Chakapuli

Chakapuli
Type Stew
Main ingredients Lamb or veal, onions, cherry plums, dry white wine, potatoes, tarragon leaves, herbs, garlic

Chakapuli (Georgian: ჩაქაფული) is a Georgian[1][2][3][4] stew. It is considered to be one of the most popular dishes in Georgia.

It is made from lamb chops or veal, onions, tarragon leaves, cherry plums or tkemali (cherry plum sauce), dry white wine, mixed fresh herbs (parsley, mint, dill, coriander), garlic and salt.[5]

See also

References

  1. ↑ Darra Goldstein, The Georgian Feast: The Vibrant Culture and Savory Food of the Republic of Georgia, p. 87
  2. ↑ Tim Burford, Georgia, p. 74
  3. ↑ Семенова С.В. Грузинская кухня, p. 16
  4. ↑ Любомирова К. Постные блюда из мультиварки, p. 14
  5. ↑ "Georgian Recipes: Chakapuli". Georgia About. Retrieved 6 March 2015.



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