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Wisconsin Master Cheesemaker Program Cheese Science & Technology: practice questions

Wisconsin Master Cheesemaker Program · USA · Center for Dairy Research, University of Wisconsin–Madison, with the Wisconsin Master Cheesemaker Board

Original practice questions written to the published syllabus · not past papers
10worked here, free
18written to this exam in this subject
7syllabus units

About Wisconsin Master Cheesemaker Program

Wisconsin Master Cheesemaker Program is a specialist certification exam run by Center for Dairy Research, University of Wisconsin–Madison, with the Wisconsin Master Cheesemaker Board, for candidates in USA. Certification by cheese variety over about 2 years 8 months: required short courses, then assessment of the candidate's cheese and knowledge.

Source: cdr.wisc.edu.

Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.

Cheese Science & Technology in Wisconsin Master Cheesemaker Program

Cheese from milk selection to ripening and processed cheese: coagulation, curd handling, salting, ripening, yield and the major variety families.

Our 18 Wisconsin Master Cheesemaker Program questions in this subject: 1 easy, 13 medium, 4 hard, across 7 units.

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

These 10 were written to Wisconsin Master Cheesemaker Program and are not shown on any other exam's page.

Question 1 · medium

A vat yielded 9.80 kg of Cheddar per 100 kg milk at 36.5% moisture. Adjust this to the plant's reference moisture of 37% using adjusted yield = actual × (1 − actual W) ÷ (1 − reference W).

  1. 9.72 kg
  2. 9.93 kg
  3. 9.88 kg
  4. 10.30 kg
Answer and working

C. 9.88 kg

9.80 × 0.635 ÷ 0.630 = 9.88: drier cheese is credited upward. 9.72 inverts the ratio; 9.93 uses moisture fractions instead of dry fractions; 10.30 adds the 0.5-point moisture difference directly to the yield.

Question 2 · medium

1,000 kg of milk with 3.6% fat gives 99 kg of cheese containing 33.5% fat. What is the fat recovery?

  1. 92.1%
  2. 93.1%
  3. 7.9%
  4. 9.9%
Answer and working

A. 92.1%

Fat in cheese = 99 × 0.335 = 33.17 kg; fat in milk = 36 kg; recovery = 33.17 ÷ 36 × 100 = 92.1%. 93.1% divides the two fat percentages (scaled), ignoring masses; 9.9% is the yield; 7.9% is the fat lost to whey.

Question 3 · medium

A young Cheddar has 38.0% moisture and 33.0% fat. What is its moisture in the non-fat substance (MNFS)?

  1. 56.7%
  2. 53.2%
  3. 115.2%
  4. 61.3%
Answer and working

A. 56.7%

MNFS = moisture ÷ (100 − fat) × 100 = 38 ÷ 67 × 100 = 56.7%. 61.3% divides by total solids (100 − moisture); 53.2% is the fat in dry matter; 115.2% compares moisture with fat directly.

Question 4 · hard

Milled Cheddar curd (900 kg) is dry-salted at 2.5% of curd weight; 40% of the added salt is lost in the press whey and the pressed cheese weighs 860 kg. What is the salt content of the cheese?

  1. 1.05%
  2. 1.57%
  3. 1.50%
  4. 2.62%
Answer and working

B. 1.57%

Added = 22.5 kg; retained = 22.5 × 0.6 = 13.5 kg; salt = 13.5 ÷ 860 × 100 = 1.57%. 2.62% ignores the loss; 1.50% uses the curd mass instead of the cheese mass; 1.05% keeps the lost fraction instead of the retained one.

Question 5 · medium

With Van Slyke, Cheddar from a given milk has (0.93F + C − 0.1) × 1.09 = 6.265. If cheese moisture rises from 37% to 38%, how much extra cheese is made from 10,000 kg of milk?

  1. 6.3 kg
  2. 0.2 kg
  3. 16 kg
  4. 100 kg
Answer and working

C. 16 kg

Yield at 37% = 6.265 ÷ 0.63 = 9.944; at 38% = 6.265 ÷ 0.62 = 10.105; difference 0.160 kg per 100 kg, so 16 kg per 10,000 kg. 0.2 kg is the per-100-kg figure; 100 kg assumes yield rises by one point of milk weight; 6.3 kg multiplies the numerator by the moisture change.

