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vekshin1
3 years ago
10

"the daily white blood cell (wbc) count in a client with aplastic anemia drops overnight from 3,900 to 2,900/µl (3.9 to 2.9 x 10

9/l). which is the appropriate nursing intervention?"
Biology
1 answer:
Gnesinka [82]3 years ago
5 0
<span>A drop in white blood cell count in someone with aplastic anemia would imply that the patient's natural ability to fight infection has greatly decreased. In this case, the nurse should notify the doctor or health provider to both isolate and monitor the patient.</span>
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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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