{"id":4277,"date":"2021-02-04T09:32:02","date_gmt":"2021-02-04T08:32:02","guid":{"rendered":"https:\/\/meddists.com\/learn\/pre-clinical\/medical-genetics\/transmission-genetics\/transmission-genetics-practice-questions\/"},"modified":"2021-02-04T11:25:41","modified_gmt":"2021-02-04T10:25:41","slug":"transmission-genetics-practice-questions","status":"publish","type":"page","link":"https:\/\/meddists.com\/learn\/pre-clinical\/medical-genetics\/transmission-genetics\/transmission-genetics-practice-questions\/","title":{"rendered":"Transmission Genetics practice questions"},"content":{"rendered":"\n

Question 1<\/h2>\n<\/span>\n\n\n

A healthy man marries a healthy woman. One of their sons has an X linked disease but the other one is healthy.<\/p>\n\n\n\n

Which of the following statements are true?<\/strong><\/p>\n\n\n\n

a. The defect is X-linked dominant<\/p>\n\n\n\n

b. The defect is X-linked recessive<\/p>\n\n\n\n

c. None of their daughters will have the disease<\/p>\n\n\n\n

d. 50% of their daughters will have the disease<\/p>\n\n\n\n

If the disease was X-linked dominant and both parents are healthy, it means that none of them possesses the trait for the disease (they would both be recessive), so they would not be able to pass it onto their children (please refer to the Transmission genetics: Single Gene disorders lesson for more details)<\/em>.
So the disease is recessive and since we were told that it is an X-linked disease, it is X-linked recessive.<\/p>\n\n\n\n

If the disease is X-linked recessive and the father is healthy, it means that he lacks the trait(he has a dominant X chromosome). This means that the mother has to be heterozygous for the trait(she has one recessive X chromosome that she gave to the sick son).<\/p>\n\n\n\n

If you make the cross, you will see that none of their daughters will have the disease because all the girls will receive one healthy, dominant X chromosome from the father irrespective of what chromosome they receive from the mother<\/p>\n\n\n\n

Making a cross of XY and Xx will give XX, Xx, XY and xy. So one of their daughters will be healthy(XX), the other is a carrier but still healthy(Xx), one son is healthy(XY) and the other son will have the disease(xy).<\/p>\n\n\n\n

The answers for this question are options b and c<\/strong><\/p>\n\n\n<\/span>\n

Question 2<\/h2>\n<\/span>\n\n\n

The pedigree below shows the inheritance of Tay-Sachs disease in a family. <\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Which of the following statements is true?<\/strong><\/p>\n\n\n\n

a. The probability of II\/3 and II\/4 to have a child with Tay-Sachs is 1\/4<\/p>\n\n\n\n

b. The probability of II\/3 to be a carrier is 1\/2<\/p>\n\n\n\n

c. The probability of II\/3 to be a carrier is 2\/3<\/p>\n\n\n\n

d. The probability of II\/3 and II\/4 to have a child with Tay-Sachs is 1\/9<\/p>\n\n\n\n

Tay-Sachs is an autosomal recessive disease. This means that I\/1, I\/2, I\/3 and I\/4 are all heterozygotes because they have children with the disease( II\/1 and II\/5).<\/p>\n\n\n\n

If we make a cross of Aa \u00d7 Aa, we see that out of all three healthy children, two are heterozygotes. So the probability of II\/3 and II\/4 to be heterozygotes is 2\/3 (we do 2 out of 3 and not 2 out of 4 because we already know that II\/3 and II\/4 are healthy so that excludes one of the options; the homozygous recessive). <\/p>\n\n\n\n

If we do not know whether or not the person is healthy or not, we do 2 out of 4 to include the homozygous recessive option. If we know that the person is healthy, we exclude the homozygous recessive option (because it is not a possible answer) and work with the 3 healthy options i.e 2 out of 3.<\/p>\n\n\n\n

If we make a cross between II\/3 and II\/4 (Aa \u00d7 Aa), we see that one out of four children will have the disease.
So the probability of II\/3 and II\/4 to have a child with the disease is 2\/3 \u00d7 2\/3 \u00d7 1\/4 = 1\/9
(2\/3 for the mother, 2\/3 for the father and 1\/4 for the child)<\/p>\n\n\n\n

