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3 years ago

imagine that one woman smokes, and another woman does not smoke. how might their likelihood of getting lung cancer differ?

1 answer:

7 0

The one who doesn’t smoke has a very low chance of getting lung cancer as the one who does make has a good change of getting lung cancer. Which makes there health different which also makes there also makes there Likelihood of getting lung cancer differ.

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_________________ results in the formation of two new cells.

Jobisdone [24]

Answer:

mitosis

Explanation:

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3 years ago

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What would be the complimentary strand for the DNA sequence ATAACGA

vladimir1956 [14]

I believe it is TATTGCT

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3 years ago

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Why did dna technology lead to more use of cladistics?a. it showed that the linnaean system was the most accurate.b. it changed

Alexandra [31]

The correct option from given options is "b", that is it changed ideas about which animals were closely related.
Cladistics was invented for the purpose of improving on taxonomy and it is a way to deal with biological classification. DNA technology lead to more use of cladistics because it changed ideas about which animals were closely related and also it showed new evolutionary relationships between animals.

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3 years ago

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Komok [63]

Similarities
Both plant and animal cells have eukaryotic origins
Both have common organelles like the nucleus, mitochondria and ribosomes
Both are surrounded by a cell membrane

differences
plant cells have a regular defined shape while animal cells have an irregular shape
plant cells have a cell wall while animal cells lack a cell wall
plant cells have a chloroplast for photosynthesis while animal cells do not have a chloroplast

5 0

3 years ago

Most black bears (Ursus americanus) are black or brown in color. However, occasional white bears of this species appear in some

Sphinxa [80]

Answer and Explanation:

The number of observed individuals:

AA 42

AG 24

GG 21

Total number of individuals, N= 87 = 42 + 24 + 21

Allelic frequencies:

f(p) = (2 x AA + AG)/ 2 x N

f (p)= (2 x 42 + 24) /2 x 87

f (p) = (84 + 24) / 174

f (p)= 108 / 174

f (p) = 0.62

f (q) = (2 x GG + AG)/2 x N

f (q) = (2 x 21 + 24 )/2 x 87

f (q) = (42 + 24)/ 174

f (q) = 66/174

f (q) = 0.38

p + q = 1

0.62 + 0.38 = 1

The expected genotypic frequency:

F (AA)= 0.62 ² = 0.3844

F (AG) = 2 x A x G = 2 x 0.62 x 0.38 = 0.4712

F (GG) = 0.38 ² = 0.1444

AA + AG + GG = 0.3844 + 0.4712 + 0.1444 = 1

The number of expected individuals:

AA= (0.62)² x 87 = 0.3844 x 87 = 33.44

AG= (0.4712) x 87 = 40.99

GG= (0.38)² x 87 = 12.563

Total number of expected individuals = 33.44 + 40.99 + 12.563 = 87

Chi square= sum (O-E)²/E

AA= (O-E)² /E

AA=(42 - 33.44) ² / 33.44

AA= 2.2

AB= (O-E)² /E

AB= (24 - 40.99)²/ 40.99

AB=7.04

BB=(O-E)² /E

BB= (21-12.563)²/12.563

BB= 5.66

Chi square= sum ((O-E)²/E) = 2.2 + 7.04 + 5.66 = 14.9

Degrees of freedom = genotypes - alleles = 3 - 1 = 2

p value less than 0.05

There is enough evidence to reject the nule hypothesis. The genotype frequencies are not in equilibrium.

5 0

4 years ago