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scoundrel [369]
3 years ago
15

1. An animal cell has a 2% salt concentration. It is placed into a solution with a 0.05% salt concentration. What would happen t

o the animal cell?
A: water will move into the animal cell causing it to shrivel
B: water will move into the animal cell causing it to swell and burst
C: water will move out of the animal cell causing it to swell and burst
D: water will move out of the animal cell causing it to shrivel
Biology
1 answer:
jenyasd209 [6]3 years ago
3 0

Answer: water will move out of the animal cell causing it to swell and burst  (Option C)


Explanation:


When a cell is immersed into a salty solution, the osmotic pressure generated by concentration differences of salt, will make the salt come in due to cell self-protection mechanism of the cell and water will come out, eventually the cell can burst .But since the animal cells has a higher concentration of salt and the solution has a lower concentration osmosis will take place and salt will move from higher salt concentration to lower.



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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
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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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