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PENN STATE / Biology / BIOL 222 / How do you know if genes are linked or unlinked?

How do you know if genes are linked or unlinked?

How do you know if genes are linked or unlinked?

Description

Bio 222


How do you know if genes are linked or unlinked?



Week 2 Notes

Chapter 3: Basic Principle of Heredity

Monohybrid Cross

∙ Analyzes only one characteristic/gene

∙ All characteristics are separate from one another

Dihybrid Cross

∙ Analyzes two characteristics/genes simultaneously o Can be linked or unlinked

∙ Usually has four possible phenotypes


How to fill out the dihybrid cross punnett square?



We also discuss several other topics like Where are hyperthermophiles found?

o Phenotypes 1 and 2 are dominant

 RRYY or RrYy  

o Phenotype 1 is dominant and phenotype 2 is  recessive

 RRyy or Rryy

o Phenotype 1 is recessive phenotype 2 is dominant  rrYY or rrYy

o Both phenotypes are recessive

 Rryy

Note this lecture is discussing unlinked genes

P (parental generation)


How to make the branch diagram?



∙ Remember these are  pure breed

F1 (comes from parental cross) ∙ Heterozygous for both  gene pairs so they self  

pollinate and result in F2  F2 (comes from F1 selfing)

Ex: Don't forget about the age old question of What is the purpose of accounting standards?
Don't forget about the age old question of Bottleneck theory is the theory of what?

R and r are two alleles for one  gene for the shape of peas R=round peas r=wrinkles Y and y are two alleles for one  gene for the color of peas Y=yellow y=green

These genes are unlinked  because the genes are on  different chromosomes We also discuss several other topics like What is the meaning of single-blind in a placebo?

P  

RRYY X rryy (crossed)

=

F1

RrYy (self pollinate)

=

F2

Makes a 9:3:3:1 phenotypic  ratio

These are what each number  stands for respectively:

 9= round and yellow peas  (homozygous and heterozygous dominant)

3=round and green  We also discuss several other topics like What is the difference between crystal and crystalline?

(homozygous and heterozygous for shape, homozygous for  color)

3=wrinkled and yellow  

(homozygous for wrinkled,  homo and hetero for yellow) 1= wrinkled and green  

peas(both genotypes are  recessive)

How to fill out the dihybrid cross Punnett Square 1. List gametes from parents on one side

2. List gametes from parents on other side

3. Fill in genetic combinations in the squares

4. Determine genetic ratios (phenotypic or  

genotypic) by counting squares

Independent Assortment

∙ Genes assort independently from one another o Ex: eye color is separately determined from hair  color

Mendel’s Second Law

∙ law of independent assortment

∙ during gamete formation the segregation of alleles of  one gene pair is independent of the segregation alleles  from another gene pair

o Different gene pairs containing there alleles  are not linked, gene pairs on different  

chromosomes assort independently during  

meiosis

We also discuss several other topics like Is water intracellular or extracellular?

Branch Diagram

∙ For F3 generation

¾ A__ : ¾ B__ and ¼ bb: 9/16 A__B__ and 3/16 A__bb ¼ aa: ¾ B__ and ¼ bb: 3/16 aaB__ and 1/16 aabb

Say Aa is a gene for hair color A is brown a is blonde Bb is a gene for eye color B is brown b is blue

This branch diagram shows that for the first thing written in  both rows is the probability that someone will get brown or  blonde hair color. The next column in both rows shows the

percentage of those that will have brown eye color  (homozygous or heterozygous) and blue hair color  (homozygous recessive). The next column in the first row  shows the probability of people who will show the dominant  phenotype of having both brown eyes and hair and also  those who will have brown hair and blue eyes. The last  column in the second row shows those who will end up  having blonde hair and brown eyes and those who show both recessive phenotypes for having blonde hair and blue eyes.  So if you think about it the first column discusses the  phenotypes for hair color and the second column shows the  phenotypes for eye color. The last column shows the  probabilities of those phenotypes occurring together.

Overall this diagram portrays how the 9:3:3:1 ratio is  established.

