Accessing the SimulationLaptops

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Accessing the Simulation-Laptops

• Log on w/student number password
• Be sure that the blue light on the right side,
  front is illuminated
• Click on My Computer
• Open the C Drive (local drive)
• Open the folder Biology labs
• Click on Neo Genetics of organisms
• The program will run. Answer the questions
  listed on the back page of the lab.
• Take the Quiz at the end and get checked off by
  me.
• Be sure to Log off return the laptops back in
  the appropriate slot in the computer cart.

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Lab 8 Life Cycle of Drosophila

• Pre-Lab from website
• Computer Simulation (multigenerational expt.)
• Key stages of life cycle
• Egg----Larval----Pupal-----Adult
• Observe Analyze the key traits
• Abdomen (blunt or pointed)
• Sex comb
• Red white eyes
• Observe autosomal sex-linked genes

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Sex-linked genes exhibit a unique pattern of
inheritance

• All genes on the sex chromosomes are said to be
  sex-linked
• In many organisms, the X chromosome carries many
  genes unrelated to sex
• Fruit fly eye color is a sex-linked
  characteristic

Figure 9.22A
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Chromosomal Linkage

• Thomas Morgan
• Drosophilia melanogaster
• XX (female) vs. XY (male)
• Sex-linkage genes located on a sex chromosome
• Linked genes genes located on the same
  chromosome that tend to be inherited together

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• Their inheritance pattern reflects the fact that
  males have one X chromosome and females have two

• These figures illustrate inheritance patterns for
  white eye color (r) in the fruit fly, an X-linked
  recessive trait

Female
Male
Female
Male
Female
Male
XrY
XRXR
XRXr
XRY
XRXr
XrY
XR
Xr
XR
XR
XR
Xr
Y
XRXr
XRXR
XRXr
Y
Y
Xr
Xr
XRY
XrXR
XRY
XrXr
XRY
XrY
XrY
R red-eye allele r white-eye allele
Figure 9.22B-D
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Statistical Tools to Analyze results

• Chi-Square Will tell you how much your data is
  different from expected (calculated) results. It
  is Non-Parametric and deals with different
  catagorical groups vs. Parametric which deals
  with numbers and which case you would use a
  T-test instead.
• Formula ? 2 (o e)2
• e
• ?2 what we are solving
• o observed value
• e expected (calculated value)

?
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Using Chi Square Analysis

• When 38 students were polled in the class if they
  liked chocolate or vanilla ice cream better. The
  following results were found
• 27 boys liked chocolate over vanilla compared
  to 11 girls. Is this difference statistically
  valid to say that boys prefer chocolate more than
  girls?
• H0 (Null hypothesis) There is no difference
  between boys and girls preferring chocolate over
  vanilla.
• Note The expected value will be 19 (for no
  difference)
• ?2 (O E)2 so. ?2 (27-19)2
  (11-19)2
• E 19
  19

6.72
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Solving Question 3

• Formula ?2 ? (o e)2
• e
• Degrees of Freedom Critical Value
• 1 3.84
• 2 5.99
• 3 7.81
• 4 9.49
• 5 11.07

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Determining Degrees of Freedom comparing to the
critical value

• In our Ice cream problem we two groups.
• Degrees of freedom is determined by subtracting
  on from the number of groups. In our case we
  will have one degree of freedom.
• Consult the probability chart under 0.05 (95
  confidence level) and look under 1 degree of
  freedom to find the critical value. In our case
  the critical value is 3.84
• Compare our chi square value to this critical
  value.
• If ?2 is less than the critical value then you
  accept the null hypothesis (H0)
• If ?2 is greater than the critical value then
  you reject the null hypothesis (H0)
• In our case the chi square value was 6.72 which
  is greater than the critical value so we will
  reject the null hypothesis of no difference
  between the values. This means that there is a
  significant difference between the observed and
  expected value and chance alone cannot explain
  the results. There seems to be a preference of
  chocolate ice cream by boys according to this
  data.

