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4 years ago

A food handler must check the temperature of tcs food being held at least every how many hours

2 answers:

5 0

Foods which need time and temperature control for safety are called TCS foods. TCS foods include milk and dairy products, eggs, meat, poultry, fish, baked potatoes, tofu, sprouts and sprout seeds, etc. Cold TCS food must be maintained at 41℉ or less and hot food must be maintained at 135℉ or above. Checking temperature by food handler every two hours would be ideal to leave time for corrective action.

Mandarinka [93]4 years ago

3 0

I have no clue but I will say 2

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Is oxygen necessary for the production of apt in your cells?

gizmo_the_mogwai [7]

Yes it is necessary for the production of apt in your cells because apt needs oxygen to form correctly.

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3 years ago

How does energy acquisition in the deep sea differ from energy acquisition near the ocean’s surface?

hram777 [196]

I think the correct answer from the choices listed above is option C. The energy acquisition in the deep sea differ from energy acquisition near the ocean’s surface by the fact that organisms in the deep sea do not have direct access to sunlight. Hope this answers the question. Have a nice day.

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3 years ago

Read 2 more answers

Most black bears (Ursus americanus) are black or brown in color. However, occasional white bears of this species appear in some

Sphinxa [80]

Answer and Explanation:

The number of observed individuals:

AA 42

AG 24

GG 21

Total number of individuals, N= 87 = 42 + 24 + 21

Allelic frequencies:

f(p) = (2 x AA + AG)/ 2 x N

f (p)= (2 x 42 + 24) /2 x 87

f (p) = (84 + 24) / 174

f (p)= 108 / 174

f (p) = 0.62

f (q) = (2 x GG + AG)/2 x N

f (q) = (2 x 21 + 24 )/2 x 87

f (q) = (42 + 24)/ 174

f (q) = 66/174

f (q) = 0.38

p + q = 1

0.62 + 0.38 = 1

The expected genotypic frequency:

F (AA)= 0.62 ² = 0.3844

F (AG) = 2 x A x G = 2 x 0.62 x 0.38 = 0.4712

F (GG) = 0.38 ² = 0.1444

AA + AG + GG = 0.3844 + 0.4712 + 0.1444 = 1

The number of expected individuals:

AA= (0.62)² x 87 = 0.3844 x 87 = 33.44

AG= (0.4712) x 87 = 40.99

GG= (0.38)² x 87 = 12.563

Total number of expected individuals = 33.44 + 40.99 + 12.563 = 87

Chi square= sum (O-E)²/E

AA= (O-E)² /E

AA=(42 - 33.44) ² / 33.44

AA= 2.2

AB= (O-E)² /E

AB= (24 - 40.99)²/ 40.99

AB=7.04

BB=(O-E)² /E

BB= (21-12.563)²/12.563

BB= 5.66

Chi square= sum ((O-E)²/E) = 2.2 + 7.04 + 5.66 = 14.9

Degrees of freedom = genotypes - alleles = 3 - 1 = 2

p value less than 0.05

There is enough evidence to reject the nule hypothesis. The genotype frequencies are not in equilibrium.

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3 years ago

Phosphorus is required to synthesize the deoxyribonucleoside triphosphates used in DNA replication. A geneticist grows some E. c

Harlamova29_29 [7]

Answer:

<h2> After one round: one strand of DNA will contain radioactive 3232P, while the other strand will not contain any radioactive phosphate. </h2><h2> After two rounds: here 50% of DNA will have 3232P in both strands, while 50% will contain 3232P in one strand and nonradioactive in the other strand. </h2>

Explanation:

1. In the initial sample which is immediately removed after addition of radioactive isotope of phosphorus ( 3232P), hence there is no incorporation of 3232P into the DNA because replication in the medium containing 3232P has not yet occurred.

2. After one round of replication in radioactive isotope of phosphorus ( 3232P) containing medium, here only one newly synthesized strand of DNA molecule will contain 3232P, while the other strand will not contain any radioactive isotope of phosphorus ( 3232P), because DNA replication occurs in semi-conservative way.

3. After two rounds of replication in medium which contains radioactive isotope of phosphorus ( 3232P), here 50% of the DNA molecules will have radioactive isotope 3232P in both strands, while the rest 50% will contain 3232P in only one strand and nonradioactive phosphorous in the other strand.

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3 years ago

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