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BIOL332 > Probabilities > Flashcards

Probabilities Flashcards

1
Q

Traits Mendle Followed

A

Each trait was controled by a single gene that has 2 alleles – he crosses the plants will studying the inhertcance patterns of traits

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2
Q

Question – What is the probability that offsrping in a tetrahybird cross is short + Yellow + Wringles + Inflated for parents TtRrYyli X TtRrYyIi

A

NOTE – for these you can split up the overall genotype by genes = can look at one gene at a time

Genotype of child = ttrrY_I_ –> inheritance of all of the genes are indepent from one another = can use multiplication rule AND YY/Yy + Ii/II are mutaually exclusive meaning you can use the addition rule for those two

P(tt) AND P(rr) AND [P(YY) OR P(Yy)] AND [P(II) OR P(Ii)]
P(tt) X P(rr) X [P(YY) + P(Yy)] X [P(II) + P(Ii)]
1/4 X 1/4 X [1/4 + 2/4] X [1/4 + 2/4] = 9/256

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3
Q

How to calculate the unique number of gametes by two parents

A

2^N

N = Number of Heterozygous Loci (ONLY use heterozygous)

N could also be the number of chromosome pairs (but only include heterozygous)

Example – Number of possible gametes crossing TtRrYyli X TtRrYyIi
2^4 = 16

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4
Q

Multiplication Rule

A

The probability of 2+ indepent events occuring together is equal to the prduct of the independent probabilities

AND – Something AND another thing = multipy the probabilities

Example – What is the probaboloyu of rolling a 4 on the first roll AND rolling a 4 on the second roll
- Want 2 AND 2 –> P(2) X P(2) = 1/6 X 1/6 = 1/36

***ONLY works if the two are independent

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5
Q

When can you use the multiplication rule

A

ONLY if the two thing are independent from each other

Example: Rolling a 4 on the first role does not affect rolling a 4 on the second roll (Both need to be able to happen)
- If they are indepent = use AND

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6
Q

Addition Rule

A

The probability of two mutaually exclusive events occuring together is equal to the sume of the proababilities of the events

***Used for OR statements

Example – what is the probability of rolling a 3 OR a 4
- Since both can’t happen at the same time they are Mutaually exlcuive = can use the addition rule
P(rolling 3) + P(rolling 4) = 1/6 + 1/6 = 2/6 = 1/3

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7
Q

When can you use the Addition Rule?

A

ONLY can use if the two are mutaully exlcuive (ONLY ONE CAN HAPPEN)

Example – rolling a 3 OR a 4 – both can’t happen at once = they are Mutually exlcusive

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8
Q

Use of monohybrid/dihybird crosses

A

Predicts genotype + phenotype rules

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9
Q

Monohybrid Cross

A

Mating between parents that are both heterozygous for a single trait
Ex. Aa X Aa

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10
Q

Dihybrid Cross

A

Mating between parents that are both heterozygous for to independent traits

Example – AaBb X AaBb

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11
Q

Monohybrid Cross Example

A
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12
Q

What is each R in RR

A

Each R = a gemete frp, yje parents

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13
Q

Genertaion nomeclature

A

Parents –> F1 (1st Fidel) – generations that results of crossing parents together – F1 X F1 –> F2

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14
Q

True breeding

A

Must be homozygous – means that when you keep crossing them you always get the same phenotype = they must be homozygous

Example RR or rr

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15
Q

Gametes in Monohybrod cross

A

Top row + the sides = all possible gamete one parent can produce (top is one parent and side is another parent)

Example Rr –> R = one gamete; r = another gamete

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16
Q

Ratios in monohybrid cross

A

Gentotypic ratio – 1AA:2Aa:1aa

Phenotypic ratio – 3 Dominent: 1 recessive

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17
Q

What did Mendle Find

A

Found the fundemental laws of inheritance by studying pea plants
- Discovered that genes come in pairs and are inherited as distict units one from each parent
- Recognized doimenet + recessive traits
- Recognized mathamzatical pattern can desribe inheritance of traits from one generation to the next

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18
Q

Traits in dihybrid cross

A

Independent from one another

Example – Yellow and green phenotype in peas is independent from the round/wrinkcled phenotype

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19
Q

Example Dyhybird Cross RRyy X rrYY

A

Cross an true breeding green round by a true breeding yellow wrinkled – RRyy X rrYY –> RrYy (F1)

F1 X F1 = RrYy X RrYy

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20
Q

How to figure out the possible gametes of individual

A

Do a monohybrid cross for gene X gene – gives the gametes an individual can make

Example RrYy – looking for gametes

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21
Q

Ratio in Dihybrid Cross (if both parents are heterozygous for both traits)

