In Caenorhabditis elegans the two sexual phenotypes that occur are males and hermaphrodites.
Males have testes and the hermaphrodites have both testes and ovaries.
They have only one type of sex chromosomes - X and have autosomes.
Sexes are determined by the ratio of sex chromosomes to autosomes. X chromosomes:autosomes
XX/AA = 1 hermaphrodite ; XO/AA =0.5 males
There are two sexes in mammals - males and female. In mammals, Y chromosome is used for
sex determination. Y chromosomes confer maleness as it has a gene responsible for testes
development of gonads. In its absence the gonads develop into ovaries conferring femaleness.
XX female; XY male
In birds, there are two sexes males and females. Here W chromosome confers femaleness.
ZZ - male; ZW - female.
2.
Reciprocal cross - the cross in which the phenotype of parents are reversed from the previous
original cross.
AA (male) x aa (female) - Original cross
aa(male) x AA(female) - Reciprocal cross
Significance of reciprocal cross - to find out the role of parental sex on pattern of inheritence.
Solution
In Caenorhabditis elegans the two sexual phenotypes that occur are males and hermaphrodites.
Males have testes and the hermaphrodites have both testes and ovaries.
They have only one type of sex chromosomes - X and have autosomes.
Sexes are determined by the ratio of sex chromosomes to autosomes. X chromosomes:autosomes
XX/AA = 1 hermaphrodite ; XO/AA =0.5 males
There are two sexes in mammals - males and female. In mammals, Y chromosome is used for
sex determination. Y chromosomes confer maleness as it has a gene responsible for testes
development of gonads. In its absence the gonads develop into ovaries conferring femaleness.
XX female; XY male
In birds, there are two sexes males and females. Here W chromosome confers femaleness.
ZZ - male; ZW - female.
2.
Reciprocal cross - the cross in which the phenotype of parents are reversed from the previous
original cross.
AA (male) x aa (female) - Original cross
aa(male) x AA(female) - Reciprocal cross
Significance of reciprocal cross - to find out the role of parental sex on pattern of inheritence..
not exaclty sure what to do here U Sexlinked Disorders Se.pdfsaravanan544051
not exaclty sure what to do here
U. Sex-linked Disorders Sex-linked traits ine controlled by genes that are found on the X
clromosome, but not on the Y chromosome in humans. The X chromosome is larger and has more
genes than the Y chromosome. This complicntes the inleritance of uraits on the X clromosome
since we expect that a normal diploid cell will have two alleles for any gene. Such is not he case
for males siuce they are XY and will only have cne copy of the genes that are locied ou the X
chromosome: Females (XX) will have two ulleles for a gene located on the X chromosome Males (
XY) will only have one allele for that trait. This, females with one dominant and one recessive
allele for a gene on the X chromosene would display the dominant phenotype. However a mate
will always displey the phenorype of the one allele that they have for the mait that is carried on the
X chroinosonie. A new way of writing alleles: XR - normal eyesiglt X-colorblind Genotypes: XRXR
XRXrXRYXRXrXY Phenotypes: Normal Eyesight and Colorblind Remember to follow the same
steps in order to solve problems: 1. Assign lefters to represent the genes 2. Determine the
parental phenotypes and genotypes. 3. Predict the outcome of a cross between the possible
gameles from the tiro parents. 4. State plenotypic and genoty pic ratios present ia the offspring. 5.
Don't forget to answer the question witlin the problem. ains. begin{tabular}{|c|c|} hlinex & x hlinexn
x & xx hlinexy & xy hline end{tabular}.
not exaclty sure what to do here U Sexlinked Disorders Se.pdfsaravanan544051
not exaclty sure what to do here
U. Sex-linked Disorders Sex-linked traits ine controlled by genes that are found on the X
clromosome, but not on the Y chromosome in humans. The X chromosome is larger and has more
genes than the Y chromosome. This complicntes the inleritance of uraits on the X clromosome
since we expect that a normal diploid cell will have two alleles for any gene. Such is not he case
for males siuce they are XY and will only have cne copy of the genes that are locied ou the X
chromosome: Females (XX) will have two ulleles for a gene located on the X chromosome Males (
XY) will only have one allele for that trait. This, females with one dominant and one recessive
allele for a gene on the X chromosene would display the dominant phenotype. However a mate
will always displey the phenorype of the one allele that they have for the mait that is carried on the
X chroinosonie. A new way of writing alleles: XR - normal eyesiglt X-colorblind Genotypes: XRXR
XRXrXRYXRXrXY Phenotypes: Normal Eyesight and Colorblind Remember to follow the same
steps in order to solve problems: 1. Assign lefters to represent the genes 2. Determine the
parental phenotypes and genotypes. 3. Predict the outcome of a cross between the possible
gameles from the tiro parents. 4. State plenotypic and genoty pic ratios present ia the offspring. 5.
Don't forget to answer the question witlin the problem. ains. begin{tabular}{|c|c|} hlinex & x hlinexn
x & xx hlinexy & xy hline end{tabular}.