The answers for this question are options c and d<\/strong><\/p>\n\n\n<\/span>\n

Question 3<\/h2>\n<\/span>\n\n\n

There are four genes; A, B, C and D present on a chromosome with gene C following gene A. The distance between the genes are given below: <\/p>\n\n\n\n

A-C: 12cM <\/p>\n\n\n\n

C-D: 57cM <\/p>\n\n\n\n

A-D: 45cM <\/p>\n\n\n\n

B-C: 10cM. <\/p>\n\n\n\n

Which of the following statements are true?<\/strong><\/p>\n\n\n\n

a. The order of the genes on the chromosome is D-A-C-B<\/p>\n\n\n\n

b. Genes B and D are linked<\/p>\n\n\n\n

c. Genes A and B are linked<\/p>\n\n\n\n

d. The order of the genes on the chromosome is B-A-C-D<\/p>\n\n\n\n

Genes are linked if the distance between them is less than 50cM<\/p>\n\n\n\n

The answers for this question are options a and c<\/strong><\/p>\n\n\n<\/span>\n

Question 4<\/h2>\n<\/span>\n\n\n

Elijah and Dora are a healthy couple who are looking to have children. Elijah\u2019s sister has an autosomal enzyme defect, Dora\u2019s maternal grandmother had the same disease and the rest of the family is healthy. <\/p>\n\n\n\n

Which of the following statements are true?<\/strong><\/p>\n\n\n\n

a. The defect is autosomal dominant<\/p>\n\n\n\n

b. The defect is autosomal recessive<\/p>\n\n\n\n

c. The probability of their child to have the same disease is \u00bc<\/p>\n\n\n\n

d. The probability of their child to have the same disease is 1\/12<\/p>\n\n\n\n

Autosomal enzyme defects are usually inherited in a recessive pattern. This means that for two healthy parents to have a child with an autosomal recessive trait, the both parents MUST be heterozygous for that trait( each parent has to contribute one mutant allele to the child)<\/p>\n\n\n\n

If Elijah’s sister has the defect, it means that his parents are heterozygous for it. Therefore the probability that Elijah is a heterozygote is 2\/3 (if you make a cross of Aa\u00d7Aa, out of three HEALTHY children, 2 are heterozygotes. We do 2 out of 3 and not 2 out of 4 because we already know that Elijah is healthy so that excludes one of the options; the homozygous recessive).<\/p>\n\n\n\n

If we do not know whether or not the person is healthy or not, we do 2 out of 4 to include the homozygous recessive option. If we know that the person is healthy, we exclude the homozygous recessive option (because it is not a possible answer) and work with the 3 healthy options i.e 2 out of 3.<\/p>\n\n\n\n

If Dora’s grandmother had the disease, it means that her mother is heterozygous for the trait(the grandmother contributed one mutant allele to the mother)
Since we know that Dora’s father is healthy and her mother is a heterozygote, the probability for her to be a carrier is 2\/4 (If you make a cross of AA\u00d7Aa, out of four HEALTHY children, two of them are carriers<\/p>\n\n\n\n

If Elijah and Dora are carriers(heterozygotes), the cross of Aa \u00d7 Aa will show that one out of all four children will have the disease.<\/p>\n\n\n\n

Therefore the probability for their child to have this disease is: 2\/3 \u00d7 2\/4 \u00d7 1\/4= 1\/12<\/p>\n\n\n\n

We include 2\/3 and 2\/4 in the calculation because the child will only have the disease if both parents are carriers(heterozygotes) so we need to account for this i.e if both parents were homozygous dominant, the probability for their child to have the disease would be 0, so we need to state that the child will only have the disease if the father AND the mother are carriers.
Hence we multiply all three chances (remember from Biostatistics that ‘AND’ means that we need to multiply.<\/p>\n\n\n\n

The answers for this question are options b and d<\/strong><\/p>\n\n\n<\/span>\n

Question 5<\/h2>\n<\/span>\n\n\n

Select the correct statements about the pedigree shown below.<\/strong> <\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The disorder is Duchenne\u2019s Muscular Dystrophy.<\/p>\n\n\n\n

a. The probability of III\/6 to be a carrier is 50%<\/p>\n\n\n\n

b. The probability of III\/6 to be a carrier is 100%<\/p>\n\n\n\n

c. The probability of IV\/2 to have the disease is 25%<\/p>\n\n\n\n

d. The probability of IV\/2 to have the disease is 12.5%<\/p>\n\n\n\n

I will explain this in a step by step manner. Before answering any pedigree question, it is important to figure out and understand the pedigree itself.<\/p>\n\n\n\n

