ch5 linkage

Copyright © 2009 Pearson Education, Inc.

5.0 Linkage

Prepared by Pratheep SandrasaigaranLecturer at Manipal International University


By the end of this chapter you should be able to:

• Identify Mendel's law of segregation and independent assortment

• Define Chromosomal linkage.• Know the importance of Chromosomal

linkage.• Make genetic map unit .• Make physical map units.

Prepared by Pratheep Sandrasaigaran

Diagram adopted fromInternet Sources


5.1 Mendel's law of segregation vs independent assortment



• Mendel mated two contrasting, true-breeding varieties, a process called hybridization.

• The true-breeding parents are the P generation.

• The hybrid offspring of the P generation are called the F1 generation.

• When F1 individuals self-pollinate or cross- pollinate with other F1 hybrids, the F2 generation is produced.

Mendel’s Experimental, Quantitative Approach



Diagram adopted fromCampbell Biology (10th Edition)

Mendel’s Experimental



Figure 14.2b

RESULTS (F1)

• When pollen from a white flower was transferred to a purple flower, the first-generation hybrids all had purple flowers.

• The result was the same for the reciprocal cross, which involved the transfer of pollen from purple flowers to white flowers


First filialgenerationoffspring(F1)

5 Examinedoffspring:all purpleflowers



RESULTS (F2)

• When Mendel crossed the F1 hybrids, many of the F2 plants had purple flowers, but some had white

• Mendel discovered a ratio of about three to one, purple to white flowers, in the F2 generation



P Generation(true-breeding

parents) Purpleflowers

Whiteflowers





parents)

F1 Generation(hybrids)

Purpleflowers

Whiteflowers

All plants had purple flowers

Self- or cross-pollination





parents)

F1 Generation(hybrids)

F2 Generation

Purpleflowers

Whiteflowers

All plants had purple flowers

Self- or cross-pollination

705 purple-flowered

plants

224 whiteflowered

plants




• Mendel reasoned that only the purple flower factor was affecting flower color in the F1 hybrids.

• Mendel called the purple flower color a dominant trait and the white flower color a recessive trait.

• The factor for white flowers was not diluted or destroyed because it reappeared in the F2

generation.

• Mendel called it as “heritable factor” is what we now call a gene.

The Law of Segregation



Figure 14.4

Allele


Allele for purple flowers

Locus for flower-color gene

Allele for white flowers

Pair ofhomologouschromosomes



• Alternative versions of genes account for variations in inherited characters, now called alleles.

• For example, the gene for flower colour in pea plants exists in two versions, one for purple flowers and the other for white flowers.

• Each gene resides at a specific locus on a specific chromosome.

The Law of Segregation-Mendel’s First Model:



• For each character, an organism inherits two alleles, one from each parent

• The two alleles at a particular locus may be identical, as in the true-breeding plants of Mendel’s P generation

• Alternatively, the two alleles at a locus may differ, as in the F1 hybrids

The Law of Segregation-Mendel’s Second Model:



• If the two alleles at a locus differ

• Then one (the dominant allele) determines the organism’s appearance.

• The other (the recessive allele) has no noticeable effect on appearance

The Law of Segregation-Mendel’s Third Model:



• law of segregation: the two alleles for a heritable character separate (segregate) during gamete formation and end up in different gametes.

• Thus, an egg or a sperm gets only one of the two alleles that are present in the organism.

• Mendel’s segregation model can be shown using a Punnett square.

The Law of Segregation-Mendel’s Forth Model:


ARepresents a dominant allele

a Represents a recessive allele


Figure 14.5-1

P Generation

Appearance:Genetic makeup:

Gametes:

Purple flowers White flowersPP pp

P p




Figure 14.5-2

P Generation

F1 Generation


Gametes:


Gametes:

Purple flowers White flowers

Purple flowersPp

PP pp

P

P

p

p1/21/2




Figure 14.5-3

P Generation

F1 Generation

F2 Generation


Gametes:


Gametes:

Purple flowers White flowers

Purple flowers

Sperm from F1 (Pp) plant

Pp

PP pp

P

P

P

P

p

p

p

p

Eggs from F1 (Pp) plant

PP

ppPp

Pp

1/21/2

3 : 1



Monohybrid crosses


TEST YOUR KNOWLEDGE 1



Define the following terms

•Homozygous An organism with two identical alleles for a character

•HeterozygousAn organism that has two different alleles for a gene

•PhenotypePhysical appearance

•Genotypegenetic makeup



Phenotype

Purple

Purple

Purple

White

3

1

1

1

2

Ratio 3:1 Ratio 1:2:1

Genotype

PP

Pp


(homozygous)

(heterozygous)

Pp(heterozygous)

pp(homozygous)


• Crossing two true-breeding parents differing in two characters produces dihybrids in the F1 generation, heterozygous for both characters

• A dihybrid cross, a cross between F1 dihybrids, can determine whether two characters are transmitted to offspring as a package or independently

