormal phenotype.
Similarly, mice homozygous for a knockout of the
gene for nidogen-2 also appeared completely normal. By
contrast, mice that were homozygous for a defined mutation
in the gene for laminin γ-1, which eliminated just the
binding site for nidogen, died at birth with severe defects
in lung and kidney formation. The mutant portion of the
laminin γ-1 chain is thought to have no other function
than to bind nidogen, and does not affect laminin structure
or its ability to assemble into the basal lamina. How
would you explain these genetic observations, which are
summarized in Table Q19–1? What would you predict
would be the phenotype of a mouse that was homozygous
for knockouts of both nidogen genes?
Table Q19–1 Phenotypes of mice with genetic defects
in components of the basal lamina (Problem 19–8).
Protein Genetic defect Phenotype
Nidogen-1 Gene knockout (–/–) None
Nidogen-2 Gene knockout (–/–) None
Laminin γ-1
Laminin γ-1
Nidogen binding-site
deletion (+/–)
Nidogen binding-site
deletion (–/–)
None
Dead at birth
+/– stands for heterozygous, –/– stands for homozygous.
19–9 Discuss the following statement: “The basal lamina
of muscle fibers serves as a molecular bulletin board,
in which adjoining cells can post messages that direct the
differentiation and function of the underlying cells.”
19–10 The affinity of integrins for matrix components can
be modulated by changes to their cytoplasmic domains:
a process known as inside-out signaling. You have identified
a key region in the cytoplasmic domains of α IIb β 3
integrin that seems to be required for inside-out signaling
(Figure Q19–3). Substitution of alanine for either D723
in the β chain or R995 in the α chain leads to a high level
of spontaneous activation, under conditions where the
wild-type chains are inactive. Your advisor suggests that
you convert the aspartate in the β chain to an arginine
(D723R) and the arginine in the α chain to an aspartate
(R995D). You compare all three α chains (R995, R995A,
and R995D) against all three β chains (D723, D723A, and
D723R). You find that all pairs have a high level of spontaneous
activation, except D723 vs R995 (the wild type) and
D723R vs R995D, which have low levels. Based on these
results, how do you think the α IIb β 3 integrin is held in its
inactive state?
extracellular
space
β 3
α IIb
plasma
membrane
cytoplasm
WKLLITIHDRKEF
WKVGFFKRNRP
COOH
COOH
Figure Q19–3 Schematic representation of α IIb β 3 integrin
(Problem 19–10). The D723 and R995 residues are indicated.
(From P.E. Hughes et al., J. Biol. Chem. 271:6571–6574, 1996.
With permission from American Society for Biochemistry and
Molecular Biology.)
19–11 The glycosaminoglycan polysaccharide chains
that are linked Problems to specific p19.16/19.15
core proteins to form the proteoglycan
components of the extracellular space are
highly negatively charged. How do you suppose these
negatively charged polysaccharide chains help to establish
a hydrated gel-like environment around the cell? How
would the properties of these molecules differ if the polysaccharide
chains were uncharged?
995
723