Duchenne’s Muscular Dystrophy(DMD) is an X-linked recessive disease. This means that if I\/1 is healthy, he has a dominant X chromosome. I\/2 can be homozygous dominant or heterozygous but we know that she is heterozygous because her son(II\/3) has DMD(if the boy gets a Y chromosome from the father, the mutant X chromosome had to have come from the mother).<\/p>\n\n\n\n

II\/2 can have a homozygous dominant or heterozygous genotype (she will get one dominant X allele from I\/1 and can get either a dominant or recessive X allele from I\/2 ). Since III\/5 has DMD and II\/1 is healthy, the mutant X came from the mother(II\/2), so we can conclude that II\/2 is a heterozygote.<\/p>\n\n\n\n

III\/6 can either be a homozygous dominant or a heterozygote because she received one dominant X allele from II\/1 and can get either a dominant or recessive X allele from II\/2 (remember that we previously concluded that II\/2 is a heterozygote.
Therefore, III\/6 has a 50% chance to be a carrier of the trait(a heterozygote). We also know that III\/7 has a genotype of XY(DMD is X linked recessive so for a male to be healthy it means he posseses the wild type X allele).<\/p>\n\n\n\n

If III\/6 is not a carrier, none of her children will have the trait. If she is a carrier and we make the cross(Xx and XY), one out of all four children will have the trait, therefore, the probability that IV\/2 will have DMD is 1\/2 \u00d7 1\/4=1\/8=12.5% <\/p>\n\n\n\n

*See question 4 for an explanation on why we multiply these two numbers.<\/p>\n\n\n\n

The answers for this question are options a and d<\/strong><\/p>\n\n\n<\/span>\n

Question 6<\/h2>\n<\/span>\n\n\n

Select the correct statements about the pedigree shown below.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The disorder is Duchenne\u2019s Muscular Dystrophy.<\/p>\n\n\n\n

a. The probability of III\/6 to be a carrier is 50%<\/p>\n\n\n\n

b. The probability of III\/6 to be a carrier is 100%<\/p>\n\n\n\n

c. The probability of IV\/2 to have the disease is 25%<\/p>\n\n\n\n

d. The probability of IV\/2 to have the disease is 12.5%<\/p>\n\n\n\n

This question is similar to question 5 but in this case, IV\/1 has Duchenne’s Muscular Dystrophy(DMD) while in question 5, IV\/1 was healthy.<\/p>\n\n\n\n

Because of this, we can conclude that III\/6 is definitely a carrier because III\/7 is healthy and the mutant X chromosome had to have come from the mother.<\/p>\n\n\n\n

Therefore the probability that III\/6 Is a carrier is 100%<\/p>\n\n\n\n

The probability for IV\/2 to also have DMD is similar to the previous question. If we make a cross between III\/6 and III\/7, one out of all four children will have DMD, so the probability of IV\/2 to have the disease is 1 \u00d7 1\/4= 25%.<\/p>\n\n\n\n

We multiply by 1 here because we are 100% sure that III\/6 is a carrier for this disease.<\/p>\n\n\n\n

The answers for this question are options b and c<\/strong><\/p>\n\n\n<\/span>\n

Question 7<\/h2>\n<\/span>\n\n\n

Select the true statements below.<\/strong><\/p>\n\n\n\n

a. For linked genes, the amount of recombinant offspring is less than 50%<\/p>\n\n\n\n

b. For linked genes, the amount of recombinant offspring is more than 50%<\/p>\n\n\n\n

c. Phenotypic ratio of the F2 generation in a mendelian monohybrid cross is 3:1 if the gene exhibits incomplete dominance<\/p>\n\n\n\n

d. Phenotypic ratio of the F2 generation in a mendelian monohybrid cross is 1:2:1 if the gene exhibits incomplete dominance<\/p>\n\n\n\n

If two genes are linked, it means that they are very close to each other on the chromosome and will most likely be inherited together. This reduces the chances of cross over between both genes and reduces the number of novel genotypes i.e recombinant offspring that will be made.<\/p>\n\n\n\n

For non-linked genes i.e independently inherited genes, the amount of recombinant offspring will be approximately half of all offspring<\/p>\n\n\n\n

