Make a cross to explain the answer:
1. Can a child having blood type A be born to parents having types AB and B respectively?
2. A Color blind man marries a woman with normal vision. Her mother was color blind. What kind of children would you expect from this marriage?
3. A boy, whose parents and grandparents had normal vision, is color-blind. What are the genotypes for his mother and his maternal grandparents. Use XB for the dominant normal condition and Xb for the recessive, color-blind phenotype.
4. A woman with red-green color-blindness has a mother with normal vision, can you determine what her father's phenotype is?
5. A woman whose maternal grandfather suffered from hemophilia has parents that are normal. The woman's husband is normal. She has a hemophiliac son. Will any of the daughters be hemophiliac? Will any be carriers?
6. A child wonders if she is adopted and therefore compares her blood type to those of her "parents".If her father is type A and her mother is type B, will blood typing help to determine if she is adopted? Explain.
7. The only member of Jane's family who is color-blind is her brother.
What is the brother's genotype? ………….
The father's genotype? ………..
The mother's genotype? ………..
Jane later has a color-blind son, what is her genotype? ………
8. A male with hemophilia marries a normal woman with no history of hemophilia in her family.
a) What is the chance that their son will have hemophilia?
………...
b) What is the chance that their daughter will have hemophilia?
………
Solve the following problems
1. In Shepherd's Purse (Capsule bursa), fruit-shape is of two types: triangular and ovoid-oblong. A cross between pure plant with triangular fruits and pure plant with ovoid-oblong fruits produces hybrid plants with triangular fruits in F1 generation. F1
hybrids on self pollination produce plants in F2 generation 945 triangular fruit and 63 ovoid ones. According to your understanding explain why you have this ratio?
2. In crossing of two plant of sweet pea one with colored flowers and the other with white flowers we have the following generation 315 colored flower plant and 245 white flowed plant. According to your understanding explain why you have this ratio?
3. Solve this CHI-Square problem if you have the following data about an experiment:
916 tall red, 295 tall white , 295 short red and 85 short white
4. Suppose you counted the second generation of flowered plant and you find that 95 of the flowers are red and 33 of them are white. Test your observation by doing the Chi-square test and decide your conclusion.
5. In pea plant pod shape can be swollen or pinched and the seed color can be either green or yellow. A plant with swollen yellow is crossed with a pinched green one and after self crossing we have the following data: 360 swollen yellow, 120 swollen green, 120 pinched yellow and 40 pinched green. Test the observation by doing the Chi- square test and decide your conclusion.
6. A cross between two type of pea plant tall with yellow seeds and short with green seed gives a tall plants in F1 generation and these plant crossed for F2 generation and we get the following data: 295 tall yellow seeds, 108 tall green seeds, 101 short yellow seeds and 32 short green seeds. Test your observation by doing the Chi- square test and decide your conclusion.
7. In drosophila fly there are three x-linkage mutations: scute bristles (sc), crossveinless wings and between of them their a locus of echinus eyes (ec). We crossed a homozygous female for the three mutations with a wild type male. If you have the following F2 generation, calculate the recombination and draw the map distant between these genes:
Phenotype Genotype F2 generation
Scute, echinus, crossveinless 1158
Wild type 1455
scute 163
ecinus,crossveinless 130
Scute,echinus 192
crossveinless 148
Scute, crossveinless 1
echinus 1
Total: 3248
8. On chromosome 3 in Drosophila, there are the following mutations: Lyra (Ly), bright red eyes (br) and between of them there is a Stubble mutation (Sb). A female homozygous for the 3 mutations was mated to a wild type male. If you have the following F2 generation, calculate the recombination and draw the map distant between these genes:
Phenotype Genotype F2 generation
Lyra, stubble, bright red 404
Wild type 422
Lyra 18
Stubble, bright red 16
Lyra, bright red 75
stubble 59
Lyra, stubble 4
bright red 2
Total: 1000
9. Cinnabar eyes (ci), smooth abdomen (sa) are on chromosome 2 of Drosophila and between of them there is a gen for curved wings (cw). The mutant alleles are all recessive to the wild-type alleles. A female homozygous for the mutant alleles at each locus is crossed to a wild type male. If you have the following F2 generation, calculate the recombination and draw the map distant between these genes:
Phenotype Genotype F2 generation
Cinnabar 300
Curved, smooth 308
Curved 150
Cinnabar, smooth 125
Wild type 50
Cinnabar, curved, smooth 45
Smooth 12
Cinnabar, curved 10
Total: 1000