It is ionic solid (compound). It will be more soluble in water as CCl4 is non-polar
Solution
It is ionic solid (compound). It will be more soluble in water as CCl4 is non-polar.
More Related Content
Similar to In Caenorhabditis elegans the two sexual phenotypes that occur are m.pdf
It is ionic solid (compound). It will be more soluble in water as CCl4 is non-polar
Solution
It is ionic solid (compound). It will be more soluble in water as CCl4 is non-polar.
Evaporation at 20°C is negligible. HgO decomposes on exposure to light, on
heating above 500°C producing highly toxic fumes including mercury and oxygen, which
increases fire hazard. Reacts violently with reducing agents, chlorine, hydrogen peroxide,
magnesium (when heated), disulfur dichloride and hydrogen trisulfide. Shock-sensitive
compounds are formed with metals and elements such as sulfur and phosphorus.
Solution
Evaporation at 20°C is negligible. HgO decomposes on exposure to light, on
heating above 500°C producing highly toxic fumes including mercury and oxygen, which
increases fire hazard. Reacts violently with reducing agents, chlorine, hydrogen peroxide,
magnesium (when heated), disulfur dichloride and hydrogen trisulfide. Shock-sensitive
compounds are formed with metals and elements such as sulfur and phosphorus..
E) Mg is the reducing agent and Cu^2+ is the oxidizing agent. is correct
Solution
E) Mg is the reducing agent and Cu^2+ is the oxidizing agent. is correct.
boiling point: KCl > CH2O > CO2 note: KCl is ionic compound, which is of high
boiling point and high melting point characters. CH2O are polar molecule, which can form
intermolecular hydrogen bonds CO2 is nonpolar molecule with a very weak intermolecular
interaction.
Solution
boiling point: KCl > CH2O > CO2 note: KCl is ionic compound, which is of high
boiling point and high melting point characters. CH2O are polar molecule, which can form
intermolecular hydrogen bonds CO2 is nonpolar molecule with a very weak intermolecular
interaction..
A is right option beacuse H2O will attack on carbonil group and formed O- will
accpet H+ from water
Solution
A is right option beacuse H2O will attack on carbonil group and formed O- will
accpet H+ from water.
Fifi eventhough is a male; it is possible to clone it to have a dupl.pdfankurelectronicsg3
Fifi eventhough is a male; it is possible to clone it to have a duplicate of it. Egg cell has to be
selected from a female pet. Nucleus of egg cell to be removed. Donor somatic cell of Fifi has to
be selected. Donor somatic cell of Fifi has the nucleus. Donor cell of Fifi (with nucleus) has to be
bombarded into an egg cell (without nucleus). The fused cell divides normally. The Embryo
developed thus has to be placed in the uterus of the female. The embryo develops in the womb of
the female into a lovable pet Fifi, which is just a copy of its father.
Solution
Fifi eventhough is a male; it is possible to clone it to have a duplicate of it. Egg cell has to be
selected from a female pet. Nucleus of egg cell to be removed. Donor somatic cell of Fifi has to
be selected. Donor somatic cell of Fifi has the nucleus. Donor cell of Fifi (with nucleus) has to be
bombarded into an egg cell (without nucleus). The fused cell divides normally. The Embryo
developed thus has to be placed in the uterus of the female. The embryo develops in the womb of
the female into a lovable pet Fifi, which is just a copy of its father..
one. If there was more than one proton, they wouldn\'t be hydrogen. (Isotopes are
varying numbers of neutrons)
Solution
one. If there was more than one proton, they wouldn\'t be hydrogen. (Isotopes are
varying numbers of neutrons).
Moles of Cu present = 0.576/63.5 = 0.00907 1L have HNO3 have = 16 moles
Thus Volume required = (1/16)*( 0.00907) = 0.000566875 L
Solution
Moles of Cu present = 0.576/63.5 = 0.00907 1L have HNO3 have = 16 moles
Thus Volume required = (1/16)*( 0.00907) = 0.000566875 L.
In R2, the only non-trivial subspaces are lines passing through the origin.
Solution
In R2, the only non-trivial subspaces are lines passing through the origin..
first one is far more acidic due to high inductive effect of three F s
Solution
first one is far more acidic due to high inductive effect of three F s.
Wait I am doing it ...please rate 5 ..Ill post the answer in comme.pdfankurelectronicsg3
Wait I am doing it ...please rate 5 ..I\"ll post the answer in comments....
Solution
Wait I am doing it ...please rate 5 ..I\"ll post the answer in comments.....