A Mendelian monohybrid F2 generation will produce 1 homozygous dominant, 2 heterozygotes and 1 homozygous recessive. In incomplete dominance, the heterozygote phenotype is different from that of the homozygous dominant phenotype.
So the phenotypic ratio will be 1:2:1<\/p>\n\n\n\n

The answers for this question are options a and d<\/strong><\/p>\n\n\n<\/span>\n

Question 8<\/h2>\n<\/span>\n\n\n

In a population where the number of males and females is the equal, a couple has two sons. <\/p>\n\n\n\n

Which of the following statements are true?<\/strong><\/p>\n\n\n\n

a. The probability that the third child will be a boy is greater than 50%<\/p>\n\n\n\n

b. The probability that the third child will be a girl is less than 50%<\/p>\n\n\n\n

c. The probability that the third child will be a boy is 50%<\/p>\n\n\n\n

d. The probability that the third child will be a girl is 50%<\/p>\n\n\n\n

The sex of a child is an independent process i.e whether the child will be a boy or a girl does not depend on previous pregnancies\/birth.<\/p>\n\n\n\n

No matter how many children a couple has had previously or the sex of the previous children, the probability for them to have a boy or a girl is the same and it is 50%<\/p>\n\n\n\n

The answers for this question are options c and d<\/strong><\/p>\n\n\n<\/span>\n

Question 9<\/h2>\n<\/span>\n\n\n
\"\"<\/figure>\n\n\n\n

Select the true statements concerning the pedigree above.<\/strong><\/p>\n\n\n\n

a. It is a maternally inherited trait<\/p>\n\n\n\n

b. It is an X-linked dominant trait<\/p>\n\n\n\n

c. If II\/3 mates with a man who lacks the trait, all their children will have it<\/p>\n\n\n\n

d. It is an autosomal dominant trait<\/p>\n\n\n\n

As I said in question 5, before answering any pedigree question, it is best to understand the pedigree first. So we will start with that.<\/p>\n\n\n\n

As seen here, all the offspring of I\/1 and I\/2 have the same trait that I\/2 has. This can be because of maternal inheritance, autosomal dominance or X-linked dominance (please refer to the Transmission genetics: Single Gene disorders lesson for more details)<\/em><\/p>\n\n\n\n

Now let us focus on generation III. As seen here, none of the children in III has this trait. This eliminates the possibility of the trait being autosomal dominant because if it were an autosomal dominant trait, at least one of the children should have it (II\/1 lacks the trait so he would be homozygous recessive, II\/2 would be heterozygous and if you make this cross, 50% of the children will be heterozygous so they would possess the trait).<\/p>\n\n\n\n

This also eliminates the possibility that it is an X-linked trait because if it is an X-linked trait, at least one of the children should have the trait(II\/1 will be recessive for the trait and II\/ 2 would be heterozygous for the trait. If you make the cross, 50% of the children would be heterozygous so they would possess the trait).<\/p>\n\n\n\n

So, we can assume that the trait is maternally inherited (please refer to the Transmission genetics: Single Gene disorders lesson for more details)<\/em><\/p>\n\n\n\n

The answers for this question are options a and c<\/strong><\/p>\n\n\n<\/span>\n

<\/h2>\n<\/span><\/span>

Question 1<\/h2>

Question 2<\/h2>

Question 3<\/h2>

Question 4<\/h2>

Question 5<\/h2>

Question 6<\/h2>

Question 7<\/h2>

Question 8<\/h2>

Question 9<\/h2>

<\/h2><\/div>","protected":false},"excerpt":{"rendered":"

Question 1 A healthy man marries a healthy woman. One of their sons has an X linked disease but the other one is healthy. Which of the following statements are true? a. The defect is X-linked dominant b. The defect is X-linked recessive c. None of their daughters will have the disease d. 50% of […]<\/p>\n","protected":false},"author":143,"featured_media":0,"parent":1346,"menu_order":6,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-4277","page","type-page","status-publish","hentry"],"yoast_head":"\nTransmission Genetics practice questions – Meddists<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/meddists.com\/learn\/pre-clinical\/medical-genetics\/transmission-genetics\/transmission-genetics-practice-questions\/\" \/>\n<meta name=\"twitter:label1\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data1\" content=\"11 minutes\" \/>\n<script type=\"application\/ld+json\" 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