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BIOL332 > Sex Link + Chi Square test > Flashcards

Sex Link + Chi Square test Flashcards

1
Q

Example link trait ratios

A

Sex linked = have different ratios than medialian proababilities

***Still dominent vs. recssive but the alleles are on sex chromosomes

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2
Q

Sex Chrosmomes

A

X and Y – everyone has an X

XX – female
XY – male

***there are times where you might have extra copies of sex chromosomes

***2 out of 46 of our chromsomes = sex chromosomes (often the last two in the karyotype)

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3
Q

Sex linked traits

A

Traits that are specifcially on sex chromosomes – Most of them are in X because teh X chromosome is larger than the Y = has more genes than Y

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4
Q

Hemophelia Example – Hemophelia = X linked recessive

A

Normal = XH
Hemophela = Xh

WOmen with no Hemophela = XHXH or XHXh –> because only needs 1 domiennet allele
- Women can be a carreier (XHXh = carrier)

Women with hemophelia = XhXh

Male with no Hemphelia = XHY
- Have nothing on Y

Male with hemophelai = XhY

***Male = only has 1 X = can’t be a carreier

***Male = more lileyt to inherit disorder because only 1 X chromsomes – this is true fro many other sex-linked recessive traits

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5
Q

Chi-Square Goodness of Fit Test

A

Determines if progeny followes Mendle’s Laws – looking to see if the observed values are consistent with teh expected values

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6
Q

Chi-Square test Example Pp X Pp

A

Start with PP X pp in parents THEN cross F1

Cross Pp X Pp –> get 3:1 ratio
- Expected in F2 = 3/4 purple and 1/4 white

Obseroved = 105 purple and 45 white
(2.33 purple: 1 White)

Question = are teh observed values consistent with expected values

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7
Q

Steps fro GOF test (Depth)

A
  1. Present the genotypes of cross that you are testing

Example – If crossimg Pp X Pp

Given the observed values
Purple = 105
White = 46

Expected values = (0.75 X 150) = 112.5 Purple
AND
(0.25 X 150) = 37.5 White

  1. State the null hypothesis – need to build a statement that can be statistically tested
  2. Calculate chi Squared
  3. State the degree of freedom
  4. Go to a Chi-squared table and determine the Chi-Squared critical value for alpha (P-value) used for hypothesis testing
    • Use P-value of 0.05
      Example – go to table – go to Df 1 for P-vlue of 0.05 – CV = 3.841
  5. Compare you X^ CV V.s your Chi-square value and state whether the null is rejected
  6. State the overall conclusion
    • If fail to reje ct the null = then the observed data are consisnet with mono/dihybrid cross
      • If reject the null then then are not consisent with mono-dihybrid cross
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8
Q

How to find expected values

A

Multiple the ratio by the total number of offspring

Example – if know you have a 3:1 ratio (know from a punet square) AND know total number of individuals is 150

3/4 –> (0.75) X 150

AND

1/4 –> 0.25 X 150

***Can get total by adding up all of the observed progeny

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9
Q

Are ratios or numbers used in Chi-Square

A

NUMNBERS – we know the rations BUT you need to sue the numbers for a chi-square

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10
Q

Null Hypothesis

A

Default position that there is no difference between the observed and expected NUMBERS and any deviations between them is due to random sampling error

Example – deviations between the observed and expected values for cross are due to random chance or sampleing error

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11
Q

Alternative hypothesis

A

Used if the data does not support the null
Ha = The data are random samples from some opther disrubution (NOT from monohybrid or duhybrid cross) – means that the differences are from something other than sampling error

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12
Q

How to calculate Chi-square

A

(O-E)^2/ E –> do this for each catagory and then add them together

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13
Q

How to find degree of freedom

A

Df = # of classes (# of phenotypes) - 1

Example – if have 2 phenoptypes = 2-1 =1

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14
Q

Chi Square CV Vs. Chi square and rejecting null

A

X^2 > CV –> then reject the null (P is <0.05)

X^2 < CV = fail to reject the null

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15
Q

Are males or females more likeley to show phenotypes of sex-linked traits?

