Question

Solve the following genetic problems. Show your workout.

1. If the parents are AO (or I^Ai) and BO (or I^Bi) genotypes for the ABO blood group, what are the possible genotypes of their children?

Clue: Blood type is determined by three alleles in humans; I^A (or A allele), I^B (or B allele), i (or O allele). Each individual receives two of these alleles; I^AI^A and I^Ai represent Blood type A; I^BI^B and I^Bi represent blood type B.  I^A and I^B alleles are dominant over allele i.  I^AI^B represents blood type AB (A and B are codominant). Two I alleles (ii) represent O type.

2. In the cross, Bb X Bb, what percentage of the offspring would have the same phenotype as the parents? what percentage of the offspring would have the same genotype as the parents?

3. A female dog with the genotype LlSs mates with a male dog with the genotype llSs.  In dogs, long hair (L) is dominant over short hair (l) and spotted coats (S) are dominant over solid coats (s).  What is the chance of getting a puppy with short hair and a solid coat?  You must show all of your work.

4. Diabetes is a recessive trait. A man and a woman who neither have diabetes each have one parent with the disorder.  If they marry and have children, what are the chances that any of their children will have diabetes? You must show your work.

5. Blue eyes appear to be recessive to dark eyes.  If two-blued people have a child, what are the chances that the child’s eyes will be dark? You must show your work.

Answer 1

1) Given,

parents- phenotype - A blood group x B blood group

genotype - IAi x IBi

Gametes - IA, i IB, i

Cross-

	IA	i
IB	IAIB - AB blood group	IBi - B blood group
i	IAi - A blood group	ii - O blood group

phenotypic ratio - 1 AB blood group : 1 A blood group : 1 B blood group : 1O blood group

so, all the four blood groups are possible with the cross of given two parents.

2) given, (*let those with B (shown by BB/ Bb) have x phenotype, and with b(shown by bb) having y phenotype)

parents- genotype - Bb x Bb

gametes - B, b B, b

	B	b
B	BB - x phenotype	Bb - x phenotype
b	Bb - x phenotype	bb - y phenotype

So phenotypic ratio - 3 x phenotype : 1 y phenotype. that is out of 4 progenies 3 will show same phenotype as parents

genotypic ratio - 1 BB : 2 Bb : 1 bb

that is out of 4 progenies 2 will have only same genotypes.

3) given, L controls long hair length, l- short hair length, S- spotted and s- solid colour

So that L_S_ will be long hair spotted

llS_ short hair spotted

llss - short hair solid

L_ ss - long hair solid

parents-- genotype LlSs x llSs

gametes-- LS, Ls, lS, ls lS, ls

cross

	lS	ls
LS	LlSS	LlSs
Ls	LlSs	Llss
lS	llSS	llSs
ls	llSs	llss - short hair solid

that is only 1 out of 8 progenies can show the trait, hence the answer is 1/8

4) it is said diabetes is a recessive trait, that means one person will only get diabetes when he/she is dd (if we consider D non-diabetes allele dominant and d diabetes allele recessive )

So, D_ is nondiabetes and dd diabetes

As the previous generation has the diabetic allele,and the parents don't have diabetes., they will be heterozygous, that is their genotype will be Dd

parents- genotype -- Dd x Dd

gametes-- D, d D, d

cross,

	D	d
D	DD	Dd
d	Dd	dd - diabetes

As there exist 1 chance of the progeny to be diabetic compared to 4 progenies resulted, the chance that the progeny is diabetic = 1/4 or 25%