A rule of It isis used to predict the probability of two or more independent events occurring at the same time, governed by the theory:
The probability of two or more independent events occurring together is equal to the product of the probabilities of their occurring separately.
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We prepared a list of exercises on the rule of thumb It isso you can test your knowledge of the probability of genetic events happening simultaneously.
You can consult the feedback and save this list of exercises in PDF at the end of the post!
1) (ENEM) Sickle Cell Anemia is one of the most prevalent hereditary diseases in Brazil, especially in regions that received massive contingents of African slaves. It is a genetic alteration, characterized by a type of mutant hemoglobin called hemoglobin S. Individuals with this disease have sickle-shaped erythrocytes, hence the name. If a person receives a gene from the father and another from the mother to produce hemoglobin S, he is born with a pair of SS genes and thus will have Sickle Cell Anemia. If she receives the gene for hemoglobin S from one parent and the gene for hemoglobin A from the other, she will not have the disease, only Sickle Cell Trait (AS), and will not need specialized treatment. However, you should know that if you have children with a person who also inherited the trait, they could develop the disease.
Two couples, both members of which are heterozygous for the AS type for the hemoglobin gene, each want to have a child. Given that one couple is composed of black people and the other by white people, the probability that both couples will have children (one for each couple) with Sickle Cell Anemia is equal to:
a) 5.05%.
b) 6.25%.
c) 10.25%.
d) 18.05%.
e) 25.00%
2) Imagine that a couple is formed by a B Rh- blood woman and an O Rh+ blood man. Knowing that this man is the son of an Rh- father and that the woman is the daughter of an O blood mother, indicate the probability of having a child with O Rh+ blood.
a) 1/2.
b) 1/4.
c) 1/6.
d) 1/8.
e) 1/12.
3) (UNIMEP) A man has the Aa Bb CC dd genotype and his wife has the aa Bb cc Dd genotype. What is the probability that this couple will have a male child carrying the bb genes?
a) 1/4.
b) 1/8.
c) 1/2.
d) 3/64.
e) None of the above.
4) A couple always dreamed of having two children: a girl and a boy. According to genetics, what is the probability that this couple's first child will be a girl and the second child will be a boy?
a) 1/2.
b) 1/3.
c) 1/4.
d) 1/5.
e) 1/6.
5) (MACKENZIE) Achondroplasia is a type of dwarfism in which the head and trunk are normal, but arms and legs are very short. It is conditioned by a dominant gene that, in homozygosity, causes death before birth. Normal individuals are recessive and those affected are heterozygous. The probability of an achondroplastic couple having a normal female child is:
a) 1/6.
b) 1/8.
c) 2/5.
d) 1/2.
e) 1/4.
6) (UFU) Albinism is conditioned by a recessive allele a. A normal couple heterozygous for albinism wants to know approximately what the probability is if they have 5 children, the first two children being albino, the third being normal heterozygotes and the last two being normal homozygotes.
a) 0.2%.
b) 1%.
c) 31%.
d) 62%.
7) (UFRN) In peas, the yellow color is dominant in relation to the green one. From the crossing of heterozygotes, 720 offspring were born. Check the option whose number corresponds to the number of yellow descendants.
a) 360.
b) 540.
c) 180.
d) 720.
8) Imagine that a person normal for melanin production and heterozygous married an albino man. What is the probability that the couple's first child will be albino?
a) 0%.
b) 25%.
c) 50%.
d) 75%.
e) 100%.
9) (UECE) The Punnett square is a resource that was devised by R. W. Punnett, a geneticist who collaborated with William Batenson (both lived in the late 19th and early 20th centuries, actively participating in the rediscoveries of the work of Mendel), which greatly facilitates the making of crosses between F1 heterozygotes, by filling in the boxes that compose it with the genotypes resulting from the cross accomplished. However, in the polyhybrid crossing, this filling is complicated due to the increase in squares. We can correctly state that, in the case of a polyhybrid cross of the type AaBbCcDdFf X AaBbCcDdFf, which presents itself with 5 (five) loci heterozygotes, located on different pairs of homologous chromosomes, the number of spaces in the Punnett square that would have to be filled with genotypes is:
a) 59049.
b) 1024.
c) 19456.
d) 32.
10) Imagine that an albino child of the couple in the previous question has a child with a normal woman who is homozygous for melanin production. What is the chance of being born an albino individual?
a) 0%.
b) 25%.
c) 50%.
d) 75%.
e) 100%.
1 – b
2 – b
3 – b
4 – c
5 – the
6 – the
7 – b
8 – c
9 – d
10 – the
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