A

Genotypic 9A_B_:3A_bb:3aaB_:1aabb

Phenotypic: 9Dom,Dom: 3Dom, rec: 3 Rec, dom: 1 Rec, Rec
- The phenotypic ration will always be the same BUT ONLY if the alles have simply dominent/recessive relationships

When simple dominent/recessive relationships exists – genotype + phenotype ratio for dihybrid corss will always be 9:3:3:1

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22
Q

Mendle’s Law of indepent assortment Schematic

A

During Meisosi – chromsomes can align in different ways
- Means that it would take 2 indepeny ,miosis events to produce all of the types of gametes that are pssible
- Arrangment of chromsomes in metaophase 1 are randome – either miotic event is equally likley –> means that the probability of any type of gamete = 1/4

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23
Q

Probability of any one gemete in RrYy X RrYy

A

Since you would need two mitotic events for all 4 gametes to be made (because any one mitotic event only makes two types of gametes) BUT can make all four in two mitotic events –> SINCE both mitotic events are equally liklet – teh probability of making any one gamete is 1/4

24
Q

How many mitotic events would it take to make all 4 gametes in RrYy X RrYy

A

Since you would need two mitotic events for all 4 gametes to be made (because any one mitotic event only makes two types of gametes) BUT can make all four in two mitotic events

25
Q

What do mono and dihybird crosses show

A

They point out some of mendles findings

26
Q

Mendle’s Laws

A
  1. Law of dominance and uniformity
  2. Law of Segregation
  3. Law of Indepent Assortment – ONLY seen in dihybrid cross
27
Q

Law of Dominence and Uniformity

A

Some alleles are dominent while others are recessive – organisms with at least 1 dominent allele will show the dominent phenotype

28
Q

Law of segregation

A

During gene formation – alles seperate from each other so that one gamete contains one allele

29
Q

How many alleles does a gemete contain

A

Gametes contain one allele – because of the law of segregation

30
Q

Law of Independent Assortment

A

Genes of different traits can segregate independently
***Seen in a dihybrid cross

31
Q

Test cross

A

Done if you do not the the genotype of an indovudal with dominent phenotype (if you do not know if they are AA or Aa)

TO KNOW = cross the dominent individual with a homozygous recessive indodual
- A_ X aa

Cross unkonw with aa –> if ANY of the offspring show the recessive phenotype then the unknown is Aa BUT if there are none with receives then unknown is AA

32
Q

Second way to test genotype of unknown indovual with dominent phenotype

A

Sequencing

33
Q

Gamete from parents in a cross

A

Go across the top and down the sides

***If there are doubles = can reduce it down – technicallly this cross would give you 2AA:4Aa:2aa –> You can reduce that down to 1:2:1

34
Q

Meiosis (overall)

A

1 diploid –> 4 haploid
***Have 4 egg or 4 sperm produced

35
Q

How do you know the possible gametes

A

From meosis

36
Q

Does one meiosis event ALWAYS produce ALL 4 gametes

A

NO – because for a hetrozygous dihybrid corss = you would need 2 miototic events for all 4 gametes to be made
- Because of independent assortment – the chromsomes can line up in two ways = have 4 possible types of gemates that can be made BUT only two types of made per mitotic event

***SLides image = shows that each mitotic event only amkes 2 types of gemates = would take two mitotic events to make all 4

37
Q

Probability of any one mitotic event occuring

A

Is equally likley as any other event occuring – makes the ration of possible gametes from two Heterozygous indidual 1:1:1:1

AaBb X AaBb

1AB:1Ab:1aB:1ab

38
Q

BUT Technically is it possible that One mitotic event produces ALL 4 gametes

A

YES – because of crossing over –> can produce all 4 gametes if have recombination –> to get all 4 possibilities if switch alleles from homologous chromsomesom during miosis 1

***Because of recombination = get all 4 possible gametes in one mitotic event
- works because the recombination only happens on one sister chromatids – other doesnt switch to give the other genotype

39
Q

Recombination

A

Cut the chromsomes and switch places and put back together

40
Q

Frequencey of recombination

A

Recombination occurs once or twice in each mitotic event – sometimes things switch positins BUT often not
- recombination = NEVER happens at a centromere – liklihood of recombination = depends on proximity to centromere

41
Q

Getting all possible gametes with and without recombination

A

No recombintion (STILL have IA) = need 2 mitotic events for all 4 possible gametes
- In this case ALL 4 gametes are equally likley (1:1:1:1)

With recombination (Still have IA) – get all 4 possible gametes in one mitotic event

42
Q

Things on tops/sides of Punnet Squares

A

All possible gamtes for individual (top is one indivdiual and side is the other indivdiual)
- they are all 4 possible egg or all 4 possible sperom