A

ANSWER: Males – because they ONLY have 1 X chromoroesm = if they have the X link trait they only need to have 1 copy of the alelle

AND X linked traits are more xommon because the X chromsome is bigger than the Y chromsome

***Even if Y linked then only amles could get it because females don’t have a Y

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16
Q

What sex chromosome often has sex linked traits

A

Sex linked traits are more often on the X chromosome because teh X xhromsome is bigger than the Y = more likley to be on X than Y

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17
Q

What predicts sex in mammales

A

Sex is typically predicted by the X and Y chromsomes

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18
Q

Homogametic Sex

A

XX – female
***When sex chromsomes are the same

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19
Q

Heterogametic Sex

A

XY – male
***When sex chromosomes are different

20
Q

Sex Vs. Gender

A

Sex – refers to the anatomy of an individuals reproductive system and secondary sex charachteristics
- Secondary Sex Charachyteristics = boobs + beards

Gender – refers to the social roles or internal awareness based on the sex of a person

21
Q

Sex determination compicatedness

A

Sex determination is way more complicated than X and Y
- Set is NOT just determined by X or Y
- Most of the genes that determine sex charachteristics are in autosomes NOT sex chromomes and they are regulated by thing on X and Y chromsomes

22
Q

Pre-1900 theories of sex determination

A

Sex is determined by:
1. heat of a male partner during intercouse
- To produce male heirs – spice it up

  1. Testes – where the larger the testicle produces bots
  2. Overies – where one produces both and the other girls
    • OWmen should sleep on one side to let gravity direct semen to the male producing overy
  3. Nutrition – boys are typically larger than girls so feed pregannet mothers a lot fo specific foods to make male heirs
  4. Envirnment – copute in the summer heat to amke males OR give birth in teh summer heat to make males
23
Q

What generally predicts sex

A

Genrally sex is predicted by X and Y

***Before 1900s they had some crazy idase on how sex is determined

24
Q

Understanding genetics history

A

Mendle – Experiments on plant hybridiztion

Hunt morgan – experimentally deomstrated chromosomal theory – connects with medelian inheritance
- All of the red eyes tgen found a white eye – made true breeding lines – found that he could preduect ratios of males to females
- knew karyotypes of Red and white eyes – males had diferent shaped chromsomes than females

1903 Sutton and Boveri – Chromsomal theory of inheritance –> found that chromsomes control heredity
- Used sea urchins –> have big cells 00 can see chromsomes under microscope
- Seprate chromomes segreagting during division
- Found that chromsomes conrtorl heredity

25
Q

X and Y chromsomes during meisosis

A

Sex chromsomes pair during Meiosis even though they are not homologous chromsomes
- We know that they pair because if tehy didn’t pair then gametes would have lots of X and Y
- They still look like normal chrosmomes

***X and Y do NOT recombine during meiosis

26
Q

What do sex chromsomes look like

A

They look like otehr chromsomes prior to interphase (one chromatid + one centromere)

27
Q

How do sex chromsomes pair up during Meiosis

A

Since X and Y are difefrent (not like Homolgous chromsomes that pair up because of sequnce similarity) – THEY STILL pair up becvause they ahve regions at there telemeres that are homolgous called “Psudoautosomal regions” that allows them to pair

Can pair because of Psudoautosomal regions

28
Q

How do autosomes pair up during meiosis

A

Homolougous chromsomes pair up by sequence similarity

29
Q

Practice – Hemophelia in humans is due to rare X chromsomeal mutations – what will be the result of mating between a non-carrier female and a hemofilic male

A

ANSWER: All of the sons are normal and all of the daughters are carriers

30
Q

Practice – A women that is carrier for hemophelia and a man that does not have hemophelia have a daughter - what is the probability that this daughter will be a carrier for hemophelia?

A

ANSWER: 50%

31
Q

Practice – A woman that is a carrier for hemophelia and a man that does not have hemophelia have 3 kids – what is teh probability that ONLY ine child will inherit the allele for hemophelia

A

ANSWER: 3/8

32
Q

Practice – A woman taht is a carrier for hemophelia and a man that does not have hemophelia have 3 kids – what is the probability that they will have an affected child and 2 unaffected children?

