BIO 182 Exam One Review PDF

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BIO 182: GENERAL BIOLOGY 182

BIO 182 Exam One Review I.

The Ideas of Lamark: i. Theory of Inheritance of Acquired Characteristics: Proposed by the French naturalist Jean-Baptiste Lamarck theorized that organisms altered their behavior in response to environmental change. Their changed behavior, in turn, modified their organs, and their offspring inherited those "improved" structures. For example, giraffes developed their elongated necks and front legs by generations of browsing on high tree leaves. The exercise of stretching up to the leaves altered the neck and legs, and their offspring inherited these acquired characteristics. Conversely, in Lamarck's view, a structure or organ would shrink or disappear if used less or not at all. Driven by these heritable modifications, all organisms would become adapted to their environments as those environments changed. ii.

II.

Lamarck believed that the environment creates needs to which organisms respond to by using features, which are then accentuated ( made more prominent) or attenuated (reduced) through use and disuse. The idea is commonly illustrated by images of giraffes stretching their necks to reach leaves high in trees, which would gradually lengthen their neck. The characteristics that an individual organism acquired through use or disuse are then passed on to its offspring.

Evolution by Natural Selection: i. Natural selection occurs when one allele (or combination of alleles of different genes) makes an organism more or less fit, that is, able to survive and reproduce in a given environment. If an allele reduces fitness, its frequency will tend to drop from one generation to the next. ii. In order for Evolution by Natural Selection to occur you must have two things: a. Heritable variation: Natural selection needs some starting material, and that starting material is heritable variation. For natural selection to act on a feature, there must already be

variation (differences among individuals) for that feature. Also, the differences have to be heritable, determined by the organisms' genes. For example, a study of natural selection on color within a population requires different individuals to have varying colors. Without a variation in characteristics, there are no traits for nature to "select" over others. b. Reproduction: Over many generations, individuals with traits most suitable for their environment tend to reproduce more than those that don't. As such, natural selection works to maximize the number of individuals with those favored traits while those with less advantageous traits slowly die off. The higher the reproduction rate of a population, the higher the competitive pressure is on an individual to survive. This pressure ensures that only the most suitable members survive while the weaker members perish. c. In other words there must be a population of replicating entities and variations between them– variation that must be heritable III.

Natural Selection and Antibiotic Resistant Bacteria: i. Antibiotic resistance is a consequence of evolution via natural selection directed by the use of antibiotics. Bacteria, that have to face the antibiotic challenge, evolve to acquire resistance and, under this strong selective pressure, only the fittest survive, leading to the spread of resistance mechanisms and resistant clones. ii. The antibiotic is an environmental pressure; those bacteria which have a mutation allowing them to survive will live on to reproduce. They will then pass this trait to their offspring, which will be a fully resistant generation

IV.

The Scientific Method: i. Control Variables: Control Variables refers to variables that are not of primary interest (i.e., neither the exposure nor the outcome of interest) and thus constitute an extraneous or third factor whose influence is to be controlled or eliminated. a. For example, In a study to determine plant growth relative to the amount of light it receives both the plant near the window and the plant in the closet need to receive the same amount of water so that testers would know that it was the differences in light and not the differences in water that made one plant grow more than the other. (Water is the Control Variable)

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(Light is the Independent Variable) (Plant Growth Dependent Variable) ii.

iii.

iv.

