Transcript Monohybrid Crosses

Genotypes & Phenotypes
• Capital letters used for dominant allele
• lowercase of the same letter used for recessive
• ex. round seeds are dominant over wrinkled.
• R = round, r = wrinkled
Genotype
• the alleles for an individual, ex. for seeds:
- RR (homozygous round)
- rr (homozygous wrinkled)
- Rr (heterozygous round)
Phenotype
• the expressed trait for an individual (what is seen) ex. for seeds:
- round
- wrinkled
• Heterozygous genotypes (Rr) always have the phenotype of the
dominant allele.
Plant Height Experiments
• F1 genotype:
- 4 Tt (heterozygous)
• F1 phenotype:
- 4 tall
• F2 genotypes:
- 1 TT (homozygous dominant)
- 2 Tt (heterozygous)
- 1 tt (homozygous recessive)
- ratio 1:2:1
P Generation
True-breeding parents
• F2 phenotypes:
Tall = dominant = T
Short = recessive = t
- 3 tall
- 1 short
F1 Generation
- ratio 3:1
Hybrid offspring
All heterozygous
Tall
TT
short
tt
all Tall
all Tt
F2 Generation
Genotypes
Monohybrid cross
from F1 generation
1 TT : 2 Tt : 1 tt
Phenotypes
Tall
TT
Tall
Tt
Tall
Tt
short
tt
3 Tall :
1 short
Probability & Punnet Squares
Probability – the likelihood of an outcome
Punnett square - used to determine the probabilities of allele
combinations when the genotypes of the parents are known
Making Punnet Squares
Practice Problem #1
• A geneticist crosses two pea plants. One of the plants is heterozygous
for the dominant inflated pea pod trait, and the other plant has
constricted pea pods. What would be the expected genotypic and
phenotypic proportions of the offspring?
Given
• inflated pod (I) - dominant
• constricted pod (i) - recessive
• II or Ii = Inflated pods
• ii = constricted pods
• heterozygous inflated x constricted = Ii x ii
Required
• Punnet square
• gametes from the cross
• expected proportions of F1 plants with genotypes II, Ii, and ii
• expected proportions of phenotypes (inflated pods and constricted)
Solution
• The Ii parent produces
gametes I and i.
• The ii parent produces
gametes i and i.
Paraphrase
1/2 of the offspring would be
heterozygous (Ii) and 1/2 would be ii.
Therefore, 1/2 of the offspring would have
inflated pods, and 1/2 would have
constricted pods.
I
i
i
Ii
ii
i
Ii
ii
Solutions #2- 4
A
a
2. Solution
• The Aa parent produces gametes A and a.
a Aa
aa
• The aa parent produces gametes a and a.
a Aa
aa
Paraphrase
• ½ the offspring would be heterozygous (Aa) and ½ would
be aa. Therefore, ½ of the offspring would be brown, and ½
would be albino
3. Solution
D
D
• The DD parent produces gametes D and D.
D DD
DD
• The Dd parent produces gametes D and d.
d Dd
Dd
Paraphrase
• ½ of the offspring would be heterozygous (Aa) and ½ would
be homozygous (AA). Therefore, all of the offspring would
have dimples.
4. Solution
T
t
• The Tt parents produce gametes T and t.
T TT Tt
Paraphrase
• ½ of the offspring would be heterozygous (Tt), ¼ would be
t Tt
tt
homozygous dominant (TT) and ¼ would be homozygous
recessive (tt). Therefore, ¾ of the offspring would taste PTC,
and ¼ would not.
Test Crosses
• how genotype can be determined from phenotype
• breed an individual with the dominant phenotype with a
homozygous recessive
• If offspring are all dominant
phenotype, the parent was
homozygous dominant
• If offspring are half dominant
phenotype, and half recessive
phenotype, the parent was
heterozygous dominant.
Dihybrid Crosses
• a cross between two individuals that differ in two traits
• Mendel used to determine whether or not the inheritance of
one characteristic influenced the inheritance of another
• ex: Does pea shape influence the inheritance of pea colour?
Method:
• choose plants homozygous for 2 traits (colour & shape)
• crossbred 2 different ones to get dihybrid F1 generation
• ex. round yellow seeds x wrinkled green
• crossed two dihybrid F1 to get F2 (dihybrid cross)
Results:
• all F1 were round yellow
• F2 had phenotypic ratio of 9:3:3:1
P Generation
F1 Generation
F2 Generation
Dihybrid Crosses
1. State allele symbols
2. Follow steps below
Step
Example
Flower location:
A = Axial
a = terminal
Height:
T = Tall
t = short
1. Draw a square
with a 4 by 4
grid.
2. Consider all
possible
gametes
produced by
the first
parent. (Use
FOIL) Write
the alleles for
these gametes
across the top
of the square.
3. Consider all
possible
gametes
produced by
the second
parent. Write
the alleles for
these gametes
down the side
of the square.
4. Complete the
square by
writing all
possible allele
combinations
from the
cross.
5. Determine the
genotypic and
phenotypic
proportions of
the offspring.
Practice Problem #1
• In mice, the normal long-tail phenotype is dominant to the short-tail
trait, and black coat colour is dominant to brown coat colour. If two
long-tailed black mice, heterozygous for both traits, are mated, what
proportion of their offspring will be brown with short tails?
Given
• Long tail (L) is dominant. Short tail (l) is recessive.
•
LL or Ll - long tails, ll - short tails.
• Black coat colour (B) is dominant. Brown coat colour (b) is recessive.
•
BB or Bb - black coat colour, bb - brown coat colour.
• The cross is BbLl X BbLl
Required
• Determine the gametes using FOIL.
• make a Punnett square
• find the proportion of offspring that will be brown with short tails
Solution
• The BbLl parents produce gametes BL, Bl, bL, bl.
BL
Bl
bL
bl
BL
BBLL
BBLl
BbLL
BbLl
Bl
BBLl
BBll
BbLl
Bbll
bL
BbLL
BbLl
bbLL
bbLl
bl
BbLl
Bbll
bbLl
bbll
Paraphrase
Mice with brown coats and short tails have the genotype
bbll. 1/16 of the offspring will be brown with short tails.
Law of Independent Assortment
• Mendel found that this phenotypic ratio (9:3:3:1) was
always found in the offspring of such dihybrid crosses.
• law states that the inheritance of alleles for one trait
does not affect the inheritance of alleles for another
trait.
• Different pairs of alleles are passed to the offspring
independently of each other
• laws apply to all living things, not just plants
• many, but not all, traits follow Mendel’s laws