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TWO TRAIT CROSSES
Mendel also studied two separate traits with a single cross. Ie.
Crossed pure breeding plants having round, yellow seeds with
pure breeding wrinkled, green seeded plants.
ROUND (R) IS DOMINANT TO WRINKLED (r) AND
YELLOW (Y) IS DOMINANT TO GREEN (y)
PURE BRED ROUND = RR
PURE BRED YELLOW = YY
PURE BRED WRINKLE = rr
yy
PURE BRED GREEN =
GENOTYPE FOR ROUND YELLOW = RRYY
GENOTYPE FOR WRINKLE GREEN = rryy
TWO TRAIT CROSSES
MENDEL CROSSED ROUND YELLOW x WRINKLE GREEN
RRYY
rryy
WHAT ARE THE POSSIBLE GAMETES FROM RRYY?
WHAT ARE THE POSSIBLE GAMETES FROM rryy?
RY
ry
REMINDER:
1. GAMETES HAVE HALF THE NUMBER OF CHROMOSOMES
(GENES) AS THE ORIGINAL BECAUSE HOMOLOGS
SEGREGATE DURING ANAPHASE I.
2. EACH PARENT CELL HAS 2 COPIES OF EACH OF THE 2
GENES FOR A TOTAL OF 4 ALLELES. GAMETES HAVE ONLY
ONE COPY OF EACH GENE FOR A TOTAL OF 2.
RY
ry
RrYy
TWO TRAIT CROSSES
MENDEL CROSSED ROUND YELLOW x WRINKLE GREEN
RRYY
rryy
WHAT ARE THE POSSIBLE GAMETES FROM RRYY?
WHAT ARE THE POSSIBLE GAMETES FROM rryy?
F1 = ALL RrYy : ALL ROUND YELLOW
WHAT ARE THE POSSIBLE GAMETES FROM RrYy?
(F.O.I.L.)
RY
Ry
rY
ry
RY
Ry
rYry
RrYy
RY
ry
TWO TRAIT CROSSES
CROSS OF RrYy x RrYy  POSSIBLE GAMETES = RY, Ry, rY, ry
PUNNETT SQUARE: ALL GAMETES ACROSS TOP; DOWN LEFT
RY
RY
RRYY
Ry
RRYy
rY
RrYY
ry
RrYy
GENO. RATIO:
1:2:1:2:4:2:1:2:1
Ry
rY
ry
RRYy
RRyy
RrYy
Rryy
RrYY
RrYy
rrYY
rrYy
RrYy
Rryy
rrYy
rryy
PHENO. RATIO:
9:3:3:1
Round Yellow:
Round Green:
Wrinkle Yellow:
Wrinkle Green
Rr x Rr
GR:
PR:
Yy x Yy
RrYy x RrYy
In humans, free ear lobes are determined by the dominant allele
E, and attached ear lobes by the recessive allele e. The dominant
allele W determines a widow’s peak hairline and the recessive
allele w determines a straight hairline. The genes for these two
traits are located on different chromosomes. Suppose a man with
the genotype EeWw and a woman with the genotype EeWw are
expecting a child. What is the probability that the child will have
a straight hairline and attached ear lobes?
a
b
c
d
Suppose a man with the genotype EeWw and a woman with the
genotype EeWw are expecting a child. What is the probability that
the child will have a straight hairline and attached ear lobes?
EW
Ew
eW
ew
EeWw x EeWw
eW
ew
EW
Ew
EEWW
EEWw
EeWW
EeWw
HOW MANY
STRAIGHT
ATTACHED?
EEWw
EEww
EeWw
Eeww
1/16 = 6%
EeWW
EeWw
eeWW
eeWw
EeWw
Eeww
eeWw
eeww
Suppose a man with the genotype EeWw and a woman with the
genotype EeWw are expecting a child. What is the probability that
the child will have a straight hairline and attached ear lobes?
EeWw x EeWw
MATHEMATICALLY:
Chances of Attached?
¼
Chances of Straight?
¼
Chances of Straight Attached?
¼ x ¼ = 1/16 = 6%
HOW MANY
STRAIGHT
ATTACHED?
