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2 years ago

10

In a certain population, the frequency of a dominant gene is represented by p = 0.7, and the frequency of the recessive gene is

represented by 0.3. what is the frequency of the heterozygous genotype?

1 answer:

ladessa [460]2 years ago

4 0

Heterozygous genotype= 2pq
2(.7*.3)= .42
ANSWER = .42

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Scientists once debated how to classify the red panda. Today, how has analysis of the genome of the red panda provided evidence

skad [1K]

Answer:

According to scientists in some countries, the latest DNA research located the red panda in its own independent family, the ailurids (Ailuridae). Ailurids are themselves part of the large superfamily Musteloidea, which also includes the Mephitidae, Mustelidae and Procyonidae families, but, unlike the giant panda, it is not a bear (Ursidae).

The taxonomic classification of red panda and giant panda has been the subject of debate for many decades, as it has characteristics of both bears and raccoons. However, they are only distantly linked by a common ancestor of the first Tertiary period. Its common ancestor dates back tens of millions of years, with a wide distribution in Eurasia.

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Musteloids (Musteloidea) are a carnivorous mammalian superfamily united by distributed characters of the skull and teeth. Musteloids share a common ancestor with pinnipeds, specifically phocids, the family to which seals belong. Musteloids consist of the families Ailuridae (red pandas), Mustelidae (mustelids: weasels), Procyonidae (protionids: raccoons and relatives) and Mephitidae (skunks). In North America, the ursoids and musteloids appear first in the Chadronian (Upper Eocene). In Europe, ursoids and musteloids first appear in the lower Oligocene immediately following the great Stehlin break. The Musteloidea superfamily may not be a monophyletic group. Some or all of the diagnostic characters may have evolved into two or more independent radiations from primitive ursoids such as Amphicynodon.

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The ATP yield for a molecule depends on where it enters glycolysis or the citric acid cycle. The yield can be compared to the ne

Andru [333]

Answer:

a) The net ATP yield from fructose 1,6‑bisphosphate is 32 ATP

b) The net ATP yield from galactose is 30 ATP

c) The net ATP yield from dihydroxyacetone phosphate is 16 ATP

d) The net ATP yield from pyruvate is 12.5 ATP

e) The net ATP yield from lactate is 14 ATP

f) The net ATP yield from phosphoenolpyruvate is 13.5 ATP

Explanation:

a) In this case, 2 ATP molecules are not used for the phosphorylation of glucose for the production of fructose bisphosphate 1.6, thus, the net ATP yield is 32 ATP.

b) Galactose is converted to glucose-1-phosphate through reactions that lead to phosphorylation. Glucose-1-phosphate is transformed into glucose-6-phosphate by the enzyme phosphoglucomutase. Later it is used in glycolysis, thus, the net ATP yield is 30 ATP.

c) The fructose 1,6 bisphosphate is decomposed in glyceraldehyde-3-phosphate and dihydroxyacetone phosphate. Dihydroxyacetone phosphate is isomerized to glyceraldehyde -3-phosphate. ATP performance is (2.5+2.5+11) = 16 ATP

d) Pyruvate is decarboxylated oxidatively to produce acetyl-CoA. ATP yields are:

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Oxidative decarboxylation = 2.5 ATP

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The net yield is 2.5 + 10 = 12.5 ATP

e) Lactate is oxidized to pyruvate by the enzyme lactate dehydrogenase with a reduction of NAD. Pyruvate is brought into the mitochondria, where it is decarboxylated oxidatively to produce acetyl-CoA. If 1.5 ATP is assumed for cytosolic NADH, therefore 1.5 is produced. So, the net yield is: 12.5 + 1.5 = 14

f) Phosphoenolpyruvate is converted to pyruvate by ATP production through phosphorylation. ATP yield is as follows:

Glycolysis = 1

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The net yield is 1 + 2.5 + 10 = 13.5 ATP

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