What properties DO NOT require a tool?

Biology

1 answer:

Nataly [62]2 years ago

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Answer:the correct answer is luster

Explanation:took the quiz made a 100

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The lymphoid tissues of the intestinal tract are collectively referred to as

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javier obsevo la cabeza de una mariposa en fase de oruga y en fase adulta para saber como se alimentan

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Answer: Las orugas, larvas de mariposas y polillas, se alimentan casi exclusivamente de plantas. La mayoría de las orugas mastican felizmente las hojas, aunque algunas se alimentan de otras partes de la planta, como semillas o flores, mientras que durante la etapa de adulto, la mariposa se alimenta de néctar y pasa de ser un herbívoro a ser un importante polinizador.

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2 years ago

What structure is found in prokaryotic cells but not eukaryotic cells?

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What structure is found in prokaryotic cells but not eukaryotic cells?

a. a single, circular DNA chromosome found in the cytoplasm

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Which of these is an anabolic reaction? a. Glycogen to glucose b. Protein to amino acids c. Amino acids to protein d. Energy rel

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Suppose "A" is a dominant gene for the ability to taste phenylthiocarbamide and "a" is a recessive gene for the inability to tas

dexar [7]

Complete question:

Suppose "A" is a dominant gene for the ability to taste phenylthiocarbamide and "a" is a recessive gene for the inability to taste it. Which couples could possibly have both a child who tastes it and a child who does not?

a. father AA, mother aa

b. father Aa, mother AA

c. father Aa, mother Aa

d. father AA, mother AA

Answer:

c. father Aa, mother Aa

Explanation:

According to the given information, the ability to taste phenylthiocarbamide is a dominant trait and is imparted by the allele "A". This phenotype would be expressed in both homozygous and heterozygous conditions. The non-taster phenotype would be expressed in the homozygous recessive genotypes only.

To have both taster and non-taster children, both the parents should have at least one copy of the recessive allele. Among the given options, the father with genotype Aa and the mother with genotype Aa have the possibility to have both taster and non-taster children.

Aa x Aa= 3/4 taster (1/4 AA and 1/2 Aa): 1/4 non-taster (1/4 aa)

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