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Dihybrid Crosses
Dihybrid Crosses• Dihybrid Cross ~ genetic cross
considering 2 gene traits at the same time, each consisting of nonidentical alleles.
• i.e. Mendel crossed numerous traits repeatedly in dihybrid crosses and found he always obtained the same ratio 9:3:3:1 ~ and thus develops Law of Independent Assortment
•Law of Independent Assortment
• During gamete formation, the two alleles for one gene segregate or assort independently of the alleles for other genes.
Diploid Cells
Gamete Cells
Law of Independent Assortment
Diploid Cells
Gamete Cells
Law of Independent Assortment
Diploid Cells
Gamete Cells
Law of Independent Assortment
Diploid Cells
Gamete Cells
Or.....
Law of Independent Assortment
Diploid Cells
Gamete Cells
Or.....
Law of Independent Assortment
Diploid Cells
Gamete Cells
Or.....Or.....
Law of Independent Assortment
Diploid Cells
Gamete Cells
Or.....Or.....
etc....
Law of Independent Assortment
Diploid Cells
Gamete Cells
So....
Law of Independent Assortment
Dad
Diploid Cells
Gamete Cells
Law of Independent Assortment
Dad Mom
• In humans, free earlobes are controlled by the dominant allele E and attached earlobes by the recessive allele e. The widow’s peak hairline is regulated by the dominant allele H, while the straight hairline by the recessive allele h.
• What would be the genotype and phenotype ratio of the F1 generation if a man heterozygous for both free earlobes and widow’s peak has children with a woman also heterozygous for both free earlobes and widow’s peak?
Sample Problem #1
• 1. Let “E” be dominant free earlobes let “e” be recessive attached earlobes let “H” be dominant widow’s peak let “h” recessive straight hair line
• 2. Parents P1 cross: EeHh x EeHh
• 3. Alleles: EH, Eh, eH, eh & EH, Eh, eH, eh How did
you do that?
EeHh x EeHhEeHhEeHhEeHhEeHh EeHhEeHhEeHhEeHh
Step 3:
EeHh x EeHhEeHhEeHhEeHh
EH
EeHhEeHhEeHhEeHh
Step 3:
EeHh x EeHhEeHhEeHh
EhEH
EeHhEeHhEeHhEeHh
Step 3:
EeHh x EeHhEeHh
eHEhEH
EeHhEeHhEeHhEeHh
Step 3:
EeHh x EeHh
eheHEhEH
EeHhEeHhEeHhEeHh
Step 3:
EeHh x EeHh
eheHEhEH
EeHhEeHhEeHh
EH
Step 3:
EeHh x EeHh
eheHEhEH
EeHhEeHh
EhEH
Step 3:
EeHh x EeHh
eheHEhEH
EeHh
eHEhEH
Step 3:
EeHh x EeHh
eheHEhEH eheHEhEH
Step 3:
EeHh x EeHh
eheHEhEH eheHEhEHx
P1 Cross:
Alleles for Punnet Square:
This is how you do step #3!
Step 3:
EH Eh eH eh
EH
eH
Eh
eh
EEHH EEHh EeHH EeHh
EeHH EeHh eeHH eeHh
EEHh EEhh EeHh Eehh
EeHh Eehh eeHh eehh
The Alleles of each parent is placed on the outside of your
4x4 punnet square
• EEHH ~ 1 eeHh ~ 2
• EEHh ~ 2 EEhh ~ 1
• EeHH ~ 2 Eehh ~ 2
• EeHh ~ 4 eehh ~ 1
• eeHH ~ 1
• Therefore the genotype ratio is 1:2:2:4:1:2:1:2:1 for above order
• Free earlobes & widow’s peak = 1 +2+2+4 = 9
• Attached earlobes & widow’s peak = 1+2 =3
• Free earlobes and straight hairline = 1+2= 3
• Attached earlobes and straight hairline = 1
• thus, the ratio of phenotypes is 9:3:3:1
• 1. State The phenotypes and genotype probability for a cross between a pure plant with round yellow seed and pure plant with wrinkled green seed. Round is dominant over wrinkled and yellow is dominant over green.
Practice Crosses
• Let “R” be dominant round trait let “r” be recessive wrinkled trait let “C” be dominant yellow trait let “c” be recessive green trait
• Cross P1: RRCC x rrcc
• Alleles: RC, RC, RC, RC x rc, rc, rc, rc
RC RC RC RC
rc RrCc RrCc RrCc RrCc
rc RrCc RrCc RrCc RrCc
rc RrCc RrCc RrCc RrCc
rc RrCc RrCc RrCc RrCc
•RrCc - 16
•Therefore, 100% of P1 are RrCc and will be round and yellow.
Pedigree Chart ~ a graphic presentation of a family tree that permits patterns of inheritance to be followed for a single gene.
chart contains a number of symbols that identify gender, relationships between individuals and whether an individual expresses a trait or carries the
Pedigree Charts
142 Chapter 4
I
II
male
female
mating
Roman numeralssymbolizegenerations.
Arabic numbers symbolizeindividuals within a givengeneration.
Birth order within each group of offspring is drawn left to right,oldest to youngest.
identicaltwins
fraternaltwins (females)
affectedindividuals
knownheterozygotesfor autosomalrecessive
31 2
Pedigree Symbols
Sample Pedigree
I
II
III
2
1 2 3
1
54
2 31
Figure 4Squares represent males and circles representfemales. Individuals that express the trait areshown in a coloured circle or square. If it isknown, individuals that carry the allele as partof a heterozygous genotype are shown by par-tial colour or shading. A vertical line connectsparents to offspring.
From the point of view of individual III - 1, the symbols represent the fol-lowing relationships:
I - 1 = grandfather I - 2 = grandmotherII - 1 and II - 2 = aunts II - 3 = uncle II - 4 = father II - 5 = motherIII - 2 = fraternal twin sister III - 3 = brother
Practice
Understanding Concepts1. Copy Figure 5 into your notebook. Indicate whether each family
member is homozygous or heterozygous for shortsightedness, orhomozygous for normal vision.
2. If couple 4 and 5 in row II had another child, what genotype mightthe child have? (Hint: What genotype is possible but not shown inthe chart?) Would the child have normal vision or be shortsighted?
142 Chapter 4
I
II
male
female
mating
Roman numeralssymbolizegenerations.
Arabic numbers symbolizeindividuals within a givengeneration.
Birth order within each group of offspring is drawn left to right,oldest to youngest.
identicaltwins
fraternaltwins (females)
affectedindividuals
knownheterozygotesfor autosomalrecessive
31 2
Pedigree Symbols
Sample Pedigree
I
II
III
2
1 2 3
1
54
2 31
Figure 4Squares represent males and circles representfemales. Individuals that express the trait areshown in a coloured circle or square. If it isknown, individuals that carry the allele as partof a heterozygous genotype are shown by par-tial colour or shading. A vertical line connectsparents to offspring.
From the point of view of individual III - 1, the symbols represent the fol-lowing relationships:
I - 1 = grandfather I - 2 = grandmotherII - 1 and II - 2 = aunts II - 3 = uncle II - 4 = father II - 5 = motherIII - 2 = fraternal twin sister III - 3 = brother
Practice
Understanding Concepts1. Copy Figure 5 into your notebook. Indicate whether each family
member is homozygous or heterozygous for shortsightedness, orhomozygous for normal vision.
2. If couple 4 and 5 in row II had another child, what genotype mightthe child have? (Hint: What genotype is possible but not shown inthe chart?) Would the child have normal vision or be shortsighted?
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