Tim and Stephanie (from #2) are pregnant again! This time with a baby girl. They are nervous that she too may get hemophilia. Based on what you know about Tim and Stephanie, what is the probability that they do have a daughter with the disease? Show a Punnett square to support your answer.

The probability that they will have a daughter with the disease is zero, 0.

Note: Further information about the couple is given in the previous question shown below:

Tim and Stephanie are devastated when they find out their newborn son has hemophilia - a sex-linked recessive disease. Tim is shocked because he doesn't have hemophilia, and figures if his son has it he would have gotten it from him. Is Tim correct in his thinking? Explain. Also explain what their genotypes must be in order for neither of them to have the disease but have a son who does.

Explanation:

Hemophilia is a sex-linked recessive disease which results in the inability of the blood to clot normally. The abnormal gene for the inheritance of this disease is present in the X-chromosome. Males are not carriers of this disease as any male who have the abnormal X-chromosome will always have the disease but women can be carriers of the disease if the abnormal gene is only on one XX-chromosome.

From the information provided above, it can be determined that the genotype of Tim and his wife is XY and XhX, such that the wife is a carrier (Xh is the abnormal gene for hemophilia while X is the normal gene) and Tim is a normal male.

In the event that Tim and Stephanie now have a baby girl, the likely genotype of the baby girl can be obtained using the Punnett square shown in the attachment below.

The probability that they will have a daughter with the disease is zero, 0.

the answer that best describes the relative refractory period in the neuron is the resting membrane potential is restored, but concentrations of sodium and potassium are not restored to their original state.

the refractory period is the time during which another stimulus given to the neuron will not lead to a second potential. thus, because na+ channels are inactivated during this time, additional depolarizing stimuli do not lead to new action potentials.  

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