
If the mother is a carrier of color blindness and the father is normal, show the possible genotype and phenotype of the offspring of the next generation, with the help of punnett square.
Answer
561.9k+ views
Hint: These are the diseases that passed down through families through one of the X or Y chromosomes. X and Y are sex chromosomes. When an abnormal gene from one parent causes disease during dominant inheritance, despite the fact that the matching gene from the other parent is normal. The abnormal gene dominates.
Complete answer:
Colour blindness is an X-linked recessive disorder. In the above question, Mother is a carrier for colorblindness. Therefore, her genotype must be ${ X }^{ c }$X. The genotype must be XY when the father is normal. With the help of punnett square, the possibility of genotypes and phenotypes in the next generation will be:
The possible genotypes are:
${ X }^{ c }$ X- Carrier daughter
${ X }^{ c }$ Y- Affected son
XX- Normal daughter
XY- Normal son
Additional information: In males, X-linked recessive diseases most often occur. In males, it has only one X chromosome. A single recessive gene on that X chromosome will be responsible for this disease.
In the XY gene pair male the Y chromosomes are the other half portion. Nonetheless, the Y chromosome doesn't contain most of the genes of the X chromosome. Therefore, it doesn't protect the male. Diseases, for example, hemophilia and Duchenne muscular dystrophy happen from a recessive gene on the X chromosome.
Note: In every pregnancy, if the mother is a carrier of a specific disease (she has just a single abnormal X chromosome) and the father isn’t a carrier for the disease, the expected result is:
$25\%$ possibility of a healthy boy
$25\%$ possibility of a boy with the disease
$25\%$ possibility of a healthy girl
$25\%$ possibility of a carrier girl without disease
On the off chance that the father has the disease and the mother isn’t a carrier, the expected results are:
$50\%$ possibility of having a healthy boy
$50\%$ possibility of having a girl without the disease who is a carrier
Complete answer:
Colour blindness is an X-linked recessive disorder. In the above question, Mother is a carrier for colorblindness. Therefore, her genotype must be ${ X }^{ c }$X. The genotype must be XY when the father is normal. With the help of punnett square, the possibility of genotypes and phenotypes in the next generation will be:
| Genotype | X | Y |
| ${ X }^{ c }$ | ${ X }^{ c }$X | ${ X }^{ c }$Y |
| X | XX | XY |
The possible genotypes are:
${ X }^{ c }$ X- Carrier daughter
${ X }^{ c }$ Y- Affected son
XX- Normal daughter
XY- Normal son
Additional information: In males, X-linked recessive diseases most often occur. In males, it has only one X chromosome. A single recessive gene on that X chromosome will be responsible for this disease.
In the XY gene pair male the Y chromosomes are the other half portion. Nonetheless, the Y chromosome doesn't contain most of the genes of the X chromosome. Therefore, it doesn't protect the male. Diseases, for example, hemophilia and Duchenne muscular dystrophy happen from a recessive gene on the X chromosome.
Note: In every pregnancy, if the mother is a carrier of a specific disease (she has just a single abnormal X chromosome) and the father isn’t a carrier for the disease, the expected result is:
$25\%$ possibility of a healthy boy
$25\%$ possibility of a boy with the disease
$25\%$ possibility of a healthy girl
$25\%$ possibility of a carrier girl without disease
On the off chance that the father has the disease and the mother isn’t a carrier, the expected results are:
$50\%$ possibility of having a healthy boy
$50\%$ possibility of having a girl without the disease who is a carrier
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