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Dependent Variables: A dependent variable is what you measure in the experiment and what is affected during the experiment. The dependent variable responds to the independent variable. It is called dependent because it "depends" on the independent variable. A dependent variable is the variable that you observe and measure. You have no control over the dependent variable; you want to observe what happens to the dependent variable when you change the independent variable a. Example: In a study to determine whether how long a student sleeps affects test scores; the independent variable is the length of time spent sleeping while the dependent variable is the test score. Independent Variables: The independent variable is the condition that you change in an experiment. It is the variable you control. It is called independent because its value does not depend on and is not affected by the state of any other variable in the experiment. a. Example: In an experiment to determine how far people can see into the infrared part of the spectrum, the wavelength of light is the independent variable and whether the light is observed (the response) is the dependent variable Replicates: A replicate is one experimental unit in one treatment. The number of replicates is the number of experimental units in a treatment. There are two primary types of replicates: technical and biological. a. Technical replicates involve taking one sample from the same source tube, and analyzing it across multiple conditions, e.g., analyzing one sample six times across multiple arrays. b. Biological replicates are different samples measured across multiple conditions, e.g., six different human samples across six arrays. c. Using replicates offers three major advantages:  Replicates can be used to measure variation in the experiment so that statistical tests can be applied to evaluate differences.  Averaging across replicates increases the precision of gene expression measurements and allows smaller changes to be detected.

v.

 Replicates can be compared to locate outlier results that may occur due to aberrations within the array, the sample, or the experimental procedure Sample Size: The number of independently assigned experimental units that receive the same treatment is the sample size. Worded another way the sample size is the number of participants or observations included in a study

V.

Heritability: i. Heritability: is a measure of how well differences in people’s genes account for differences in their traits. Traits can include characteristics such as height, eye color, and intelligence, as well as disorders like schizophrenia and autism spectrum disorder. In scientific terms, heritability is a statistical concept (represented as h²) that describes how much of the variation in a given trait can be attributed to genetic variation. An estimate of the heritability of a trait is specific to one population in one environment, and it can change over time as circumstances change. a. Heritability estimates range from zero to one. A heritability close to zero indicates that almost all of the variability in a trait among people is due to environmental factors, with very little influence from genetic differences. Characteristics such as religion, language spoken, and political preference have a heritability of zero because they are not under genetic control. A heritability close to one indicates that almost all of the variability in a trait comes from genetic differences, with very little contribution from environmental factors.

VI.

Mutations: i. Mutations are "random" in the sense that the sort of mutation that occurs cannot generally be predicted based upon the needs of the organism. Only hereditary mutations which occur in egg or sperm cells, can be passed to future generations and potentially contribute to evolution. Some mutations occur during a person’s lifetime in only some of the body’s cells and are not hereditary, so natural selection cannot play a role. Also, many genetic changes have no impact on the function of a gene or protein. and are not helpful or harmful. In addition, the environment in which a population of organisms lives is integral to the selection of traits. Some differences introduced by mutations may help an organism survive in one setting but not in another—for example, resistance to a certain bacteria is only

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advantageous if that bacteria is found in a particular location and harms those who live there VII.

Mendelian Genetics: i. Law of Segregation: When an organism makes gametes, each

ii.

iii. iv. v. vi.

vii.

VIII.

gamete receives just one gene copy, which is selected randomly. This is known as the law of segregation. A Punnett square can be used to predict genotypes (allele combinations) and phenotypes (observable traits) of offspring from genetic crosses. A test cross can be used to determine whether an organism with a dominant phenotype is homozygous or heterozygous. A dominant allele hides a recessive allele and determines the organism's appearance. Pure-breeding means that the plant will always make more offspring like itself, when self-fertilized over many generations A homozygous dominant allele combination contains two dominant alleles and expresses the dominant phenotype (expressed physical trait). A homozygous recessive allele combination contains two recessive alleles and expresses the recessive phenotype. a. Example: The gene for seed shape in pea plants exists in two forms, one form (or allele) for round seed shape (R) and the other for wrinkled seed shape (r). The round seed shape is dominant, and the wrinkled seed shape is recessive. A homozygous plant contains either of the following alleles for seed shape: (RR) or (rr). The (RR) genotype is homozygous dominant and the (rr) genotype is homozygous recessive for seed shape. Heterozygous refers to having different alleles for a particular trait. When alleles are heterozygous one allele is dominant and the other is recessive.

Sex-Linked Inheritance: i. When a gene is present on the X-chromosome it is said to be Xlinked. X-linked genes have different inheritance patterns than genes on non-sex chromosomes (autosomes). That's because these genes are present in different copy numbers in males and females. ii.