***To know all possible gametes for each individual do a cross of gene X gene

43
Q

Example question – what is the probability that a kitten would be striped + dillute for AaDd X AaDd

A

3/16

Striped + dillute = A_dd
- based on dihybrid cross gentype ratio is 3A_dd

OR can do P(A_) X P(dd) –> 3/4 X 1/4 = 3/16

44
Q

When would phenotype ratio for a dihybrid cross NOT be 9:3:3:1

A
  1. If you did not have heterozygius parents (ex. llss X llss)
  2. IF BOTH STILL HETEOZYGOUS –> Would not be the same ratio if it was not completley dominent/recessive relationships
    • The phenotypic ratio would only work for classic dom/rec relationships (BUT the genotyoic ratio would stay the same for two heteozygous parents no matter the allele relationships)

***For two heteozygous = the genotyoic ratio will always be the same BUT the phenotyoic ratio might not be

45
Q

Predicted phenotype vs. genotype ratios

A

The predicted genotype ratios for corsses wil always be the same BUT the predicted phenotype ratios will depend on the relationship of the alleles

***Phenotypic will only work for normla dominent/recessive realtionships

46
Q

How to calculate the number of unique gametes

A

2^N

N= # of Pairs of HC –> ISSUE = only include heterozygous pairs

Ex. 3 HC –> 2^3 = 8 unique gametes

Example AabbCc –> 2^2 = 4

47
Q

Example Question – Assuming that each gene is on a different chromsome – how many unique types of gametes can the following cay produce AaDdSs?

A

2^3 = 8

N=3 –> 3 pairs of HC

48
Q

Example Question – What is the probability that a kitten will be Striped + Dilute + No white

AaDdSs X AaDdSs

A

Cat wanted = A_ddss – P(A_ddss)

P(A_) X P(dd) X P(ss) = 3/4 X 1/4 X1/4 = 3/64
P(A_) = P(AA) + P(Aa) = 1/4 + 2/4 = 3/4 –> use addition because OR

***Other way to do this is 8X8 Punnent square (KNOW is 8X8 because each individual can make 8 unique gamtes because 2^3)

49
Q

Example Question – What is the probability that a kitten will be striped and dilute AaDd X AaDd

A

Genoype needed = A_Dd –> P(A_Dd)

Option 1 – look at dihybrid cross for A_Dd –> get 3/16

Option 2 – P(A_) X P(Dd) –> [P(AA) + P(Aa)] X P(Dd) = 3/16

Striped AND dilute –> Multiply
Strpied = AA OR Aa = ADD

50
Q

What is the probabilty that a cat will be striped and diliute or solid and non-dilute

A

Two Genotypes:
1. A_dd –> P(A_dd)
2. aaD_ –> p(aaD_)

P(A_dd) + p(aaD_) –> OR = +

P(A_dd) –> [P(AA) + P(A_)] X P(dd) = 3/16
P(aaD_) –> [P(DD) + P(Dd)] X P(aa) = 3/16

3/16 + 3/16 = 6/16

51
Q

Dominent White Allele (W) is dominente to w and leads to loss of pigmentation + diffreent colored eyes + deafness –> Dominent White is an exmaple of

A

Epestais + Pleitropty + Dominent vs. recessive

52
Q

Epistatic alleles

A

Mask the effects of alleles of other genes – when an allele at one locus prevent you from knowinh the alleles at another
-Mask phenotypic effect of alleles at another locus
- Non-additive – one gene masks other genes

Example – When have the W allele in cate = the cat is ALL white BUT you can’t konow if the cat would be agouti (if A_ or aa) OR can’t know if the cat is organge or black (Xo or Xb) – the cat HAS alleles at the locus for those genes BUT we don’t know what they are because the phenotype that is suaoly expressed because of them is masked by the W allele
- Don’t know if Xb or Xo – because W has masked it

53
Q

Pleitropic genes

A

ONE gene leads to many different phenotypes –> W genes leading to white hair and deaf and eye color
- 3 phenotypes (Hair color + deafness + eye color) ALL due to one gene

54
Q

Context of allele relationships

A

Allele relationships ate context specific – genetic dominance relationships are context specific
- Allele might not be dominent ALL of the time –> Might be dominet to one thing and not another

Example – Blood Type
iA – codiminent with iB
iA is dominet to i
iB is dominet to i
iAiA or iAi = Type A
iBiB or iBi = Type B
IAiB = Type AB
ii = Type O
- Shows that allele relationships are conetxt specific to the OTHER allele you are compining it to – the relationship between iA and iB says nothing about the relationship between iA and i

55
Q

What alleles can we compare

A

We can only compare the realtionships between one allele in relationship to other alleles at the same locus

BIOL332 (25 decks)

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