: Albinstn is caused by deficiency of tyrosinase. In albinistm clisorder body makes little or tone of a substance called melarin. Melarin determines the colour of skin hair and eye.
Haryana PMT-2005
Principle of Inheritance and Variation
184700
A woman has haemophilic son and three normal children. Her genotype and that of her husband with respect to this gene would be
1 \(X X\) and \(X^4 Y\)
2 \(\mathrm{X}^{\mathrm{t}} \mathrm{X}^4\) and \(\mathrm{X}^{\mathrm{t}} \mathrm{Y}\)
3 \(\mathrm{X}^{\mathrm{H}} \mathrm{X}^{\mathrm{h}}\) and \(X Y\)
4 \(\mathrm{X}^{\mathrm{H}} \mathrm{X}\) and \(X Y\)
Explanation:
: A woman has haemophilic son and three tormal children. Her genotype and her husband with tespect to this gene would be \(\mathrm{X}^{\mathrm{t}} \mathrm{X}\) and XY because, there are three normal children, so it is clear that the lusband is not haemophilic (XY). Haemophilic son inclicates that mother is a camier for the gene ( \(\mathrm{X}^{\mathrm{t}} \mathrm{X}\) ).
MGIMS Wardha-2009 / CG PMT-2009
Principle of Inheritance and Variation
184701
A normal woman is married with a man having hypertrichosis condition. They got one daughter and one son. What is the possibility of this daughter to have hypertrichosis condition?
1 \(0 \%\)
2 \(25 \%\)
3 \(50 \%\)
4 \(100 \%\)
Explanation:
: A tormal woman is married with a man having hypettrichosis condition. The possibility of their daughter to have hypertrichosis condition is \(0 \%\). Because hypertrichosis is Y-linked clisease, only males tormally have a Y-chromosome, Y-linked genes can only be transmitted from father to son.
CG PMT-2009 / MGIMS Wardha-2009
Principle of Inheritance and Variation
184702
A haemophilic man marries a carrier woman, their daughter progenies are
1 all hemophilic
2 halfhaemophilic
3 half hemophilic and half carrier
4 all nomal
Explanation:
: The hasmophilia is an X linked recessive gene. It is of two types haemophilia A which occurs due to low amount of clotting factor VII, haemophilia B which occurs due to low level of clotting factor IX. A hemophilic man marries a carrier woman, their daughter progeries are half haemophilic and half carrier.
: Albinstn is caused by deficiency of tyrosinase. In albinistm clisorder body makes little or tone of a substance called melarin. Melarin determines the colour of skin hair and eye.
Haryana PMT-2005
Principle of Inheritance and Variation
184700
A woman has haemophilic son and three normal children. Her genotype and that of her husband with respect to this gene would be
1 \(X X\) and \(X^4 Y\)
2 \(\mathrm{X}^{\mathrm{t}} \mathrm{X}^4\) and \(\mathrm{X}^{\mathrm{t}} \mathrm{Y}\)
3 \(\mathrm{X}^{\mathrm{H}} \mathrm{X}^{\mathrm{h}}\) and \(X Y\)
4 \(\mathrm{X}^{\mathrm{H}} \mathrm{X}\) and \(X Y\)
Explanation:
: A woman has haemophilic son and three tormal children. Her genotype and her husband with tespect to this gene would be \(\mathrm{X}^{\mathrm{t}} \mathrm{X}\) and XY because, there are three normal children, so it is clear that the lusband is not haemophilic (XY). Haemophilic son inclicates that mother is a camier for the gene ( \(\mathrm{X}^{\mathrm{t}} \mathrm{X}\) ).
MGIMS Wardha-2009 / CG PMT-2009
Principle of Inheritance and Variation
184701
A normal woman is married with a man having hypertrichosis condition. They got one daughter and one son. What is the possibility of this daughter to have hypertrichosis condition?
1 \(0 \%\)
2 \(25 \%\)
3 \(50 \%\)
4 \(100 \%\)
Explanation:
: A tormal woman is married with a man having hypettrichosis condition. The possibility of their daughter to have hypertrichosis condition is \(0 \%\). Because hypertrichosis is Y-linked clisease, only males tormally have a Y-chromosome, Y-linked genes can only be transmitted from father to son.
CG PMT-2009 / MGIMS Wardha-2009
Principle of Inheritance and Variation
184702
A haemophilic man marries a carrier woman, their daughter progenies are
1 all hemophilic
2 halfhaemophilic
3 half hemophilic and half carrier
4 all nomal
Explanation:
: The hasmophilia is an X linked recessive gene. It is of two types haemophilia A which occurs due to low amount of clotting factor VII, haemophilia B which occurs due to low level of clotting factor IX. A hemophilic man marries a carrier woman, their daughter progeries are half haemophilic and half carrier.
