Question

Which of the following additive makes Gulabjamun relatively resistant to microbial spoilage

• A. use of nuts
• B. soaking in sugar syrup
• C. use of condensed milk
• D. added essence

Hint:

High sugar = High osmotic pressure = Low water activity ($a_{w}$). This is the same principle used to preserve jams and honey!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Preservation of traditional Indian dairy sweets (mithai) relies on techniques that lower the water activity ($a_{w}$) of the product.
Water activity is the amount of "free" water available for microbial growth.

Step 2: Detailed Explanation:
Gulabjamun consists of fried balls of khoa/dough that are subsequently soaked in a hot sugar syrup.
1.Sugar Syrup: The syrup usually has a high sugar concentration (around 60 - 70 % TSS). This creates high osmotic pressure.
2.Mechanism: When the Gulabjamun is soaked, the sugar syrup penetrates the product. The high concentration of sugar binds water molecules, effectively reducing the water activity ($a_{w}$).
3.Result: Most bacteria cannot grow in environments with very low $a_{w}$ because the osmotic pressure causes water to leave the bacterial cells (plasmolysis). This makes the product relatively resistant to spoilage compared to fresh khoa.
Additives like nuts or essence do not significantly alter the microbial stability of the product.

Step 3: Final Answer:
Soaking in sugar syrup makes Gulabjamun resistant to microbial spoilage.