How to make the branch diagram

1. List the probabilities for one even happening (getting  the dominant phenotype for one gene and the recessive phenotype for that same gene)

2. List probabilities of second event next to those of the  next

(these are the last to columns and this is when you  include the probabilities of the second gene)

3. Next list the probabilities of those genes alleles  occurring together (blonde hair and blue eyes, brown  hair blue eyes, etc..) and use the product rule to  determine these genetic ratios

refer to branch diagram above

for steps 1 and 2 you can draw to separate monohybrid crosses and then figure out the probabilities for step 3

To calculate genotypic frequency

EX:

F1

AaBbCcDdEe X AaBbCcDdEe

1. Determine the probability of getting each allele for each separate gene  

∙ AA

o (1/2A)(1/2A)=1/4

 Each parent has a 50% chance of passing  down an A  

 Remember to use multiplication rue

∙ bb

o (1/2b)(1/2b)=1/4

 Each parent has a 50% chance of passing  down a b  

∙ Cc

o (1/2C)(1/c)+(1/2C)(1/2c)=1/2

 This is different because this is a  

heterozygote. Each parent has 2 alleles  

they can pass down so it has to be taken  

into consideration the probability of each  

of the alleles each parent can pass down.

 Both parents have a 50% chance of  

passing down both alleles

 Since there are two alleles being taken into consideration the addition rule and  

multiplication rule are used

Now by applying these examples and knowledge you  can find probabilities for trihybrid, tetrahybrid crosses  and so on

Using these parents  

F1

AaBbCcDdEe X AaBbCcDdEe

Finding the probability of getting AAbbCcDdee Respectively

(1/4)(1/4)(1/2)(1/2)(1/4)=1/256

P=1/256

The ¼ are always homozygotes and the ½ is the  probabilities of the heterozygotes

To calculate phenotypic frequency

F1

AaBbCcDdEe X AaBbCcDdEe

Probability of getting A__, bb, C__, D__, ee

 A__ (dominant phenotype)

o ¾ because its dominant  

 bb

o ¼ chances of getting a recessive is less because it needs to be homozygous

 C__

o ¾

 ee

o ¼  

Respectively: (3/4) (1/4) (3/4) (1/4)= 27/1024

Probability of getting these phenotypes is 27/1024

Human Pedigree Analysis

Only ones you need to know are  

 Unaffected person

 Obligate carrier

 Family

 Consanguinity

Autosomal Recessive Disorders

 Mendalian inheritance of an autosomal recessive  disorder is revealed by the appearance of the  phenotype in male and female progeny of unaffected  individuals

 The gene is recessive so the only way someone would  exhibit the disorder or trait is if the genotype is  homozygous recessive

 A person who is heterozygous for the disorder is a  carrier

 If both parents are homozygous recessive for the  disorder all the children will have the disorder  It tends to skip generations

Autosomal dominant disorders

 Mendalian inheritance of an autosomal dominant  disorder show affected males and females in each  generation and also show affected males and females

transmitting the condition to sons and daughter in  equal proportions

 If the family size is large enough you can see the  disorder in every generation

 Dominant disorders can never be a carrier because if  you have the dominant trait you definitely get the  disorder

Chapter 4: Extensions and Modifications  of Basic Principles

Sex Determination

XX-XO

 Females have 2 X chromosomes (XX): homogametic sex  Males have 1 X chromosome (XO) heterogametic sex  During Meiosis in the male ½ of the gametes get an X  giving a 1:1 sexual frequency

XX-XY

 Human situation

 Females 2X chromosomes (XX): homogametic sex  Males have 1X and 1Y chromosome (XY):  

heterogametic sex

ZZ-ZW

 Females have 1 Z and 1 W chromosome (ZW)  heterogametic sex

 Males have 2 Z chromosomes (ZZ) homogametic sex

Genic Sex Determination

 No sex chromosomes. Specific genes on the autosomes  determine the sex

 Genes on various autosomes in the genome. Genes  spread out in a variety

 Ex: plants and protozoans

Environmental Sex Determination

 Environmental factors determine the sex  

 Ex: temperature

Drosophila

 X: a ratio dictates the sex THIS IS TRIVIA

Humans

 The presence of the Y chromosome dictates maleness o SRY gene is on the y chromosome, this gene is  what establishes maleness. If SRY is deleted the  person can be female

SRY Gene

 Male determining gene in humans on Y chromosome   Females develop if this gene is missing from the Y in an  XY individual

