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loris [4]
4 years ago
12

In the Krebs cycle:

Biology
1 answer:
Aleksandr [31]4 years ago
7 0
Answer no 1 is correct :)
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Incomplete Dominance Predicting Flower Color in Snapdragons
Mashutka [201]

Answer:

In snapdragons, the color of the flower is controlled by incomplete dominance. The two alleles are red (R) and white (r).

Explanation:

4 0
3 years ago
NO FAKE ANSWERS AND NO LINKS. Why did countries work together to ban CFCs to help restore the ozone layer?
ira [324]

CFCs were phased out via the Montreal Protocol due to their part in ozone depletion. The atmospheric impacts of CFCs are not limited to their role as ozone-depleting chemicals. Infrared absorption bands prevent heat at that wavelength from escaping earth's atmosphere

No Problem

6 0
2 years ago
What most likely caused the change in the distribution of fur colors in the graphs shown?
weeeeeb [17]

Answer: D

Explanation: I think it is d because of process of elimination. A is not reasonable because the graph 2000 goes up and down so not being able to reproduce would not fluctuate. The other two choices the. Are close but not sensible to how the graphs are structured. Therefore best choice is option is D.

6 0
2 years ago
Help me figure out the right answer please
Lady bird [3.3K]
The answer is A: Synaptic cleft, motor end plate
6 0
3 years ago
Suppose a species of bird called the red‑crested warbler has a plumage length that is controlled by a single gene. The P????mPlm
andreev551 [17]

Answer:

Check the explanation

Explanation:

North American:

112 birds total => 224 alleles

112 - 55 birds = 57 short plume birds.

q^2= (2*57) / 224 = 0.5089

Freq(short plume allele) = q = 0.7134

Freq(long plume allele) = p = 1 - q = 0.2866

From this North American population we get

0.2866 * 224 = 64 long plume alleles

0.7134 * 224 = 160 short plume alleles

South American:

452 birds total => 904 alleles

452 - 75 birds = 377 short plume birds.

q^2 = (2*377) / 904 = 0.8341

Freq(short plume allele) = q = 0.9133

Freq(long plume allele) = p = 1 - q = 0.0867

From this South American population we get

0.0867 * 904 = 78 long plume alleles

0.9133 * 904 = 826 short plume alleles

Blended population has

64 + 78 = 142 long plume alleles

160 + 826 = 986 short plume alleles

142 + 986 = 1128 total alleles

p = 142/1128 = 0.1259

q = 986/1128 = 0.8741

All right. The new population has the p and q from the blended population. The new population has 1000 individuals. The portion of long plumed will be homozygote long plumage + heterozygote, which is p^2 + 2pq, and you multiply that by the population size 1000 to get the final answer.

population size * (p^2 + 2pq) = 1000 * (0.1259^2 + 2*0.1259*0.8741) = 236 birds

4 0
3 years ago
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