# Genetics Test Questions

Topics: Blood type, Allele, Genetics Pages: 5 (1453 words) Published: December 12, 2012
http://faculty.clintoncc.suny.edu/faculty/michael.gregory/files/bio%20100/bio%20100%20lectures/genetics-genes/mendelia.htm Challenging Crosses (punett squares) Questions?

1. What are the expected offspring produced by a cross between a heterozygous black short-haired guinea pig and a homozygous white, long-haired guinea pig. Assume Black color (B) and short-hair (S) are dominant traits.

2. The ability to roll the tongue into almost a complete circle is conferred by a dominant trait, while its recessive allele fails to confer this ability. A man and his wife can both roll their tongues and are surprised to find that their son cannot! Explain this by showing the genotypes of all three persons.

3. In rabbits, spotted coat (S) is dominant to solid, and black (B) is dominant to brown. In a large population, brown spotted rabbits are mated to solid black ones and all the offspring are black spotted. What are the genotypes of the parents? What would be the appearance of the F2 if two of these F1 black spotted rabbits were mated?

4. In pea plants, tall plants (T) are dominant to dwarf, and yellow color (Y) is dominant to green, and smooth seed (S) is dominant to wrinkled seed. What would be the phenotypes of the following matings?

TtYySs x ttyyss

1. This can be solved using probability, but you may find it easier to draw a Punnett square. The parental genotypes are BbSs and bbss. The way I would solve this is to draw two separate Punnett squares, one for each trait (B and S). The B square will show you that half the offspring will be black and half will be white. The same is true for the S square (half long hair, half short hair). Multiply these probabilities to get your offspring ratios: 0.5 x 0.5= 0.25. There will be 0.25 with short black fur, 0.25 with long black fur, 0.25 with long white fur and 0.25 with short white fur. This is a 1:1:1:1 ratio. 2. The parents would have the genotypes Aa and Aa. The son would have the recessive genotype aa. 3. The genotypes must have been SSbb and ssBB. All of the F1 offspring would be SsBb. Crossing the F1s will give you a 9:3:3:1 phenotypic ratio. You can draw a Punnett square to work out the genotypes. 4. This is a trihybrid testcross, so you can expect a 1:1:1:1:1:1:1:1 phenotypic ratio. In a testcross, you can basically ignore the pure-breeding parent. The first parent will produce the following gametes: TYS, tYS, TyS, TYs, Tys, tYs, tys and tyS.

Codominance and Multiple Alleles- Example: ABO blood group
Up to this point, we have discussed two possible alleles for any gene locus. For example, at the flower color locus, there is either the red or the white allele (A or a). With human blood types, there are three alleles: A, B, or O. This is referred to as multiple alleles. I is dominant to i.

There are two forms of I: IA and IB but only one form of i.
6 possible genotypes, 4 phenotypes: Multiple alleles:- more than two alternative forms of a gene -normally, each locus (the place on the chromosome where a gene for a particular trait is located) contains only two contrasting forms of one trait - example is human blood types. There are three allele possibilities (A, B and O) that produce 4 blood types (A, B, AB and O) - The IA allele codes for the production of the A antigen (a glycoprotein found on the surface of RBC’s) - The IB allele codes for the production of the B antigen

- The i allele codes for no antigen production on the RBC (neither A nor B) - Note that IA and IB are co dominant (both traits are expressed in a heterozygous organism {Type AB blood}) - IA and IB are dominant traits, i is a recessive trait

Blood Type| PossibleGenotypes| Antigens on the RBC| Antibodies in the Blood Serum (Plasma)| A| IAIAIAi| A| Anti-B|
B| IBIBIBi| B| Anti-A|
AB| IAIB| A and B| none|
O| ii| None| Anti-AAnti-B|
IAIA and IAi = blood type A
IBIB and IBi = blood type B
IAIB = blood type AB
i i = blood type O
People with blood type A have a...