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Anna11 [10]
3 years ago
11

Describe at least one behavioral trait and one anatomical trait were selected for among the first dog.

Biology
1 answer:
Veseljchak [2.6K]3 years ago
3 0

Answer:

an astonishing amount of behavioral variation is captured within the more than 350 breeds of dog recognized worldwide. Inherent in observations of dog behavior is the notion that much of what is observed is breed specific and will persist, even in the absence of training or motivation.

Explanation:

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In snapdragons, flower color is determined by a gene with two alleles. Red flower color (R) shows incomplete dominance over the
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C) The allele R' is becoming more common in the population.

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Mary has type 2 diabetes. What can you deduce about Mary from this information? Mary doesn’t get regular exercise. Mary does not
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<span>Mary’s body is unable to use insulin properly. </span>
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Which group of organism is always found at the base of a food chain or food web?
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"PRODUCERS" are always found at the base of a food chain.
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6. You analyzed a protein sample and found that it was methionine, leucine,
kotegsom [21]

Answer:

In the analysis, it is necessary to have into account the characteristics of each of these amino acids:

Lysine (Lys) and arginine (Arg) are charged (basic) amino acids. On the other hand, glycine is an amino acid generally found in the surface of proteins. Finally, methionine (Met) is known to have both polar and non-polar properties .

The codons involved in the formation of these amino acids are:

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Arginine: CGT; CGC; CGA; CGG

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7 0
3 years ago
Deer ticks can carry both Lyme disease and human granulocytic ehrlichiosis (HGE). In a study of ticks in the Midwest, it was fou
Nostrana [21]

Answer:

(a) 0.16

(b) 1

Explanation:

Let Probability that ticks in the Midwest carried Lyme disease, P(L) = 0.16

Probability that ticks in the Midwest carried HGE disease, P(H) = 0.10

Probability that ticks in the Midwest carried either Lyme disease or HGE disease, P(L \bigcup H) = 0.10

(a) Probability that a tick carries both Lyme disease (L) and HE (H) is given by

    P(L \bigcap H);

As we know that P(A \bigcup B) = P(A) + P(B) - P(A \bigcap B)

So, in our question;

       P(L \bigcup H) = P(L) + P(H) - P(L \bigcap H)

        0.10 = 0.16 + 0.10 - P(L \bigcap H)

       P(L \bigcap H) = 0.16 + 0.10 - 0.10 = 0.16

Therefore, the probability that a tick carries both Lyme disease (L) and HE (H) is 0.16 or 16% .

(b) <em>Conditional Probability P(A/B) is given by</em> = \frac{P(A\bigcap B)}{P(B)}

So, the conditional probability that a tick has HE given that it has Lyme disease is given by = P(H/L)

    P(H/L) = \frac{P(H\bigcap L)}{P(L)} = \frac{0.16}{0.16} = 1 .

3 0
3 years ago
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