Females: Since females have two X chromosomes, she will have two copies of each X-linked gene. For instance, in the fruit

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fly Drosophila (which, like humans, has XX females and XY males), there is an eye color gene called white that's found on the X chromosome, and a female fly will have two copies of this gene. If the gene comes in two different alleles, such as XW (dominant, normal red eyes) and Xw (recessive, white eyes), the female fly may have any of three genotypes: XW XW(red eyes), XW Xw (red eyes), XwXw (white eyes). iii.

Males: Males has different genotype possibilities than a female. Since he has only one X chromosome (paired with a Y), he will have only one copy of any X-linked genes. For instance, in the fly eye color example, the two genotypes a male can have are XWY text (red eyes) and XwY (white eyes). Whatever allele the male fly inherits for an Xlinked gene will determine his appearance, because he has no other gene copy—even if the allele is recessive in females. Rather than homozygous or heterozygous, males are said to be hemizygous for X linked genes. a. Example: We can see how sex linkage affects inheritance patterns by considering a cross between two flies, a white-eyed female (XwXw) and a red-eyed male (XWY). If this gene were on a non-sex chromosome, or autosome, we would expect all of the offspring to be red-eyed, because the red allele is dominant to the white allele. What we actually see is the following:

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b. Because the gene is X-linked, and because it was the female parent who had the recessive phenotype (white eyes), all the male offspring—who get their only X from their mother—have white eyes (XwY). All the female offspring have red eyes because they received two X chromosomes, with the XW from the father concealing the recessive Xw from the mother.

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IX.

Chi-Square: i. Formula:

ii.

iii.

iv.

Expected

Example: In a cross between two heterozygote (Tt) plants, the offspring should occur in a 3:1 ratio of tall plants to short plants. Next, you cross the plants, and after the cross, you measure the characteristics of 400 offspring. You note that there are 305 tall pea plants and 95 short pea plants; these are your observed values. Meanwhile, you expect that there will be 300 tall plants and 100 short plants from the Mendelian ratio. You are now ready to perform statistical analysis of your results, but first, you have to choose a critical value at which to reject your null hypothesis. You opt for a critical value probability of 0.01 (1%) that the deviation between the observed and expected values is due to chance. This means that if the probability is less than 0.01, then the deviation is significant and not due to chance, and you will reject your null hypothesis. However, if the deviation is greater than 0.01, then the deviation is not significant, and you will not reject the null hypothesis. So, should you reject your null hypothesis or not? Here's a summary of your observed and expected data: Tall Short 300

100

Observed 305 95 Now, let's calculate Pearson's chi-square:  For tall plants: Χ2 = (305 - 300)2 / 300 = 0.08  For short plants: Χ2 = (95 - 100)2 / 100 = 0.25  The sum of the two categories is 0.08 + 0.25 = 0.33  Therefore, the overall Pearson's chi-square for the experiment is Χ2 = 0.33 Next, you determine the probability that is associated with your calculated chisquare value. To do this, you compare your calculated chi-square value with theoretical values in a chi-square table that has the same number of degrees of

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freedom. Degrees of freedom represent the number of ways in which the observed outcome categories are free to vary. For Pearson's chi-square test, the degrees of freedom are equal to n - 1, where n represents the number of different expected phenotypes (Pierce, 2005). In your experiment, there are two expected outcome phenotypes (tall and short), so n = 2 categories, and the degrees of freedom equal 2 - 1 = 1. Thus, with your calculated chi-square value (0.33) and the associated degrees of freedom (1), you can determine the probability by using a chi-square table (Table 1). Table 1: Chi-Square Table Degre Probability (P) es of 0.99 0.99 0.97 Freed 5 5 om (df)