When ever you impelemented an interface we must define the abstract .pdfankurelectronicsg3
When ever you impelemented an interface we must define the abstract methods in that interface
in inheritance class. so that we must define the all methods in that interface (collection)
solved problem:
import java.util.Collection;
import java.util.Iterator;
public class PartialArrayMultiset implements Collection
{
protected E[] _store;
protected int _size;
protected long _modCount;
public PartialArrayMultiset()
{
_modCount=0;
clear();
}
public void clear()
{
_store=(E[])(new Object[16]);
_size=0;
_modCount+=1;
}
public boolean add(E e)
{
if(_size==_store.length)
return false;
_store[_size]=e;
_size++;
return true;
}
public void removeAtIndex(int i)
{
if(i<0 || i>= _size)
throw new ArrayIndexOutOfBoundsException(i);
E oldValue=_store[i];
for(int hole=i;hole<_size-1;hole++)
_store[hole]=_store[hole+1];
_store[_size-1]=null;
_size--;
}
public boolean contains(Object obj)
{
return false;
}
public boolean addAll(Collection arg0) {
// TODO Auto-generated method stub
return false;
}
public boolean containsAll(Collection arg0) {
// TODO Auto-generated method stub
return false;
}
public boolean isEmpty() {
// TODO Auto-generated method stub
return false;
}
public Iterator iterator() {
// TODO Auto-generated method stub
return null;
}
public boolean remove(Object arg0) {
// TODO Auto-generated method stub
return false;
}
public boolean removeAll(Collection arg0) {
// TODO Auto-generated method stub
return false;
}
public boolean retainAll(Collection arg0) {
// TODO Auto-generated method stub
return false;
}
public int size() {
// TODO Auto-generated method stub
return 0;
}
public Object[] toArray() {
// TODO Auto-generated method stub
return null;
}
public T[] toArray(T[] arg0) {
// TODO Auto-generated method stub
return null;
}
}
now use this program and test run y
Solution
When ever you impelemented an interface we must define the abstract methods in that interface
in inheritance class. so that we must define the all methods in that interface (collection)
solved problem:
import java.util.Collection;
import java.util.Iterator;
public class PartialArrayMultiset implements Collection
{
protected E[] _store;
protected int _size;
protected long _modCount;
public PartialArrayMultiset()
{
_modCount=0;
clear();
}
public void clear()
{
_store=(E[])(new Object[16]);
_size=0;
_modCount+=1;
}
public boolean add(E e)
{
if(_size==_store.length)
return false;
_store[_size]=e;
_size++;
return true;
}
public void removeAtIndex(int i)
{
if(i<0 || i>= _size)
throw new ArrayIndexOutOfBoundsException(i);
E oldValue=_store[i];
for(int hole=i;hole<_size-1;hole++)
_store[hole]=_store[hole+1];
_store[_size-1]=null;
_size--;
}
public boolean contains(Object obj)
{
return false;
}
public boolean addAll(Collection arg0) {
// TODO Auto-generated method stub
return false;
}
public boolean containsAll(Collection arg0) {
// TODO Auto-generated method stub
return false;
}
public boolean isEmpty() {
// TODO Auto-generated method stub
return false;
}
public Iterator iterator() {
// TO.
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In Caenorhabditis elegans the two sexual phenotypes that occur are m.pdf
1. In Caenorhabditis elegans the two sexual phenotypes that occur are males and hermaphrodites.
Males have testes and the hermaphrodites have both testes and ovaries.
They have only one type of sex chromosomes - X and have autosomes.
Sexes are determined by the ratio of sex chromosomes to autosomes. X chromosomes:autosomes
XX/AA = 1 hermaphrodite ; XO/AA =0.5 males
There are two sexes in mammals - males and female. In mammals, Y chromosome is used for
sex determination. Y chromosomes confer maleness as it has a gene responsible for testes
development of gonads. In its absence the gonads develop into ovaries conferring femaleness.
XX female; XY male
In birds, there are two sexes males and females. Here W chromosome confers femaleness.
ZZ - male; ZW - female.
2.
Reciprocal cross - the cross in which the phenotype of parents are reversed from the previous
original cross.
AA (male) x aa (female) - Original cross
aa(male) x AA(female) - Reciprocal cross
Significance of reciprocal cross - to find out the role of parental sex on pattern of inheritence.
Solution
In Caenorhabditis elegans the two sexual phenotypes that occur are males and hermaphrodites.
Males have testes and the hermaphrodites have both testes and ovaries.
They have only one type of sex chromosomes - X and have autosomes.
Sexes are determined by the ratio of sex chromosomes to autosomes. X chromosomes:autosomes
XX/AA = 1 hermaphrodite ; XO/AA =0.5 males
There are two sexes in mammals - males and female. In mammals, Y chromosome is used for
sex determination. Y chromosomes confer maleness as it has a gene responsible for testes
development of gonads. In its absence the gonads develop into ovaries conferring femaleness.
XX female; XY male
In birds, there are two sexes males and females. Here W chromosome confers femaleness.
ZZ - male; ZW - female.
2.
Reciprocal cross - the cross in which the phenotype of parents are reversed from the previous
original cross.
AA (male) x aa (female) - Original cross
2. aa(male) x AA(female) - Reciprocal cross
Significance of reciprocal cross - to find out the role of parental sex on pattern of inheritence.