: Albinstn is caused by deficiency of tyrosinase. In albinistm clisorder body makes little or tone of a substance called melarin. Melarin determines the colour of skin hair and eye.
Haryana PMT-2005
Principle of Inheritance and Variation
184700
A woman has haemophilic son and three normal children. Her genotype and that of her husband with respect to this gene would be
1 \(X X\) and \(X^4 Y\)
2 \(\mathrm{X}^{\mathrm{t}} \mathrm{X}^4\) and \(\mathrm{X}^{\mathrm{t}} \mathrm{Y}\)
3 \(\mathrm{X}^{\mathrm{H}} \mathrm{X}^{\mathrm{h}}\) and \(X Y\)
4 \(\mathrm{X}^{\mathrm{H}} \mathrm{X}\) and \(X Y\)
Explanation:
: A woman has haemophilic son and three tormal children. Her genotype and her husband with tespect to this gene would be \(\mathrm{X}^{\mathrm{t}} \mathrm{X}\) and XY because, there are three normal children, so it is clear that the lusband is not haemophilic (XY). Haemophilic son inclicates that mother is a camier for the gene ( \(\mathrm{X}^{\mathrm{t}} \mathrm{X}\) ).
MGIMS Wardha-2009 / CG PMT-2009
Principle of Inheritance and Variation
184701
A normal woman is married with a man having hypertrichosis condition. They got one daughter and one son. What is the possibility of this daughter to have hypertrichosis condition?
1 \(0 \%\)
2 \(25 \%\)
3 \(50 \%\)
4 \(100 \%\)
Explanation:
: A tormal woman is married with a man having hypettrichosis condition. The possibility of their daughter to have hypertrichosis condition is \(0 \%\). Because hypertrichosis is Y-linked clisease, only males tormally have a Y-chromosome, Y-linked genes can only be transmitted from father to son.
CG PMT-2009 / MGIMS Wardha-2009
Principle of Inheritance and Variation
184702
A haemophilic man marries a carrier woman, their daughter progenies are
1 all hemophilic
2 halfhaemophilic
3 half hemophilic and half carrier
4 all nomal
Explanation:
: The hasmophilia is an X linked recessive gene. It is of two types haemophilia A which occurs due to low amount of clotting factor VII, haemophilia B which occurs due to low level of clotting factor IX. A hemophilic man marries a carrier woman, their daughter progeries are half haemophilic and half carrier.
: Albinstn is caused by deficiency of tyrosinase. In albinistm clisorder body makes little or tone of a substance called melarin. Melarin determines the colour of skin hair and eye.
Haryana PMT-2005
Principle of Inheritance and Variation
184700
A woman has haemophilic son and three normal children. Her genotype and that of her husband with respect to this gene would be
1 \(X X\) and \(X^4 Y\)
2 \(\mathrm{X}^{\mathrm{t}} \mathrm{X}^4\) and \(\mathrm{X}^{\mathrm{t}} \mathrm{Y}\)
3 \(\mathrm{X}^{\mathrm{H}} \mathrm{X}^{\mathrm{h}}\) and \(X Y\)
4 \(\mathrm{X}^{\mathrm{H}} \mathrm{X}\) and \(X Y\)
Explanation:
: A woman has haemophilic son and three tormal children. Her genotype and her husband with tespect to this gene would be \(\mathrm{X}^{\mathrm{t}} \mathrm{X}\) and XY because, there are three normal children, so it is clear that the lusband is not haemophilic (XY). Haemophilic son inclicates that mother is a camier for the gene ( \(\mathrm{X}^{\mathrm{t}} \mathrm{X}\) ).
MGIMS Wardha-2009 / CG PMT-2009
Principle of Inheritance and Variation
184701
A normal woman is married with a man having hypertrichosis condition. They got one daughter and one son. What is the possibility of this daughter to have hypertrichosis condition?
1 \(0 \%\)
2 \(25 \%\)
3 \(50 \%\)
4 \(100 \%\)
Explanation:
: A tormal woman is married with a man having hypettrichosis condition. The possibility of their daughter to have hypertrichosis condition is \(0 \%\). Because hypertrichosis is Y-linked clisease, only males tormally have a Y-chromosome, Y-linked genes can only be transmitted from father to son.
CG PMT-2009 / MGIMS Wardha-2009
Principle of Inheritance and Variation
184702
A haemophilic man marries a carrier woman, their daughter progenies are
1 all hemophilic
2 halfhaemophilic
3 half hemophilic and half carrier
4 all nomal
Explanation:
: The hasmophilia is an X linked recessive gene. It is of two types haemophilia A which occurs due to low amount of clotting factor VII, haemophilia B which occurs due to low level of clotting factor IX. A hemophilic man marries a carrier woman, their daughter progeries are half haemophilic and half carrier.
