Genetic Crosses

Genetic Terms

• Somatic cells are all body cells except reproductive cells.

• Gametes: are haploid cells that are capable of fusion

• Fertilisation is the union of two gametes to form a zygote.

Fusion of gametes

Genetic Crosses terms

Scenario:

In cats, black coat (B) is dominant over white coat (b)

Genes are represented by letters

Alleles are different forms of same gene

Example: B = black coat, b = white coat

Genetic crosses terms

• Locus: position of gene on a chromosome

Genetic Crosses terms

• Dominant: this allele prevents the recessive allele from working (expressing itself).

Example: B (black) is dominant to small b (white)

• Recessive: this allele is prevented from working by dominant allele.

Example: b (white) is recessive to B (black)

Genetic Crosses Terms

• Genotype: the genetic make up of an organism (genes that are present)

Example: BB, Bb, or bb

• Phenotype: the physical make up (appearance) of an organism.

Example: BB = black, Bb = Black, bb = white

Physical make up

Genetic Crosses Terms

• Homozygous: the two alleles are the same

Example: Homozygous dominant = BB

OR

Homozygous recessive = bb

Heterzygous

• Heterozygous: means that the alleles are different.

Example: Bb

Progeny: refers to the offspring produced

In cats, black coat (B) is dominant over white coat (b).Give the phenotypes and genotypes for the offspring of a cross involving

two cats whose genotypes are (BB) and (bb).

B = Black coat b = white coat

Parental phenotypes Black x white

Parental genotypes BB bb

Gametes B b

Offspring genotypes Bb

Offspring phenotypes All black

In a fruit fly, Drosophilia, the allele for grey (G) is dominant to the allele for black body (g). If two heterozygous flies are crossed,

what is the ratio of grey bodies to black bodies among the offspring.

G = grey g = black

Parental genotypes Gg x Gg

G g

Gametes G GG Gg

g Gg gg

Progeny genotypes GG Gg Gg gg

Progeny phenotypes 3 Grey and 1 black

Incomplete dominance

• Neither allele is dominant or recessive with respect to the other.

Examples: Shorthorn cattle

RR = Red coat, rr = white coat, Rr = roan

Incomplete dominance

Example: Snap Dragons

RR = Red flowers, rr = white flowers,

Rr = pink flowers

Flower colour in snapdragons shows incomplete dominance (heterozygous condition Rr is pink).

Give the phenotypes and genotypes for the progeny of the

following:

A white flowered plant and a red flowered plant

RR = Red flower, Rr = pink flower, rr = white flower

Parents rr x RR

Gametes r x R

Offspring genotype Rr

Offspring phenotype All pink flowers

Incomplete dominance Two pink flowered plants

RR = Red flower, Rr = pink flower, rr = white flower

Parents Rr x Rr

R r

Gametes R RR Rr

r Rr rr

Offspring genotype RR, Rr, Rr, rr

Offspring phenotypes 1 red: 2 pink,: 1 white flower

Incomplete dominance in cattle

Parent phenotypes: Red male x white female

Parent genotypes: RR x rr

Gametes R x r

Offspring phenotypes Rr

Offspring genotypes All roan cattle

Pedigree studies The ability to produce the skin pigment melanin is controlled by a dominant allele (N). Lack of pigment (albinism) is controlled by recessive allele (n). Two parents that are heterozgous and have normal skin are crossed, show the offspring genotype and phenotype:

Parent genotype: Nn x Nn

N n

Gametes N NN Nn

n Nn nn

Parent genotype NN, Nn, Nn, nn

Parent phenotype 3 normal, 1 albino

• Somatic cell: all body cells except the reproductive cells.

• Autosomes: non sex chromosomes (44 in humans)

Sex Chromosomes in Humans

Female Male

Why in humans do males determine the sex of a child:

Female Male

Parent genotype XX XY

X Y

Gametes X XX XY

Female Male

I

f a sperm containing an X chromosome fertilises an egg, a girl is produced.

