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antoniya [11.8K]
2 years ago
13

Suppose that of a cohort of 150 mice in a mouse colony born in February, 125 are still at the end of March and 115 are still ali

ve at the end of April. What is the survivorship up to the start of April? Round to the nearest hundredth.
Biology
1 answer:
kogti [31]2 years ago
5 0

Answer: Survival rate up to start of April = 76.67%

Explanation: size of colony = 150mice.

Size of colony in February =125mice.

Size of colony in march = 115 mice.

1. Survivorship rate in February

= Size of colony / size in February *100.

= 122 / 150 * 100

= 83.33% to the nearest hundredth

2. Survivorship rate in march

= Size of colony / size in march *100.

= 115/150 * 100

= 76.67% to the nearest hundredth.

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<u>Complete Question:</u>

Which statements describe proteins? Check all that apply.

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0.2404

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Tom and Terri each have elliptocytosis (they are E_), and each is Rh+ (they are R_).

Tom's mother has elliptocytosis (E_) and is Rh- (rr), so she has the genotype Er/_r. His father is healthy (ee) and has Rh+ (R_), so he has the genotype eR/e_. Tom must have inherited his E allele from his mother and his R allele  from his father, so he has the genotype eR/Er.

Terri's father is Rh+ (R_) and has elliptocytosis (E_), while Terri's mother is Rh- (rr) and is healthy (ee) with the genotype er/er. Terry could only receive the chromosome <em>er </em>from her mother, and because she is heterozygous for both genes the dominant alleles were both received from her father. Terri's genotype is ER/er.

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<u>Tom will produce the following gametes:</u>

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And the total probability of having a rrE_ child will be 0.2304 + 0.0096 + 0.0004 = 0.2404

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