0.95

0.90

0.10

0.05

0.025 0.01

0.005

1

---

---

0.00 1

0.00 4

0.01 6

2.706 3.841 5.024 6.635 7.879

2

0.01 0

0.02 0

0.05 1

0.10 3

0.21 1

4.605 5.991 7.378 9.210 10.59 7

3

0.07 2

0.11 5

0.21 6

0.35 2

0.58 4

6.251 7.815 9.348 11.34 12.83 5 8

4

0.20 7

0.29 7

0.48 4

0.71 1

1.06 4

7.779 9.488 11.14 13.27 14.86 3 7 0

5

0.41 2

0.55 4

0.83 1

1.14 5

1.61 0

9.236 11.07 12.83 15.08 16.75 0 3 6 0

6

0.67 6

0.87 2

1.23 7

1.63 5

2.20 4

10.64 12.59 14.44 16.81 18.54 5 2 9 2 8

7

0.98 9

1.23 9

1.69 0

2.16 7

2.83 3

12.01 14.06 16.01 18.47 20.27 7 7 3 5 8

8

1.34 4

1.64 6

2.18 0

2.73 3

3.49 0

13.36 15.50 17.53 20.09 21.95 2 7 5 0 5

9

1.73 5

2.08 8

2.70 0

3.32 5

4.16 8

14.68 16.91 19.02 21.66 23.58 4 9 3 6 9

10

2.15 6

2.55 8

3.24 7

3.94 0

4.86 5

15.98 18.30 20.48 23.20 25.18 7 7 3 9 8

11

2.60

3.05

3.81

4.57

5.57

17.27 19.67 21.92 24.72 26.75

9

3

3

6

5

8

5

12

3.07 4

3.57 1

4.40 4

5.22 6

6.30 4

18.54 21.02 23.33 26.21 28.30 9 6 7 7 0

13

3.56 5

4.10 7

5.00 9

5.89 2

7.04 2

19.81 22.36 24.73 27.68 29.81 2 2 6 8 9

14

4.07 5

4.66 0

5.62 9

6.57 1

7.79 0

21.06 23.68 26.11 29.14 31.31 4 5 9 1 9

15

4.60 1

5.22 9

6.26 2

7.26 1

8.54 7

22.30 24.99 27.48 30.57 32.80 7 6 8 8 1

16

5.14 2

5.81 2

6.90 8

7.96 2

9.31 2

23.54 26.29 28.84 32.00 34.26 2 6 5 0 7

17

5.69 7

6.40 8

7.56 4

8.67 2

10.0 85

24.76 27.58 30.19 33.40 35.71 9 7 1 9 8

18

6.26 5

7.01 5

8.23 1

9.39 0

10.8 65

25.98 28.86 31.52 34.80 37.15 9 9 6 5 6

19

6.84 4

7.63 3

8.90 7

10.1 17

11.6 51

27.20 30.14 32.85 36.19 38.58 4 4 2 1 2

20

7.43 4

8.26 0

9.59 1

10.8 51

12.4 43

28.41 31.41 34.17 37.56 39.99 2 0 0 6 7

21

8.03 4

8.89 7

10.2 83

11.5 91

13.2 40

29.61 32.67 35.47 38.93 41.40 5 1 9 2 1

22

8.64 3

9.54 2

10.9 82

12.3 38

14.0 41

30.81 33.92 36.78 40.28 42.79 3 4 1 9 6

23

9.26 0

10.1 96

11.6 89

13.0 91

14.8 48

32.00 35.17 38.07 41.63 44.18 1 8 6 2 7

24

9.88 6

10.8 56

12.4 01

13.8 48

15.6 59

33.19 36.41 39.36 42.98 45.55 6 5 4 0 9

25

10.5 20

11.5 24

13.1 20

14.6 11

16.4 73

34.38 37.65 40.64 44.31 46.92 2 2 6 4 8

26

11.1 60

12.1 98

13.8 44

15.3 79

17.2 92

35.56 38.88 41.92 45.64 48.29 0 2 3 5 3

27

11.8 08

12.8 79

14.5 73

16.1 51

18.1 14

36.74 40.11 1 3

10

5

0

5

7

43.19 46.96 49.64 5 3 5

28

12.4 61

13.5 65

15.3 08

16.9 28

18.9 39

37.91 41.33 44.46 48.27 50.99 6 7 1 8 3

29

13.1 21

14.2 56<...