Intermediate of Krebs cycle that can form Chlorophyll?
The standard treatment of this topic is covered in Section 24.4.3 of Berg et al., available freely online:

Succinyl CoA is a precursor for porphyrin in mammalian cells by condensing with glycine to form δ-aminolevulinate as shown in this diagram from that book:



This reaction is catalyzed by mitochondrial δ-aminolevulinate synthase. 

To quote from that reference:


  Two molecules of δ-aminolevulinate condense to form porphobilinogen, the next intermediate. Four molecules of porphobilinogen then condense head to tail to form a linear tetrapyrrole in a reaction catalyzed by porphobilinogen deaminase. The enzyme-bound linear tetrapyrrole then cyclizes to form uroporphyrinogen III, which has an asymmetric arrangement of side chains. This reaction requires a cosynthase. In the presence of synthase alone, uroporphyrinogen I, the nonphysiologic symmetric isomer, is produced. 


This is illustrated in Fig. 24.35 of that section, which also shows that uroporphyrinogen I gives rise to protoporphyrin IX, the precursor of haem:



As regards the synthesis of plant chlorophyl (which also contains a porphyrin ring), the review quoted by March Ho contains the following statement:


  The porphyrin ring with its conjugated double bonds is assembled in the chloroplast from eight molecules of 5-aminolevulinic acid…


(5-aminolevulinic acid is the chemically preferred name for δ-aminolevulinic acid.)

However the review goes on to explain that although 5-aminolevulinic acid synthesized in plant mitochondria uses the same pathway as mammals, that produced in the chloroplast and used for the synthesis of haem that gives rise to chlorophyl uses a different pathway, termed the C5 pathway, in which — as March Ho stated — glutamate is the precursor:



I suspect that the person who set the question is an animal or biochemist who thought he would try a variant of the standard question “which intermediate is a precursor of haem”, and ventured into waters where he was out of his depth.