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kompoz [17]
3 years ago
7

During cellular respiration, NADH:________.a. is converted to NAD+ by an enzyme called dehydrogenase. b. none of the choices are

correct. c. is chemically converted into ATP. d. is reduced to form NAD+ delivers its electron load to the first electron carrier molecue.
Biology
1 answer:
skad [1K]3 years ago
5 0

Answer:

a. is converted to NAD+ by an enzyme called dehydrogenase

Explanation:

The electron transport chain of cellular respiration is the final step that oxidized NADH and FADH2. These reducing powers are formed during glycolysis and Kreb's cycle. Complex I of the electron transport chain present in the inner mitochondria membrane is NADH dehydrogenase. This protein complex accepts electrons from NADH and oxidizes it into NAD+. NADH dehydrogenase couples oxidation of NADH with the pumping of proton towards the intermembrane space.

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In dogs, wire hair is due to a dominant gene (W), smooth hair id due to its recessive allele.a. If a homozygous wire-haired dog
Vsevolod [243]

Answer:

a) All the offspring produced would be a heterozygous in the first generation (F1 generation) with genotype: Ww, Ww, Ww, and Ww. Since wire hair is dominant, all would be wired haired.

b) In the F2 (second generation), we breed the heterozygous offspring from the F1 generation together (Ww x Ww). This will produce 3:1 ratio of dominant:recessive phenotypes having 3/4 of the offspring wire-haired (1/4 WW homozygotes and 1/2 Ww heterozygotes) and 1/4 will be smooth-haired (ww). Also the genotype would have the ratio 1:2:1 (i.e 1 homozygote WW, 2 heterozygote Ww and 1 smooth hair)

c) the chances of producing a smooth-haired pup is 1/4, and the chances of producing a wire-haired pup are 3/4.

d) Therefore, 1/2 of the offspring will have the genotype Ww and be wire-haired, and 1/2 of the offspring will be ww and be smooth-haired.  Also the phenotype ratio is 1:1 (1/2 is heterozygote wired hair and 1/2 is smoth haired)

Explanation:

Since the wire hair is the dominant gene (W) and smooth hair (w) is the recessive allele

a)  If a homozygous wire-haired dog is mated with a smooth-haired dog, All the offspring produced would be a heterozygous in the first generation (F1 generation) with genotype: Ww, Ww, Ww, and Ww. Since wire hair is dominant, all would be wired haired.

b) In the F2 (second generation), we breed the heterozygous offspring from the F1 generation together (Ww x Ww). This will produce 3:1 ratio of dominant:recessive phenotypes having 3/4 of the offspring wire-haired (1/4 WW homozygotes and 1/2 Ww heterozygotes) and 1/4 will be smooth-haired (ww). Also the genotype would have the ratio 1:2:1 (i.e 1 homozygote WW, 2 heterozygote Ww and 1 smooth hair)

c) If two wire-haired dogs produce a smooth-haired pup, that means that both parents must be heterozygotes (Ww) having a pair of dominant W allele and recessive w allele to pass on to the offspring. Therefore, if these two dogs were to mate again (Ww x Ww), the chances of producing a smooth-haired pup is 1/4, and the chances of producing a wire-haired pup are 3/4.

d) If the mother of the wire-haired male was smooth-haired, that means that the recessive allele w had been passed on to the male making the male a   heterozygote (Ww). When this male mates with a smooth-haired female (ww), the cross is Ww x ww. Therefore, 1/2 of the offspring will have the genotype Ww and be wire-haired, and 1/2 of the offspring will be ww and be smooth-haired.  Also the phenotype ratio is 1:1 (1/2 is heterozygote wired hair and 1/2 is smoth haired)

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Explanation:

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Are there any choices to choose from?

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