1/16 = 6%
Calculate the probability that the couple will have a child with
(a) a widow’s peak and free ear lobes
¾ x ¾ = 9/16
¼ x ¾ = 3/16
(b) a straight hairline and free ear lobes
¾ x ¼ = 3/16
(c) a widow’s peak and attached ear lobes
GENE INTERACTION
• Genetic traits in organisms can be discrete
(aka. discontinuous) OR continuous
Complications & Gene Interactions
• PLEIOTROPIC – condition when single
gene simultaneously affects many traits….(not
really gene interactions) … one gene , many traits
• POLYGENIC – phenotypic expression of
trait depends on additive effect of a number
of genes … many gene , one traits
– like eye colour, skin color and height!!
Gene Interactions
• EPISTATIC – nonhomologous genes that
interfere with the expression of others
– STOP…go to p. 617 for an example!!
• COMPLIMENTARY – when 2 genotypes
together give a new phenotype that neither
alone produces
GENE INTERACTION
Some (most) traits are controlled by more than one gene (at the
same time). Ex. Skin colour, eye colour, height =>
POLYGENIC
EX. DOG COAT COLOUR: BLACK (B) > BROWN (b) BUT…
A SECOND GENE (W) MASKS ANY OTHER COLOUR. THE
BROWN OR BLACK IS ONLY EXPRESSED IF THE WHITE
GENE IS ww.
WWBB = WHITE
WwBb = WHITE
wwBB =
wwBb =
BLACK
BLACK
Wwbb =
wwbb =
BROWN
WHITE
Genes that interfere in the expression of other genes are called =>
EPISTATIC
GENE INTERACTION
EX. DOG COAT COLOUR: BLACK (B) > BROWN (b) BUT… A
SECOND GENE (W) MASKS ANY OTHER COLOUR. THE
BROWN OR BLACK IS ONLY EXPRESSED IF THE WHITE GENE
IS ww.
WHAT IS THE PHENO. RATIO OF CROSSING WwBb x wwBb?
GAMETES FROM WwBb: WB, Wb, wB, wb
GAMETES FROM wwBb: wB, wb
WB
wB
WwBB
Wb
WwBb
wB
wwBB
wb
wwBb
GR.: 1:2:1:1:2:1
PR.: 4:3:1
wb WwBb
Wwbb
wwBb
wwbb
WHITE:BLACK:BROWN
GENE INTERACTION
Other genes can have COMPLIMENTARY
INTERACTION Where the combined phenotype is
different than either of the individual ones.
EX. In chicken: the R Allele makes a ROSE comb. The P ALLELE
(On a different chromosome) causes a PEA comb. R and P
together make a WALNUT comb. the absence of both makes a
SINGLE comb
ROSE
PEA
SINGLE
WALNUT
GENE INTERACTION
P=
RRpp
ROSE
F1 =
F2 =
rrpp
x
rrPP
PEA
ALL RrPp
WALNUT
9
: 3
: 3 : 1
R _ P _ : R _ pp : rrP _ :
WALNUT : ROSE : PEA : SINGLE
GENE INTERACTION
Chemical reactions are catalysed by enzymes which are
proteins coded for by genes. The production of cyanide in
clover takes place through 2 chemical steps, each with an
enzyme
E1
E2
GG or Gg
HH or Hh
CYANIDE
If either of the enzymes is absent, cyanide won’t be
produced.
GGhh
NO CN
x
ggHH
NO CN
GgHh
CYANIDE
GENE INTERACTION
Some single genes can cause many different effects in the body
(many “phenotypes”).These genes are called
PLEIOTROPIC.
EX. Marfan syndrome
EX. Sickle cell anemia => normal hemoglobin carries oxygen on
red blood cells (HAHA)
A point mutation (HS) Causes mis-shaped hemoglobin = sickle
cell. Because of this, hemoglobin can’t carry oxygen very well.
Many effects come from the homozygous condition (HSHS):
- Fatigue, weakness, enlarged spleen, rhumatism,
pneumonia, heart/kidney/lung damage, stroke.
Interestingly, sickle cell gives resistence to malaria. People who are
HAHS have an advantage.
EFFECTS OF THE
ENVIRONMENT ON PHENOTYPE
The environment can cause genes to express themselves in
different ways.
