Mendelian Genetics Flashcards

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1
Q

What three scientists did Mendel learn under at the University of Vienna and what did he learn?

A

-Particulate theory of inheritance from Doppler
-Plant hybridization from Unger
-Probability theory from Ettinghausen

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2
Q

what were the five key features of Mendel’s experiments?

A
  1. controlled crosses: identity of parents were known
  2. pure breeding strain of pea plants used : to idental homozygous alleles
  3. dichotomous traits used: two possible phenotypes
  4. Quantified results: patterns of heredity could be predicted
  5. reciprocal crosses and test crosses used
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3
Q

Parental Crosses

A

Experiment:
Mendel crosses two pure-breeding parents(Homozygous Dominant and Recessive).
One was violet one was white
The F1 generation was all purple.

Conclusions:
revealed the principle of dominance
rejected blending inheritance–> inheritance is particulate not blending

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4
Q

Reciprocal Crosses:

A

Experiment:
switched the sex of the purple and white flowers and the f1 generation was still all purple

(pollen from white flowers fertilized ovule from purple flowers= pollen from purple flowers fertilized ovules from white flowers)

Conclusions:

revealed that trait dominance is not influenced by the sex of the parent

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5
Q

Test Crosses:

A

Experiment:
Crossing the unknown dominant genotype with a homozygous recessive genotype

Conclusions:
Can determine if the dominant phenotype is coming from a Pure breeding genotype or heterozygous

1:1 phenotypic ration if heterozygous
1:0 phenotypic ration of homozygous

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6
Q

Monohybrid Crosses:

A

Experiment:
All purple F1 generation were self-crossed
Conclusions:
3:1, Dom: rec phenotypic ratio observed
F1 generation is heterozygous dominant
the reappearance of recessive phenotype reveals that:
alleles segregate during gamete formation ( law of segregation)
two alleles are present in each plant

alles exist in pairs (one comes from sperm the other room egg)

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7
Q

F3 self crosses

A

Experiment:
Round vs Wrinkled determind by the R and r alleles
Conclusions:
all F1 were heterozygous dominant
F2s ha 3:1 round vs wringkled ration
F3s demonstrates that 1/3 of the round plants were homozygous dominant(RR) and 2/3 were heterozygous dominant (Rr)

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8
Q

Dihybrid Crosses

A

Experiment:
Crosses dihybrid pure breeds( ggWW- green and round x GGww- yellow and wrinkled) and (GGWW-yellow and round x ggww-green and wrinkled)
Conclusion:

F1S ALL GgWw ( yellow and round)

self cross of F1s–>
9:3:3:1 phenotypic ratio

reveals the law of independent assortment (during gamete formation the segregation of allels of one gene is independent of the segregation of alleles of another gene)

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9
Q

What is the law of segregation?

where does it occur in meiosis

A

alleles separate During gamete formation in a 1:1 ratio.

Occurs in anaphase I during the separation of homologous pairs

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10
Q

What is the law of independent?assortment?

where does it occur in meiosis?
occurs during metaphase I as the homologous pairs ranadomoly and independently align on metaphase plate

A

The segregation of alleles for one gene is independent of the segregation of alleles for another gene

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11
Q

Tri-hybrid cross

A

Conclusion:
all F1 are AaBbCc
each F1 can make 8 gametes
Phenotypic ratio: 27:9:9:9:3:3:3:1

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12
Q

What is the product Rule

A

if two events are independe, the the joint probability of both events occurring is the product of their individual probabilities

KEYWORD: AND–> multiply probabilities

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13
Q

What is the sum rule

A

when the outcomes are mutually exclusive the sum of the individual probailities calculates the joint probability

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14
Q

What is conditional probability rule

A

conditioning on one outcome what is the probability of another outcome.

E.g: Probability that F2 purple flower peas(dominant are heterozygous)

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15
Q

What is the binomial probability rule?

A

given two possible outcomes performed many times the probability of each combination is (p+q)^n

P= probability of dominant trait for respective cross
q= probability of recessive triat for respective cross

if n=3 and find rate of dom chosen 2 times and recssive chosen 1
(p+q)^3= p^3+3p^2q+3pq^2+q^3

choose factor: 3p^2q–> 3*(probability of dom)^2(probability of recessive) ^1

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16
Q

How do you do the chi square test to determine if something is consistent with mendelian inheritance?

A

(observed-expected)^2/observed

expected= probability of phenotype * total offspring
df=number of phenotypes-1
p= box that correlates df row to 0.05 column in chi square table

if SUM> P does not support
if SUM<P does support

17
Q

Is all inheritance fundamentally Mendelian?

A

Yes, Continuous characters (such as height) also have a particulate inheritance

18
Q

What is not fundementally mendelian

A

extranuclear inheritance (mitochondria and chloroplasts)

meiotic drive
linkage