 Males develop if this gene is trans located to an X in an  XX individual  

Sex Linked Inheritance

 Inheritance pattern of genes on sex chromosomes  Sex chromosomes carry many genes unrelated to male  and female development

X-linked Recessive Inheritance

 Many more males than females have it

 Mutation is on a gene on the X chromosome  o Females need the mutation on both X  

chromosomes  

o Males need the mutation on only one X  

chromosome because there genotype is XY

 Ex gene for colorblindness is on the X  

chromosome

Experiment 1  

X+ is dominant for red eye color

Xw is recessive for white eye color

Remember X is what determines eye color not Y

Parental Generation (P)

 X+X+ (red eye female) X w Y (white eyed male) F1 Generation

 All females are red: X+Xw because X+ is dominant over Xw

 All males are also red X+Y because males pass on  the Y chromosome and the female was  

homozygous dominant so the males can only get  an X+

 X+Xw times X+Y gives F2 generation

F2 Generation

 All females get red eyes because X+ is passed from  male and X+ is dominant

 50% of males have white eyes because it’s a 50%  chance females pass the Xw or X+ genotype

 Phenotype frequency for these flies is 3:1

 Experiment 3

 Reciprocal cross of experiment 1

X+ is dominant for red eye color

Xw is recessive for white eye color

Parental Generation

XwXw (white female) times X+Y (red male)  

F1 Generation

 Red Female XwX+ because they definitely get the X+  from the male

 White Male XwY because male passes down the Y and  female is homozygous recessive so it has to get Xw   These cross

F2

1:1:1:1  

Red Female: White Female: Red Male: White Male  (X+Xw) (XwXw) (X+Y) (XwY)  Because female give 50% of one of the alleles and the  male gives the recessive type but the females still give

50%. Think in a way that the female overall determines  what happens here.

Experiment 2 : Test Cross

F1

Red Female (X+Xw) X white male

Get the same result from experiment for F2 as you did in the  parental generation (refer to that above)

X-Linked Recessive

 Many more males than females show the recessive  phenotype  

 If it is rare almost all affected people are male o If rare none of the offspring of an affected male  are affected but all daughters are carriers

o Female Carriers will pass condition to ½ her sons  and ½ her daughters will be carriers

o Recessive will skip generations

X-linked Dominant

 Affected males pass condition to all daughters never to  sons because sons get Y from dads not X

 Affected females pass condition to ½ of sons and ½ of  daughters  

Y-linked Inheritance

 Other than the SRY gene found on the Y chromosome in males very few human phenotypes are known to be Y linked

Dosage Compensation in mammals: X Inactivation  Female mammals inherit 2 X chromosomes

 Early in development one of the 2 X chromosomes is  randomly inactivated in each cell

 The same X chromosome remains inactive in all of the  daughter cells of each progenitor

Barr Body

 The inactive X chromosome

Pseudominance

 If the functional X has a recessive allele the recessive  phenotype is expressed. Since the inactivation process  is random all female mammals are genotypes  corresponding to inactivation alternatives.

Chapter 4: Extensions and Modifications  Dominance  

∙ Phenotype of the heterozygote is the same as one of  the homozygotes

Codominance

∙ Phenotype of heterozygote includes phenotypes from  both homozygotes

∙ When two alleles are both shown

∙ Ex: A fish being colored blue and red

Incomplete Dominance

∙ Phenotype of the heterozygote is intermediate between the two homozygotes

∙ white and red flowers make a pink flower  

∙ The pink flower illustrate incomplete dominance Human Hemoglobin

∙ HbA

o One allele that codes for normal blood cells o It is completely dominant with respect to anemia o Incomplete dominance to sickles

∙ HbS

o Abnormal “sickle” shaped RBCs= anemia = fatal o HbA is dominant to it

Examples

∙ HbAHbA

o this is for normal blood cells

∙ HbSHbs

o Abnormal sickle shaped RBCs, anemia

∙ HbAHbS

o Since there is HbA it is dominant to getting  anemia but since HbS is present at low oxygen  levels sickles can form

o So no anemia but can get sickle shaped cells

Pleiotropic Allele

∙ An allele that affects more than one characteristic ∙ HbS is an allele that is an example of this

