1. Dihybrid cross DIHYBRID CROSS

2. To this point we have followed the expression of only one gene
To this point we have followed the expression of only one gene. Mendel also performed crosses in which he followed the segregation of two genes.
These experiments formed the basis of his second law, the law of independent assortment.

3. Dihybrid cross
In general, it is a cross following the simultaneous expression of TWO genes (traits)
specifically, a cross between two individuals heterozygous for both traits (AaBb x AaBb)

4. example:
You have become rabbit breeder. From your extensive research on rabbit genetics you know that in rabbits, the allele for short hair is dominant over the allele for long hair. Also, black fur is dominant to brown fur. One horrible morning you discover that Bugs, your prized short-haired, black rabbit, (heterozygous for both traits) has escaped! Luckily he has not gone far. You find him with Jessica, another short-haired, black rabbit (also heterozygous for both traits).

5. As the days pass you discover that Jessica is pregnant
As the days pass you discover that Jessica is pregnant. You are very excited to know what the babies will look like. You decide that you can't wait for them to be born. Instead, you apply your knowledge of Mendelian genetics to determine the genotypic and phenotypic ratios of the baby rabbits.

6. Key t Key to solving dihybrid genetics problems using a Punnett Square:
 1. READ THE PROBLEM. You are being asked to find the genotypic and phenotypic ratios in the litter that would result from a cross between Bugs and Jessica.

7. Let B represent the dominant allele for black fur.
2. Write the INFORMATION GIVEN TO WRITE A KEY.
Define the letters to be used.
Let B represent the dominant allele for black fur.
Let b represent the recessive allele for brown fur
Let H represent the dominant allele for short hair.
Let h represent the recessive allele for long hair.

8. 3. Use your key to determine the genotypes of the parents you want to cross.
Bugs is heterozygous for short hair (Hh) and heterozygous for black fur (Bb), thus his
genotype is HhBb.
Jessica is also heterozygous for short hair (Hh) and heterozygous for black fur (Bb), hence her
genotype is also HhBb.
           

9. 4. WRITE OUT THE CROSS.
BbHh x BbHh

10. Gametes: BH, bH, Bh, bh x BH, bH, Bh, bh
  5. DETERMINE THE GAMETES OF THE PARENTS.
Remember that gametes are haploid. This means that each gamete must have HALF of the genetic information of the parents (ie. only one allele for each trait).
Gametes: BH, bH, Bh, bh x BH, bH, Bh, bh
Notice that there are four different combinations of alleles for each parent and therefore 4 different gametes for each parent.

11. SQUARE. 6. DRAW AND SOLVE THE PUNNETT
Since there are four different gametes for each parent, the Punnett square will be 4 x 4 Punnett Square:

12. BbHH bbHh bbHH BbHh BBHh BbHh BBhh Bbhh BbHh Bbhh bbhh bbHh BbHH BBHh

13. Make sure you actually answer the question.
How many different genotypes are possible?
How many different phenotypes are possible?

14. Find genotype and Phenotype ratios
GR PR
BBHH 1/ Black with short hair
BBHh 2/ Black with short hair
BBhh 1/ Black with long hair
BbHH 2/ Black with short hair
BbHh 4/ Black with short hair
Bbhh 2/ Black with long hair
bbHH 1/ Brown with short hair
bbHh 2/ Brown with short hair
bbhh 1/ Brown with long hair

15. Genotypic Ratios Phenotypic Ratios
1 BBHH
2 BbHH
2 BBHh
4 BbHh
9 black, short-haired
2 Bbhh
1 BBhh
3 black, long-haired
2 bbHh
1 bbHH
3 brown, short-haired
1 bbhh
1 brown, long-haired

16. What is the probability that Bugs and Jessica could have a brown, long-haired offspring?
Looking at the previous Punnett square, the chance of getting bbhh is one in sixteen.
We could have approached the problem by following the two traits independently (as two separate monohybrid crosses), then calculating the probability of them occurring at the same time

17. There is a 1 in 4 chance of getting bb (brown)
There is a 1 in 4 chance of getting hh (long)
So the chance of these two independent events occurring simultaneously (bbhh) is:

18. Mendel's Second Law
The law of independent assortment: Because the genes are on separate chromosomes, the segregation of the alleles of one gene is independent of the segregation of the alleles of the other gene. Recall that during metaphase I, a pair of homologous chromosomes can align two different ways at the equator. (see page 217)

19. http://www. sumanasinc
mations/content/independentassortment.ht ml
videos/2011/9/14/029-mendelian- genetics.html

20. Linked Genes
Genes that are on the same chromosome are inherited together.
Linked genes do not follow Mendel’s second law
See pages
nt/chp10/ html

21. HOMEWORK Page 216 Practice Problems #11 to 20 Page 218 #6 to 12
Check answers at back of textbook