If a sperm with a Y chromosome ferilises an egg, a boy is produced

Gregor Mendel • Father of genetics

• He studied the garden pea plant

• He is known for his two laws:

1) Mendel’s first law

2) Mendel’s second law

Law of Segregation • Inherited characteristics are controlled by pairs of

factors which separate from each other at gamete formation where only one is found in each gamete.

NOTE: these factors are now known as alleles.

Pair of alleles Tt

meiosis

Gametes formed T t

Law of independent assortment

• When gametes are formed either of a pair factors is equally likely to combine with another pair of factors.

Genotypes of parents AaBb

Meiosis

Gametes AB Ab aB ab

Dihybrid Crosses

Demonstration of Mendel’s law of Independent Assortment A a B b

Meiosis

A B A b a B a b

In the fruit fly, the allele for grey body (G) is dominant to the allele for ebony body (g) and the allele for long wings (L) is

dominant to the allele for vestigial wings (l). These two pairs of alleles are located on different chromosome pairs.

Determine all the possible genotypes and phenotypes of the progeny of the following cross:

Grey body, long wings (heterozygous for both) x ebony body, vestigial wings

Grey body, long wings (heterozygous for both) x ebony body, vestigial wings

G = Grey g = ebony, L = long wings l = vestigial wing

Parent genotype: GgLl x ggll

gl

Gametes GL GgLl

Gl Ggll

gL ggLl

gl ggll

F1 genotype GgLl Ggll ggLl ggll

F1 phenotype Grey body, Long wings

Grey body, vestigial wing

Ebony body, Long wing

Ebony body, vestigial wing

What is the significance of the fact that the two allele pairs are located on different chromosome pairs?

• They assort independently (obey the law of independent assortment)

• It allows for more variation

In dogs, black coat (B) is dominant to white coat (b). Also short hair (S) is dominant to long hair (s)

Show the genotypes and phenotypes of the F1 progeny for a cross involving a black coated, short haired dog (heterozygous for both traits) and a white coated, long haired animal.

B = black coat b = white coat, S = short hair s = long hair

Parent genotype BbSs x bbss

bs

Gametes BS BbSs

Bs Bbss

bS bbSs

bs bbss

Offspring genotype BbSs Bbss bbSs bbss

Offspring phenotype Black coat, short hair

Black coat, long hair

White coat, short hair

White coat, long hair

Purple flower is dominant over red flower. Short stemmed plants are dominant over long

stemmed plant.

Give the expected genotype and phenotype of the F1 progeny if a purple flowered, short stemmed plant is self pollinated.

P = purple p = white, S = short s = long

Parent genotypes PpSs x PpSs

PS Ps pS ps

Gametes PS PPSS PPSs PpSS PpSs

Ps PPSs PPss PpSs Ppss pS PpSS PpSs ppSS ppSs

ps PpSs Ppss ppSs ppss

Parent phenotype 9 = Purple flowered, Short stemmed

3 = Purple flowered, long stemmed

3 = Red flowered, Short stemmed

1 = Red flowered, long stemmed

When has independent assortment occurred?

• When one parent is heterozygous for both traits (AaBb) and the second parent is homozgous recessive (aabb), the resulting offspring are produced in the ratio 1:1:1:1

• If both parents are heterozygous (AaBb x AaBb) and the offspring are in the ratio 9:3:3:1.

Summary of Dihybrid genetic crosses

Ratio Example Explantation

1:1:1:1 All offspring genotype BbLl x bbll

formed in equal numbers

9:3:3:1 Black coat, long tail = 176 BbLl x BbLl

Black coat, short tail = 62

White coat, long tail = 59

White coat, short tail = 20

In the fruit fly, the allele for full wing is dominant to the allele for vestigial wing. One parent was homozygous in respect to full wing and the other parent was heterozygous. What is the %

chance of obtaining offspring with full wing?