The Law of Independent Assortment



Figure 14.8

P Generation

F1 Generation

Predictions

Gametes

EXPERIMENT

RESULTS

YYRR yyrr

yrYR

YyRr

Hypothesis ofdependent assortment

Hypothesis ofindependent assortment

Predictedoffspring ofF2 generation

Sperm

Spermor

EggsEggs

Phenotypic ratio 3:1

Phenotypic ratio 9:3:3:1

Phenotypic ratio approximately 9:3:3:1315 108 101 32

1/21/2

1/2

1/2

1/41/4

1/41/4

1/4

1/4

1/4

1/4

9/163/16

3/161/16

YR

YR

YR

YRyr

yr

yr

yr

1/43/4

Yr

Yr

yR

yR

YYRR YyRr

YyRr yyrr

YYRR YYRr YyRR YyRr

YYRr YYrr YyRr Yyrr

YyRR YyRr yyRR yyRr

YyRr Yyrr yyRr yyrr




• The law of independent assortment states that each pair of alleles segregates independently of each other pair of alleles during gamete formation

• Strictly speaking, this law applies only to genes on different, nonhomologous chromosomes or those far apart on the same chromosome*

• Genes located near each other on the same chromosome tend to be inherited together (linkage)*

law of independent assortment


Diagram adopted from Human Genetics concepts and Application 9th ed.


• Linkage refers to the transmission of genes on the same chromosome.

• Linked genes do not assort independently and do not produce Mendelian ratios.

• linkage has been used as a mapping tool in identifying disease-causing genes, and helped pave the way for genome-wide association studies.

Linkage




1. Imagine that you encounter creatures in outer space whose traits are inherited according to Mendel's laws. In these creatures, purple eyes (E) are dominant to yellow eyes (e). Two purple-eyed creatures mate and produce six offspring (four purple-eyed and two yellow-eyed). What are the genotypes and phenotypes of the parents? What are the genotypes of the offspring?

• Two creatures with purple eye (EE or Ee) mate each other. If the parents were homozygous dominant (EE and EE), the offspring cannot be with yellow eye.


E eE EE Eee Ee ee

• Genotype of Parent = Ee and Ee• Phenotype of Parents = Purple colored eye• Genotype of offspring = 1:2:1• Phenotype of offspring = 3:1


2. In fruit flies, long wings (L) are dominant to short wings (l). Two long-winged flies produced 49 short-winged and 148 long-winged offspring. What were the probable genotypes of the parents? About how many of the long-winged offspring should be heterozygous?

• 49 are short wings• 148 are long wings

• 148:49 = 3:1


• In order to obtain 3:1 phenotype the parents should be heterozygous (Ll and Ll).

• 2/3 of 148 will be heterozygous (from punnet table) = 99

L lL LL Lll Ll ll


3. In humans, brown eyes (B) are dominant to blue eyes (b). A brown-eyed man marries a blue-eyed woman. They have eight children (all are brown-eyed). What are the possible genotypes of both parents and their offspring?

• For the entire child to be brown colored eye, the father must be homozygous dominant (BB).


• Hence man (BB), female (bb) while the offspring all will be heterozygous (Bb) and brown eyed.

B Bb Bb Bbb Bb Bb


4. A black, smooth guinea pig was mated with an albino, rough guinea pig. Their offspring were black, rough and black, smooth. These were the only offspring types produced over a period of several years after multiple mattings. Black color and rough fur are the dominant traits for guinea pigs. What was the probably genotype of each parent?

Parents•Black = B_•White = bb•Rough = R_•Smooth = rr


In F1 generation

•For the offspring to be constantly black, rough and black, smooth for over the period of several years, Black must be homozygous dominant (BB).•Since the offspring are mixture of smooth and rough, the rough trait must be heterozygous (Rr)•Hence Black-smooth genotype is BBrr while the Albino-rough genotype is bbRr.•Gametes are Br from Black-smooth parent and bR and br for Albino-rough parent.•Offspring will produce BbRr (black, rough) and Bbrr (black, smooth).


5. In moose, brown coat color (B) is dominant to albino (no pigment) (b) and rough coat (R) is dominant to smooth coat (r). Two moose are selected for breeding and their genotypes are BBRR (female) and bbrr (male). Determine the expected genotypic and phenotypic ratios for the:

a. F1 generation

b. F2 generation

c. Cross between an F1 moose and a moose with the genotype BBRr

• Gametes from BBRR is BR and Gametes from bbrr is br.• F1 generation: Genotype (BbRr) and phenotype (Brown-rough)• F2 generation: Genotype (Punnet table 1) and phenotype (9:3:3:1)• F1 BbRr vs BBRr (Punnet table 2) and phenotype (3:1)


BR Br bR brBR BBRR BBRr BbRR BbRrBr BBRr BBrr BbRr BbrrbR BbRR BbRr bbRR bbRrbr BbRr Bbrr bbRr bbrr

Punnet table 1

BR Br bR brBR BBRR BBRr BbRR BbRrBr BBRr BBrr BbRr Bbrr

Punnet table 2


5.2 Chromosomal linkage




• Observed the unexpected ratios indicating linkage in the early 1900s.