A

ANSWER: 27/64

33
Q

Example – You join a Fly lab and are given stocks of true breed red eyed and true breeding white yes flies
- You know that red eyes are Wild-Type and White eyes can be controled by an X-linked trait
- You ALSO know that there are autosomal mutations that can lead to white eyes

What experiment can you do to determine if your white eyed flies are due to the sex linked hgene or due to the autosomal mutation

A

ANSWER: can cross a white eyes female + red eye male AND cross a Red eye female with a White eyed male

DO A RECIPRICAL CROSS
- if you have an X linked trait –> then a mutatant male is XwY IF its autosomal then the mutant male is ww and normal male is ++

THEN you do an opposite cross and have XwXw and XY if sex-linked or ww and ++ if autosmoal

Outcomes of teh cross = depends on difefrent results
1. Autsomal = would give mendles ratios
2. Sex linked = gives different ratios

***When doing a reciprical cross you get diferent ratios in the F2 depending on if the male or female has the WT or recessive IF it was autosomal then you would not get different ratios in if males or females have tge WT or recessive (because have equal chance of being male or female)

34
Q

How do you test for sex linkage

A

Do a reciprocal cross
***Can look at slides for more information

***If it was autosmale trait you would just get 3:1 ratio in F2
- not the ratio you get in sex linked

***When doing a reciprical cross you get diferent ratios in the F2 depending on if the male or female has the WT or recessive IF it was autosomal then you would not get different ratios in if males or females have tge WT or recessive (because have equal chance of being male or female)

35
Q

Example using data

Wt female X Mutatnt male
Given that in F2 Generation have:
428 red eye females
192 red eye males
180 white eye males

A

Ratioes =
2.4 red female:1.1Red male: 1 White male
***This looks similar to a 2:1:1 ratio (that we see in sex linked – ask is this sex linked)

36
Q

Why we use statsticis in general

A

We need to know if something thats is difefrent from our expecatation is only different because of random chance or because of something else

***We expect something BUT we now that things don’t always happen how we expect

Example:
1 family with all boys = kinda weird but not that weird
- We expect a family with 4 kids to be 2 boys and 2 girls
- Wierd if all boyes but not crazy
50 families with all boys = weird
- that is very strange
Is the weird because of random chnace or something specific?

37
Q

How do we get predictions for geneotypes

A

You can preidct phenotype and genotype ratio from genetic cross

38
Q

Use of Chi-square

A

Shows if your predeictions are correct – shows if the observations are consistemt with whay we expect to see `

39
Q

Steps for a chi-square test

A
  1. Write down your genotypes of the cross that you are predicting + formally state what you are testing
    • If have a WT female and a Mutant male

XX by XwY

F1 = Red Femal XXw and Red male XY
F2 = 2:1:1
“I am testing whether the white eye phenotype is the result of an X linked recessive alelel. If X- linked I expect different outcomes in F2s than outcomes in F1s. Do my observations meet my expectation”
- Did what I expext to see fit what I actually see

  1. Calculate the expected values
    IF have total of 800 AND know its a 2:1:1 ratio:
    For red eyed female – 800 X 0.5 = 400
    For Red males – 800 X 0.25
    For white makes – 800 X 0.25

Observed = given
For red eyed female – 428
For Red males – 192
For white makes – 180

  1. State the Null hypothesis and althernaatiuve hypothesis
  2. Calculate teh Chi-squared statistic
  3. State the degree of freedome (# of phenotypic classes -1)

Here – have 3 phenotypic classes = 3-1 = 2

  1. Determine teh Vhi-squared critical value for the alapha (p-value) used for hypothesis tesing
    • Based on Degree of freedom
      Here for 3 df at alspha of 0.05 critical value = 5.99
  2. Compare the critical valie to chi-square value and state whether null is rejected or not
  3. State overall conclusion
40
Q

Do you use ratios or numbers in chi-square

A

For a chi-square we work with NUMBERS NOT ratios

41
Q

General Null hypothesis

A

Deveiations between the expected and observed NUMBERS are due to random chance/sampling error
***Always numbers

***Stays the same for all Chi-square

42
Q

General Alternative Hypothesis

A

Random chance does not explain the diffreences between the observed and expected opbservtions for a ___ (sex linked) trait therefore ______

Example – Random chance does not explain the diffreences between the observed and expected opbservtions for a sex linked trait therefore the trait is due to something else

43
Q

Calculating Chi Square example
***Numbers of Obs and expected on slide 35

A
44
Q

General conclusion

A
  1. The differences between the observed and exoected numbers are consistent/inconsistent with random sampleing error
  2. These data support/do not support our hypothesis that white eyes is due to an X-linked recesive trait
45
Q

What does reciprical cross give you

A

Gives you expected values for what you woudl see if the trait IS sex lnked – gievs expected rations –> Can calculate expected numbers

BIOL332 (25 decks)

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