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Sladkaya [172]
3 years ago
7

About how many total kilocalories does one serving of crackers provide if it contains 5 g of fat, 22 g of carbohydrates, and 3 g

of protein?
Health
2 answers:
andrew-mc [135]3 years ago
4 0

Answer:

The correct answer will be-0.145 kcal/serve.

Explanation:

The Carbohydrates on oxidation produces 4 calories per gram, the proteins provide 4 calories per gram and the Fats provides 9 calories per gram.

If one serve of crackers contains

1. Carbohydrates- 22 g, will produce  calories- 22 x 4= 88 calories

2. Proteins- 3 g will produce  calories- 3x4= 12 Calories

3. Fats- 5 g will produce  calories- 5 x 9= 45 calorie  

Therefore, the total calories in a single serve will be  88 +12 + 45

Total calorie = 145 calorie  

                     OR

Kilocalorie= 145/1000 = 0.145 kcal/serve.

Thus, 0.145 kcal/serve is the correct answer.

Zinaida [17]3 years ago
4 0

Answer:

145 total kilocalories.

Explanation:

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Two parents who are heterozygous for type A blood and have sickle cell trait have children.
GuDViN [60]

This question is incomplete, but here is the complete question below.

In humans , blood type is a result of multiple alleles: I^{A}, I^{B}, i^{o}.

A few simple rules of blood type genetics are that:

I^{A} is dominant over i^{o}

I^{B} is dominant over i^{o}, and

I^{A}I^{B} are codominant

Two plants who are heterozygous for type A blood and have sickle cell trait have children. Answer the following questions:

a. What is the genotype of the parents?

b. What are the genetic make ups of all the possible gametes they can produce.

c. Complete the dihybrid Punnet square to determine the frequency of the different phenotypes in the offspring.

(NOTE: Consider blood type and normal versus mutant hemoglobin in the various phenotypes.)

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a) I^{A}i^{o}AS

b) I^{A}A,I^{A}S,i^{o}A,i^{o}S

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a)

From the question the genotype of the parents can be determined:

Given that Two parents are heterozygous for  type  A and have sickle cell trait.

∴ heterozygous for  type  A will be  = I^{A}i^{o} &

heterozygous for having sickle cell traits = AS

Therefore, the traits can be determined the genotype of the parents can be determined as: I^{A}i^{o}AS

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the genetic make ups of all the possible gametes they can produce can be determined if the parent are self crossed.

If I^{A}i^{o}AS self crossed, we have:

                I^{A}                 i^{o}

A              I^{A}A               i^{o}A

S              I^{A}S                i^{o}S

∴ The genetic makeups of all the possible gametes they can produce are:

I^{A}A, I^{A}S, i^{o}A, i^{o}S

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Below shows the dihybrid Punnet square and  the frequency of the different phenotypes in the offspring

               I^{A}A               I^{A}S                    i^{o}A                 i^{o}S

I^{A}A     I^{A}I^{A}AA     I^{A}I^{A}AS        I^{A}i^{o}AA        I^{A}i^{o}AS

I^{A}S     I^{A}I^{A}AS      I^{A}I^{A}SS         I^{A}i^{o}AS        I^{A}i^{o}SS

i^{o}A     I^{A}i^{o}AA      I^{A}i^{o}AS         i^{o}i^{o}AA        i^{o}i^{o}AS

i^{o}S     I^{A}i^{o}AS       I^{A}i^{o}SS         i^{o}i^{o}AS         i^{o}i^{o}SS

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\frac{6}{16}= 37.5% : Blood type A, normal and mutant hemoglobin (sickle cell trait) = I^{A}I^{A}AS, I^{A}i^{o}AS, I^{A}I^{A}AS, I^{A}i^{o}AS, I^{A}i^{o}AS, I^{A}i^{o}AS

\frac{3}{16}= 18.75% : Blood type A, mutant hemoglobin (sickle cell anemia) = I^{A}I^{A}SS, I^{A}i^{o}SS, I^{A}i^{o}SS

\frac{1}{16}= 6.25% : Blood type O, normal hemoglobin (normal RBCs) = i^{o}i^{o}AA

\frac{2}{16}= 12.5% : Blood type O, Blood type A, normal and mutant hemoglobin (sickle cell trait) = i^{o}i^{o}AA, i^{o}i^{o}AS

\frac{1}{16}= 6.25% : Blood type O, mutant hemoglobin (sickle cell anemia) = i^{o}i^{o}SS

I hope that helps alot!

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