Genetic Problem – Assignment Example
1. a) Gametes produced from AABB: AB b) Gametes produced from AaBB: AB and aB c) Gametes produced from Aabb: Ab and ab d) Gametes produced from AaBb: AB, Ab, aB, and ab
2. a) AABB produces AB, aaBB produces aB
Offspring Produced: AaBB
b) AaBB produces AB, aB and AABb produces AB, Ab
Offspring produced: AABB, AABb, AaBB, and AaBb
c) AaBb produces AB, Ab, aB, ab and aabb produces ab
Offspring produced: AaBb, Aabb, aaBb, and aabb
d) AaBb produces AB, Ab, aB, ab
Offspring produced: AABB, AABb, AAbb, AaBB, AaBb, Aabb, aaBB, aaBb, and aabb
3. Because the father has pointed eyebrows (in which pointed is dominant) and the daughter has smooth eyebrows (which is recessive). The father must be heterozygous. He has both the pointed eyebrow and smooth eyebrow alleles (Pp, where P = pointed eyebrows and p = smooth eyebrows).
4. For bb x Bb (B = brown eyes, b = blue eyes), 1/2 of the children will have blue eyes (bb). The children will all get the b allele from their father. Half will get the B allele from their mother and the other half will get the b allele from their mother.
5. cross genotypic ratio phenotypic ratio
P: AA x aa
F1: Aa 1 0 will be phenotypically recessive
- All offspring will get A from one parent and a from the other parent
F2: AA:Aa:aa 1:2:1 1:3 will be phenotypically recessive
(aa : AA and Aa)
- AA: (1/2*1/2)=1/4, Aa: (1/2*1/2)+(1/2*1/2)=1/2, aa: (1/2*1/2)=1/4
- 1/4:1/2:1/4 = 1:2:1
6. The children have blood types A, B, and AB. In order to get children with just the A and B blood types, they cannot get B or A respectively from both parents since neither parent has the AB blood type. Therefore, both parents must have one copy of the O-type allele (i). Therefore, the father is IAi and the mother is IBi. The genotypes of the children are IAi, IAi, IBi, and IAIB.
7. The man would have to have either the O blood type or the A blood type in order for him to not have any B blood type children with an A blood type woman. If he has the O blood type, all of the children have the A blood type (IAi). If he has the A blood type, all of the children will be either AB bloodtype (IAIB) or A blood type (IAi). If he has the B or AB blood type, he can have B blood type children if the mother has the genotype IAi.
8. Because colorblindness is X-linked recessive and the mother is colorblind (both X-chromosomes have the colorblind allele), there is a 100% chance that a son will be colorblind because he will get one of the colorblind alleles from one of his mothers X-chromosomes and his fathers Y-chromosome (which doesnt complement the mothers colorblind allele). There is a 0% chance that a daughter will be colorblind but she will be a heterozygous carrier for colorblindness since she get a colorblind allele from here mother and a wild type allele from her father (who is not colorblind and has a wild type allele on his X-chromosome).
9. The correct order is as follows:
C -(7 map units)- A -(6 map units)- B -(5 map units)- D (or conversely D-B-A-C)
10. If these two genes are not linked, the expected phenotypic ratio is 1:1:1:1 (all the offspring will get the recessive allele from one parent and have an equal chance of getting a dominant or recessive allele at the two loci from the other heterozygous parent). The phenotypic ratio generated by the cross is 118:122:32:28 or 1.0 : 1.0 : 0.26 : 0.23. Because the phenotypic ratio is not what is expected, the two genes must be on the same chromosome causing some genetic linkage. The recombinant phenotypes make up 20% ((32+28)*100/(32+28+118+122) = 20) of the progeny, so the two genes are 20 map units apart.
11. This disorder is autosomal recessive. We know it is recessive since 3 offspring in the second generation have the disorder while their parents do not. We know it is autosomal and not sex-linked because two siblings of opposite sex in the second generation have the disorder. Person #4 must be a carrier for one of this persons offspring to have this recessive disorder and has the genotype Aa. Person #7 also has to be a carrier and has the genotype Aa since one of this persons parents has the disorder and the other parent is assumed to not be a carrier since that individual does not have the disorder and is not closely related to anyone in the pedigree involving the disorder).