F = full wing f = vestigial wing

Parent genotype FF x Ff

F f

Gametes F FF Ff

F1 genotype FF, Ff

F1 phenotype All full wings (100%)

In roses there is incomplete dominance between the allele governing red petals and the allele governing white petals.

Heterozygous individuals have pink petals. A plant with pink petals was crossed with a plant with white petals. What is the %

chance of obtaining offspring with white petals:

R = Red r = white (BUT Rr = pink)

Parent genotype Rr x rr

r

Gametes R Rr

r rr

F1 genotype Rr , rr

F1 phenotype 1 pink petal, 1 white petal (50%)

In Dalmatian dogs the allele for brown spots is recessive to the allele for black spots. The two parents were heterozygous in

respect of spot colour. What is the % chance of obtaining offspring with black spots:

B = black spots b = brown spots

Parent genotype Bb x Bb

B b

Gametes B BB Bb

b Bb bb

F1 genotypes BB, Bb, Bb, bb

F1 phenotype 3 black spots, 1 brown spot

Red hair in humans is recessive to all other hair colours. A red haired woman and a black haired man, whose own father was red haired, started a family. What is the % chance of obtaining

offspring with red hair:

B = Black hair b = red hair

Parent genotype bb x Bb

B b

Gametes b Bb bb

F1 genotype Bb bb

F1 phenotype 1 black hair, 1 red hair (50%)

In guinea pigs the allele for black hair (B) is dominant to the allele for brown hair (b) and the allele for short hair (S) is dominant to the allele for long hair (s). The allele governing hair colour are located on different chromosome pair to those governing hair length.

Determine the possible genotypes and phenotypes of the offspring of a cross between the following guinea pigs:

Brown hair, heterozygous short hair x Heterozygous black hair , long hair

Brown hair, heterozygous short hair x Heterozygous black hair , long hair

B = Black hair, b = brown hair S = short hair, s = long hair

Parent genotype bbSs x Bbss

Bs bs

Gametes bS BbSs bbSs

bs Bbss bbss

Offspring Genotype BbSs bbSs Bbss bbss

Offspring phenotype Black hair, Short

Brown hair, Short

Black hair, long

Brown hair, long

In pea plants, the trait tall (T) is dominant to short (t) with regard to plant height. Yellow (Y) is dominat to green (y) with regard to seed colour.

T = tall, t = short Y = yellow, y = green

Parent genotype Tt Yy x Ttyy

Ty ty

Gametes TY TTYy TtYy

Ty Ttyy Ttyy

tY TtYy ttYy

ty Ttyy ttyy

F1 phenotypes 3 Tall and yellow, 3 tall and green

1 small and yellow, 1 small and green

In the flour beetle, black eye (P) is dominant over pearl eye (p) and brown body (B) is dominant over sooty body (b). These genes are located on different chromosomes.

P = Black eye, p = pearl body B = brown eye, b = sooty body

Parent genotype ppbb x PpBb

PB Pb pB pb

Gametes pb PpBb Ppbb ppBb ppbb

F1 Genotype PpBb, Ppbb, ppBb, ppbb

F1 phenotype 1 Black eye, brown body

1 Black eye, sooty body

1 Pearl body, brown body

1 Pearl body, sooty body

What is linkage:

• Linkage means the genes are located on the same chromosome.

A a

B b

Draw a simple chromosome diagram to illustrate the following cells and the gametes produced.

The genes are not linked and the genotype is AaBb

The genes are linked (A to B and a to b) and the genotype is AaBb

A a

B b

A a B b

Linked genes

• When genes are linked, there is only two gametes formed (AB) and (ab).

• This is a contradiction of the law of independent assortment. (contradicts Mendel’s second law).

Draw simple chromosome diagrams to show each of the following cells. In each case indicate the gametes that each cell might produce:

Genotype RrSs, the genes are not linked:

Parent Cell R r S s

meiosis

Gamete

R S R s r S r s

Genotype RrSs, the genes are linked

Genotype RrSs, the genes are linked (R to S and r to s).