• They crossed true-breeding plants with purple flowers and long pollen grains (genotype PPLL ) to true-breeding plants with red flowers and round pollen grains (genotype ppll ).

• The dihybrid F1 self crossing did not yield the expected 9:3:3:1 phenotypic ratio.

William Bateson and R. C. Punnett’s experiments




• Those with the parental phenotypes P_L_ and ppll were more abundant than predicted.

• Other two progeny classes, ppL_ and P_ll were less common.

• Bateson and Punnett hypothesized genes are transmitted on the same chromosome and they do not separate during meiosis.





• Why the other two progeny classes, ppL_ and P_ll were still observed even though their number is less?

• If a small but distinct distance exist between gene loci, cross over occurs and can generate few recombinants.

• But many are still parental gametes.





• How to distinguish parental chromosomes from recombinant chromosomes?

• Only if the allele configuration of the two genes is known• Cis (a)• Trans (b).

Allele configuration




• Explain in your own word, what is linkage?

• How do you expect the offspring of F1 generation cross of a linked allele compared to Mendelian hypothesis?

• What is a Cis and Trans allele mean?



5.3 Linkage Maps



• Crossing over happens in a way that does not disrupt a particular gene (allele).

• In 1911, Alfred Sturtevant (Thomas Hunt Morgan’s student) from Columbia University made an important hypothesis.

• The farther apart two genes are on a chromosome, the more likely they are to cross over simply because more physical distance separates them.

The frequency of crossover




• The correlation between crossover frequency and the distance between genes is used to construct linkage maps, which show the relative positions of genes in chromosomes.

• The distance is represented using “map units” called centimorgans, where 1 centimorgan equals 1 percent recombination.

• These units are used today to construct SNP maps; an estimate of genetic distance along a chromosome.






• When two genes are widely separated on a chromosome, the likelihood of a crossover is so great



• The recombinant class may approach up to 50%

• Which may appear to be the result of independent assortment (the result is 1:1:1:1 of the four types).


• Determining the degree of linkage by percent recombination.

• Lets consider the traits of Rh blood type and a form of anemia called elliptocytosis.

• An Rh + phenotype corresponds to genotypes RR or Rr.

• The anemia corresponds to genotypes EE or Ee.

Solving a Problem- Rh and anemia



• In 100 one-child families, one parent is Rh negative with no anemia (rree), and the other parent is Rh positive with anemia (RrEe), and alleles are in cis.

• Of the 100 offspring, 96 have parental genotypes ( re/re or RE/re ) and four are recombinants for these two genes ( Re/re or rE/re ).

• Percent recombination is therefore 4 percent, and the two linked genes are 4 centimorgans apart.

Solving a Problem- Rh and anemia


r r

e e

r

e

R

Evs


• Nailpatella syndrome is a rare autosomal dominant trait that causes absent or underdeveloped fingernails and toenails, and painful arthritis in the knee and elbow joints.

• The gene is 10 map units from the/ gene that determines the ABO blood type on chromosome 9.

• Geneticists determined the map distance by pooling information from many families to predict genotypes and phenotypes in offspring,.

Solving a Problem- Nail patella syndrome


Diagram adopted from Internet source.


• Greg and Susan each have nail patella syndrome. Greg has type A blood. Susan has type B blood.

• What is the chance that their child inherits normal nails and knees and type O blood?

Nail patella syndrome




• Human blood type is determined by co-dominant alleles.

• Each of us has two ABO blood type alleles, because we each inherit one blood type allele from our biological mother and one from our biological father.

The Human ABO markers: The A, B, and O alleles


Diagram adopted from Internet source.


• Greg’s mother has nail-patella syndrome and type A blood.

• His father has normal nails and type O blood.

• Therefore, Greg must have inherited the dominant nail-patella syndrome allele ( N ) and the IA allele from his mother, on the same chromosome.

Analysis for Greg


n

i

N

IA

Mother Father


• Susan’s mother has nail-patella syndrome and type O blood, and so Susan inherited N and i on the same chromosome.

• Because her father has normal nails and type B blood, her homolog from him bears alleles n and IB.

• Her alleles are in trans.

Analysis for Susan


n

IB

N

i

Mother Father


Nail patella syndrome gene mapping




• Determining the probability that Susan and Greg’s child could have normal nails and knees and type O blood.

• The only way this genotype can arise from theirs is if an ni sperm fertilizes an ni oocyte.

• 45% x 5% = 2.25% nnii genotype.

• Calculating other genotypes for their offspring is more complicated For example, a child with nail-patella syndrome and type AB blood?

Greg and Susan offspring gene mapping



• A man who has type O blood has a child with a woman who has type A blood.

• The woman’s mother has AB blood, and her father, type O.

• What is the probability that the child is of blood type:• O• A• B• AB

The Human ABO blood markers


ch5 linkage

Science

pearson education

pratheep sandrasaigarandiagram

purple flower color

purple flower factor

purple flowersself

white mendel

fromcampbell biology

white flower color