Question 6 · medium

Why is the temperature raised slowly at the start of cooking?

  1. To make sure all lactose is fermented before the whey is drained
  2. To prevent the starter bacteria from producing acid too early
  3. To avoid case hardening, which traps whey inside the curd
  4. To keep the coagulant active until the curd is fully drained off
Answer and working

C. To avoid case hardening, which traps whey inside the curd

Rapid heating case-hardens the particle surface, sealing in moisture. Slow cooking is not about keeping the coagulant active or delaying acid.

Question 7 · medium

A Cheddar maker drains whey at a lower pH than usual. What change in the cheese is expected?

  1. Less calcium and more retained coagulant: a softer, more crumbly body
  2. A higher pH in the final cheese because lactose drains with the whey
  3. More calcium and less coagulant, giving a firmer, more elastic body
  4. No change in texture, because pH only affects flavour development
Answer and working

A. Less calcium and more retained coagulant: a softer, more crumbly body

Lower draining pH dissolves more colloidal calcium phosphate into the whey and retains more coagulant in the curd, giving more acid, crumbly cheese. Draining pH is a key texture control.

Question 8 · medium

The rinds of cheeses in a calcium-free brine become soft and slimy. What is the usual remedy?

  1. Lower the brine salt to below 5% so that less water leaves the cheese
  2. Raise the brine temperature to 35 °C so that salt penetrates faster
  3. Add calcium to the brine and keep its pH close to that of the cheese
  4. Replace the brine with plain water for the first day of salting
Answer and working

C. Add calcium to the brine and keep its pH close to that of the cheese

Cheese loses calcium to a calcium-free brine and its surface softens; brines are kept with calcium and a matched pH. Weak or warm brines and plain water make spoilage and soft rinds worse.

Question 9 · medium

A Cheddar shows many small gas holes and a spongy body within the first few days. The most likely cause is:

  1. Early gas from coliforms or yeasts: poor hygiene or slow acid
  2. Calcium lactate crystallisation from racemisation by NSLAB bacteria
  3. Late gas from clostridial spores, pointing to silage feeding of cows
  4. Eye formation by propionibacteria added accidentally with the starter
Answer and working

A. Early gas from coliforms or yeasts: poor hygiene or slow acid

Early blowing appears within days and is caused by coliforms or yeasts. Clostridial late blowing appears after weeks; calcium lactate gives crystals, not holes.

Question 10 · medium

What is the function of a block-forming tower in a Cheddar plant?

  1. Separating cream from milk before the milk is pumped to the vats
  2. Fusing salted curd under vacuum into blocks of uniform weight
  3. Brining Gouda wheels by moving them through a tall column of brine
  4. Stretching curd in hot water to align protein fibres for pizza cheese
Answer and working

B. Fusing salted curd under vacuum into blocks of uniform weight

Towers take milled, salted curd and form uniform blocks under vacuum. Cooker-stretchers handle pasta filata; separators skim milk.

A study plan for Cheese Science & Technology in Wisconsin Master Cheesemaker Program

This plan follows where our questions actually fall, so the time goes where the marks are likely to be. It is our reading of the syllabus, not the exam body's weighting.

  1. First, the heavy units (72% of the 18 questions): Cheese yield: formulas and factors (7, 39%); Cheese defects and grading (3, 17%); Salting: dry, brine and its effect (3, 17%).

    Read the manual chapters for these first and do every worked question you can find on them.

  2. Then the middle band: Cutting, cooking and syneresis (2, 11%); Cheddaring, pasta filata stretching and other curd treatments (1, 6%); Mechanised and UF-based cheese making (1, 6%).

    Aim for one sitting per unit, then mixed questions across them.

  3. Then the lighter units: Rennet coagulation and gel formation (1, 6%).

    Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.

  4. Leave time for the hard ones: Cheese defects and grading (1 hard of 3), Salting: dry, brine and its effect (1 hard of 3), Cheese yield: formulas and factors (2 hard of 7).

    Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.

  5. Bank the easy marks: Cheese yield: formulas and factors (1 easy of 7).

    Do these early in revision so they are fast on the day.

  6. Finish timed. Sit the subject as a timed set without notes, mark it, and send each wrong answer back to its unit above.

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