Parent Cell R r

S s

Gamete R r

S s

The genes are linked, the cell is homozygous for R and heterozygous for S

Parent cell R R S s

meiosis

Gametes R S R s

The genes are linked and the cell is homozygous dominant for bothe genes

Parent cell R R

S S

meiosis

R

S

Sex Linkage

• Sex linkage means that a characteristic is controlled by a gene on the X chromosome.

X Y

X chromosome contain many genes

Y chromosome contain few genes (shorter).

Sex Linkage

• Males suffer more often from sex linked characteristics. The recessive phenotype is more likely to occur in males

Examples of Sex (X) linked characteristics are:

• Colour blindness

• Haemophilia (not able to clot blood)

• Duchenne muscular dystrophy (muscles waste away).

Colour blindness Test

Colour Blindness

Normal individuals can detect red, green and blue.

Colour blindness cannot distinguish between red and green.

N = Normal n = colour blindness

The gene for colour blindness is located on the X chromosome.

Sex Linkage

• Since males only have one X chromosome, all genes on it, whether dominant or recessive, are expressed

• In contrast, a mutant gene on an X chromosome in a female is usually covered up by the normal allele on the other X.

• Most mutations are recessive. So most people with sex-linked genetic conditions are male.

Colour Blindness in Females

The genes for colour blindness in female

X X X X X X N N N n n n

Normal vision Normal vision Colour blind (homozygous dominant) ( Heterozygous, called a carrier)

XNXN XNXn XnXn

Colour blindness in males

X Y X Y

N n

Normal vision Colour blind XNY- XnY-

Colour blindness in Males

• Males have only one allele for colour vision on the X chromosome.

• The Y chromosome has no allele for colour vision.

• A male is either colour blind or Normal, males cannot be carriers.

• Males only need one recessive allele to be colour blind.

Haemophilia

• Is a bleeding disorder.

• Caused by the lack of a certain protein.

• Suffer from frequent bleeding often in the joints.

• May bleed to death without treatment.

• More common in men than women.

Haemophilia

The gene for haemophilia is located on an X chromosome. Normal blood clotting (N) is dominant over haemophilia (n).

Show the genotypes and phenotypes of the offspring of a cross between a mother who is a carrier and a father who is normal for this trait.

N = Normal n = haemophilia

Parent genotype XNXn x XNY-

XN Y-

Gametes XN XNXN XNY-

Xn XNXn XnY- Offspring genotype XNXN, XNY-, XNXn, XnY-

Offspring phenotype Female, normal clotting

Female, normal clotting (carrier)

Male, normal clotting

Male, haemophilia

Haemophilia in humans is governed by a sex linked allele. The allele for normal blood clotting (N) is dominant to the allele for

haemophilia (n):

Determine the possible genotypes and phenotypes of the progeny of the following cross:

Haemophilic male x heterozygous normal female

N = Normal n = haemophila

Parent genotypes XnY- x XNXn

XN Xn

Gametes Xn XNXn XnXn

Y- XNY- XnY-

Offspring genotypes XNXn, XnXn, XNY-, XnY-

Offspring phenotypes Female, normal clotting (carrier)

Female, haemophilia

Male, normal

Male, haemophilia

Exampaper 2012 Q10 (b)

S = smooth s = wrinkled Y = yellow y = green

(ii)

1) Give the genotype of a pea plant who is homozygous in respect to seed texture and heterozygous in respect to seed colour.

SSYy OR ssYy

2) State the its phenotype:

Smooth + Yellow OR wrinkled + yellow

Exampaper 2012 Q 10b

(iii)

What phenotype will be produced from SsYy:

Smooth + yellow

Give another genotype that will produce the same phenotype.

SSYY

Exampaper 2012 Q10 b

(iv) If the alllele for smooth were linked to green and the alllele for wrinkled were linked to yellow. Give the two gametes that the parent SsYy would produce:

Sy and sY