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Sample Problem using Chi square

• Two hybrid Tall plants are crossed. If the F2
  generation produced 787 tall plants and 277 short
  plants. Does this confirm Mendels explanation?
• What is the expected value? This is your null
  hypothesis (HO)
• Total number of plants 1064
• 31 Phenotypic ratio
• Expected value should be 798 tall and 266 short
• (75) (25)

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Calculation of Chi Square Value

• ?2 ? (O E)2
• E
• ?2 (787 798)2 (277 266)2
  0.61
• 798 266
• There are two categories and therefore the
  degrees of freedom would be 2-1 1 .
• Look up the critical value for 1 degree of
  freedom 3.84 (next slide-always given)
• If your value is LARGER than the critical value
  then you reject the null hypothesis and assume
  that there is a significant difference between
  the observed value and the expected. Values are
  statistically different.
• 0.61 is less than 3.84 therefore we accept the
  null hypothesis and accept that our values are
  similar enough Theres no significant difference
  between the observed expected values. Values
  are similar enough.

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When to use Chi-Squared Test

• Can only be used with raw counts (not
  measurements)
• Comparing Experimental expected (theo.) values
• Sample size must be more than 25 to be reliable
• Aims to test the null hypothesis (H0)
• (H0) the hypothesis that theres no difference
  between the data sets
• Alternative hypothesis there is a significant
  difference
• Compare with a critical value table (p values)
• To reject the (H0) value must be GREATER than
  the critical value favor the alternative
  hypothesis.
• Accepting the null means that theres no
  significant difference between the data sets.
  The observed and expected values are similar
  enough.

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Genetic Crosses to do for Lab 8

• 1. F1 cross Gg X Gg (G green, g albino)
• Observed Offspring 72 green
• 12 albino
• A study of Incomplete dominance in tobacco
  seedlings generated these results Green (GG)
  22
• Yellow (Gg) 50
• Albino (gg) 12
• Determine the chi-square value and determine if
  this data fits the expected 121 ratio
• 3. Cross between pure bred, long wing
  Drosophila w/pure bred short wing produces all
  offspring F1 to have intermediate wing lengths.
  A cross between two heterozygotes from F1
  produces the following results
• 230 long wings
• 510 intermediate-length wings
• 260 short wings
• What is the genotype of the F1 intermendiate
  winged flies?
• Write a hypothesis describing the mode of
  inheritance of wing length in Drosophila (this is
  your null hypothesis)
• Calculate the chi-square value for these results.
• According to the critical value can you accept or
  reject the null hypothesis.

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Comparing red eyed and sepia eyed Flies.

• 4. If red Eyes are dominant to sepia eye color
  and if the parental cross involved a pure red
  eyed female with a sepia eyed male. The result
  will be a. ______________. (F1 generation)
• B. While crossing two flies from the F1
  generation, the results showed 146 red eyed
  flies and 110 sepia eyed flies. Compare these
  values with Mendels prediction (null hypothesis)
  and determine if these results correspond using
  a chi square analysis.

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Sex Linked Inheritance of Red eyes and white
eyes.

• 5. F1 generation All females Heterozygous red
  eyed and all the males were white eyed.
• F2 generation produced these results
• Red eyed females 78
• White eyed females 62
• Red-eyed Males 69
• White eyed males 73
• From these results Write a null hypothesis,
  calculate a Chi-square value, compare it to the
  critical value, and determine if you accept or
  reject the null hypothesis.

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• Table 7.1 F1 Generation Data

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F2 Generation Sex-Linked trait(Quest. 5)
Assume that the following Data was obtained and
determine if it is significant or not by
calculating a Chi-Square value.
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Question 4-

• Red-Eyed Dominant S
• Sepia eyes Recessive s
• Parental cross SS X ss
• F1 Cross Ss X Ss
• F1 All Ss
• F2 31 (redsepia)

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Data for Question 4