Question

ПОМОГИТЕ РЕШИТЬ ЗАДАЧУ ПО БИОЛОГИИ АА - родится мертвый Аа - лёгкая форма анемии. Женщина с нормальным зрением и лёгкой формой анемии родила сына от здорового мужчины дальтоника. У сына лёгкая анемия и лёгкий дальтонизм. Определите шанс рождения здорового ребёнка.

Answer 1

Ответ:АА-Родился мёртвый

Answer 2

Inheritance of Anemia and Color Blindness

To determine the chance of having a healthy child, we need to understand the inheritance patterns of anemia and color blindness. In this scenario, the woman has a mild form of anemia, and the man is color blind. Let's break down the inheritance of these traits and calculate the probability of having a healthy child.

Inheritance of Anemia

Anemia is a condition characterized by a decrease in the number of red blood cells or a decrease in the amount of hemoglobin in the blood. In this case, the woman has a mild form of anemia, which we'll represent as "Aa." The lowercase "a" represents the normal allele, while the uppercase "A" represents the allele for anemia.

The woman's genotype is Aa, meaning she carries one normal allele and one allele for anemia. Since anemia is a recessive trait, individuals need to inherit two copies of the allele for anemia (aa) to express the condition. Therefore, the woman does not have anemia herself but is a carrier.

Inheritance of Color Blindness

Color blindness is a condition that affects the ability to perceive certain colors. It is more common in males because the genes responsible for color vision are located on the X chromosome. In this case, the man is color blind, which we'll represent as "XcY." The lowercase "c" represents the allele for color blindness.

The man's genotype is XcY, meaning he has one normal allele and one allele for color blindness. Since color blindness is a recessive trait, males only need to inherit one copy of the allele for color blindness (Xc) to express the condition.

Determining the Child's Genotype

To determine the child's genotype, we need to consider the inheritance of both anemia and color blindness. Since the woman is a carrier for anemia (Aa) and the man is color blind (XcY), there are four possible combinations of alleles for the child:

1. AXc: The child inherits the normal allele for anemia from the mother (A) and the allele for color blindness from the father (Xc). 2. AY: The child inherits the normal allele for anemia from the mother (A) and the normal allele for color vision from the father (Y). 3. aXc: The child inherits the allele for anemia from the mother (a) and the allele for color blindness from the father (Xc). 4. aY: The child inherits the allele for anemia from the mother (a) and the normal allele for color vision from the father (Y).

Probability of Having a Healthy Child

Out of the four possible combinations, two result in a healthy child (AY and aY), while the other two combinations result in a child with both anemia and color blindness (AXc and aXc). Therefore, the probability of having a healthy child is 2 out of 4, which can be simplified to 1 out of 2 or 50%.

It's important to note that this calculation assumes independent assortment and does not take into account other genetic factors that may influence the inheritance of these traits. Additionally, the actual probability may vary depending on the specific genetic makeup of the individuals involved.

I hope this helps! Let me know if you have any further questions.