Penetrance and Expressivity

Penetrance

∙ Percentage of individuals having a particular genotype  that expresses the expected phenotype  

∙ Ex: B is an allele for brown hair. BB genotype is  expected to have the phenotype of brown hair. So, this  is the percent of individuals from a selected group,  population, etc who actually show brown hair with this  genotype

Incomplete Penetrance

∙ A genotype that does not always produce the expected  phenotype  

∙ Ex: Human polydactyly: it is a dominant allele but not  everyone with the allele has extra digits

Expressivity

∙ Degree to which a phenotype is expressed

Incomplete Penetrance and Expressivity are  caused by other genes and environmental factors

Lethal Alleles

∙ Alleles of a gene that cause death

∙ Death may occur during early embryonic development,  after birth, or later in life

Mouse Coat Color

∙ Yellow coat is a variant

∙ Dominant yellow is lethal and the mice die in utero  before birth (YY)

Ex:

Parental Generation

Yellow X Yellow

F1

2:1 Yellow: non yellow

¼= YY (die in utero)

Normally ratios should add to four but this one adds to  three because one of the mice were not born which  suggest that the mice presenting the yellow phenotype  have to be heterozygous because if they were  homozygous they would die before they were born Manx Cat

∙ Has no tail when it is heterozygous  

∙ When it is homozygous it is lethal

∙ Factor C clotting mutation

These two examples of lethality (Mouse and cat)  are pleiotropic

o Pleotropic: Allele that affects more than one  characteristic

Multiple Alleles

∙ A gene that has more than two alleles  

∙ EX: ABO blood groups

Blood Groups

∙ IA, IB, i

o IAi and IBi are completely dominant to IAi and Ibi  and codominant in IAIB

o A and B are codominant

Universal donor

Univers

al  

Accept

AB blood has no antibodies because it can accept both  antigen types hence why this is the universal acceptor

Pepper Color

9:3:3:1 ratio

Two genes affect one characteristic

Y+, y and C+, c affects the color of

a pepper

Parental Generation

Y+Y+ C+C+ makes a pepper red

∙ Homozygous Dominant for

both gene pairs

yycc makes a pepper cream

∙ Homozygous recessive for

both gene pairs

F1 Generation

Y+yC+c

∙ Heterozygous for both gene

pairs

F2 Generation

 Y+__C+__

∙ Having one dominant allele

for both gene pairs makes a

red pepper

Y+__ cc

∙ Having one dominant Y and a recessive cc makes the  color peach

yyC+__

∙ Having recessive yy and one dominant C makes the  color orange

yycc

∙ Having homozygous recessive for both gene pairs  makes the color cream

Epistasis

∙ When the effect of one gene masks the effect of  another gene at a different locus

Recessive Epistasis

∙ Need to be homozygous recessive for one of the gene  pairs to hide the expression

∙ Ex: Labrador Retriever

Labrador Retriever

Bb: pigment production

Ee: Pigment disposition

B: black color

B: brown color

Parental Generation

BBEE (black) X bbee (yellow)  

∙ Yellow is pigment disposition  

∙ Recessive ee affects all coloration

∙ Dominant EE or Ee has no impact on color  

only look at Bb alleles to see the color

F1 Generation

BbEe(black) X BbEe

∙ Both turn out to be heterozygotes so you still  only look at the Bb alleles because there is no

recessive ee to dominate the colors

F2 Generation

Gets a 9:3:3:1 ratio

∙ B_E_ = 9 black

∙ bbE_= 3 brown

∙ B_ee= 3 yellow

∙ bbee= 1 yellow

Recessive Epistasis Requires ee

Dominant Epistasis

∙ Only need one dominant allele for epistasis

∙ Ex: Summer squash

Summer Squash

W: the dominant allele for epistasis prevents conversion of  compound A to compound B to make the squash a WHITE  COLOR

Parental Generation  

WWYY (white) X wwyy(green)  

F1

WwYy (white) X WwYy (selfing)

F2

W_Y_: 9 white

W_yy: 3 white

wwY_: 3 yellow

wwyy:1 green

W is epistatic to Y and y  

References

<Dr. Paul Babitzke > “Chapter 4àExtensions and Modifications of Basic  Principles (9-2-16)” BIO 222. 2016.” The Pennsylvania State University, Pa.

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