Ex. Freckles – only expressed after sun exposure
- Some water plants grow different shaped leaves under water than
above water.
PAGE 619, 18.5
QUESTIONS #1-4
Page 619, #1-4
1. GUINEA PIGS:
BLACK (B) > WHITE (b);
SHORT (S) > LONG HAIR (s)
GIVE GENOTYPES AND PHENOTYPES:
a) HOMO. BLACK/HETERO. SHORT x WHITE/LONG GENOTYPE?
BBSs
x
bbss
POSSIBLE GAMETES? FROM BBSs = BS or Bs
POSSIBLE GAMETES? FROM bbss = ONLY bs
PUNNETT SQUARE:
BS
bs
BbSs
Bs
Bbss
GR.: 1:1
PR.: 1:1
(BLACK/SHORT:
BLACK/LONG)
Page 619, #1-4
1. GUINEA PIGS:
BLACK (B) > WHITE (b);
SHORT (S) > LONG HAIR (s)
GIVE GENOTYPES AND PHENOTYPES:
b) HET. BLACK/HET. SHORT x WHITE/LONG GENOTYPE?
BbSs
x
bbss
POSSIBLE GAMETES? FROM BbSs = BS, Bs, bS, bs
POSSIBLE GAMETES? FROM bbss = ONLY bs
PUNNETT SQUARE:
BS
bs
`BbSs
GR.: 1:1:1:1
Bs
bS
bs
Bbss
bbSs
bbss
PR.: 1:1:1:1
(BL./SH.: BL./L.:
WH./SH.:WH./L.)
Page 619, #1-4
1. GUINEA PIGS:
BLACK (B) > WHITE (b);
SHORT (S) > LONG HAIR (s)
GIVE GENOTYPES AND PHENOTYPES:
c) HOMO. BLACK/LONG x HET. BLACK/HET. SHRT GENOTYPE?
BBss
x
BbSs
POSSIBLE GAMETES? FROM BBss = Bs
POSSIBLE GAMETES? FROM BbSs = BS, Bs, bS, bs
PUNNETT SQUARE:
BS
Bs
`BBSs
Bs
bS
bs
GR.: 1:1:1:1
BBss
BbSs
Bbss
PR.: 1:1
(BL./SH.: BL./L.)
Page 619, #1-4
2. COCKER SPANIELS: BLACK (B) > WHITE (b);
SOLID (S) > SPOTTED (s)
GIVE GENOTYPES:
BLACK/SOLID x
B __
b S __
s
a) WHITE/SOLID  2 BL/SOL: 2 WH/SOL
bb S __
B b S __
bb S __
b) BLACK/SOL
B __
b S __
s
 1 WH/SPOT
bbss
c) WHITE/SPOT
 1 WH/SOL:1 WH/SPOT:
1 BL/SOL:1 BL/SPOT
bbSs: bbss:BbSs:Bbss
bbss
GENOTYPES?
Page 619, #1-4
3. HUMANS: (A BLOOD (IA) = B BLOOD (IB))> O BLOOD (i)
POSITIVE (R) > NEGATIVE (r)
GIVE POSSIBLE PHENOTYPES FROM O, Rh- x A, Rh+:
WHAT ARE THE POSSIBLE GENOTYPES?
O, Rh- = iirr
x A, Rh+ = IAIARR, IAiRR, IAIARr, or
IAiRr
POSSIBLE GAMETES? FROM iirr =
POSSIBLE GAMETES? FROM IAiRr =
PUNNETT SQUARE:
ir
IAR
IAr
iR
`IAiRr
IAirr
iiRr
ir
iirr
ONLY ir
IAR, IAr, iR, ir
GR.: 1:1:1:1
PR.: 1:1:1:1
(A+,A-,O+,O-)
Page 619, #1-4
4. A trait that is controlled by more than one gene pair could have
many different possible phenotypes. One gene pair could produce
smaller numbers of phenotypes.
1 pair (only 2 alleles): Dominant/Recessive = 2 traits
1 pair (only 2 alleles): Blending = 3 traits
2 pair (only 2 alleles each): Dominant/Recessive = 4 traits
2 pair (only 2 alleles each): Blending = 9 traits ??
etc.