UNIT 2 Mendelian Genetics

Question 1

An allele is:

A. another word for a gene
B. a homozygous genotype
C. a heterozygous genotype
D. one of several possible forms of a gene

Answer 1

D. one of several possible forms of a gene

Question 2

True or False

Sister chromatids are exact replicas of each other, and so with homologous chromosomes.

Answer 2

False

Sister are exact replicas, homologous are not.

Question 3

The DNA sequence for the trait is physically found in?

Answer 3

Gene

Question 4

True or False

Homologous chromosomes are connected at the centromere during the interphase.

Answer 4

False

Explanation: Sister chromatids are connected at the centromere. The chromosomes in interphase are not doubled yet.

Question 5

Which of the following statements is true about Mendel?

A. His discoveries concerning genetic inheritance were generally accepted by the scientific community when he published them during the mid-19th century.
B. He believed that the genetic traits of parents will usually blend in with their children.
C. His ideas about genetics apply equally to plants and animals.

Answer 5

C. His ideas about genetics apply equally to plants and animals.

Explanation:
A (his discoveries were not believed until chromosomal theory were surfaced)
B (he did not follow the blending theory, he followed the particulate theory)

Question 6

The molecular/cellular basis of inheritance that was drawn from Mendel’s work and is now correlated with what happened in chromosomes is known as?

Answer 6

Sutton-Boveri Theory (of Chromosomal Inheritance)

Question 7

The idea that for any particular trait, the pair of alleles of each parent separate and only one allele for each parent passes to an offspring is Mendel’s principle of:

A. independent assortment
B. hybridization
c. segregation
d. dominance

Answer 7

C. segregation

Question 8

During ____, alleles separate, such that each ____ is equally likely to receive either of of the two alleles present in the diploid individual.

A. mitosis, somatic cell
B. independent assortment, stem cell
C. S phase, gamete
D. meiosis, gamete

Answer 8

D. (meiosis, gamete)

Question 9

True or False

Law of Segregation is easily proven by a monohybrid cross.

Answer 9

True

Explanation: Monohybrid cross is used for both dominant-recessive principle and segregation principle.

Question 10

In guinea pigs, black hair (B) is dominant to brown hair (b) and short hair (H) is dominant to long hair (h). A black, long-haired guinea pig (Bbhh) is crossed with a brown, short-haired guinea pig (bbHh). What percentage of the offspring will be black with long hair.

A. 50%
B. 12.5%
C. 25%
D. 75%
E. 100%

Answer 10

C. 25%

Explanation: Each of them have four possibilities of offspring, so it is 1/4.

Question 11

Dihybrid Cross is related to the law of:

A. Law of Dominance
B. Principle of Segregation
C. Principle of Independent Assortment

Answer 11

C. Principle of Independent Assortment

Question 12

Both loci assort independently; One allele at each locus is completely dominant; and each of four possible phenotypes can be distinguished unambiguously, with no interactions between the two genes that would alter the phenotypes. The usual phenotypic ratio is?

Answer 12

9:3:3:1

Question 13

Which is true of Mendelian genetics?

A. Mendelian inheritance is seen in the way several genes interact to produce eye color.
B. The traits that strictly follow Mendelian inheritance are rarer than polygenic or gene interactive kinds.
C. Mendel’s ratios holds true even in the case of linkage because it is universal.
D. Mendelian genetics follow the blending in hypothesis of trait inheritance

Answer 13

B. The traits that strictly follow Mendelian inheritance are rarer than polygenic or gene interactive kinds.

Explanation:
A (Gene interaction)
C (False, linkage HIDES independent assortment and segregation)
D (False, Mendelian counteracts Blending Theory and followed the Particulate Theory)

Question 14

_____ disease is a term for diseases that can be accurately predicted via Mendelian genetics alone.

A. Polygenic
B. Single-gene
C. Pleiotropic

Answer 14

B. Single-gene

Question 15

Which is an example of a possible single-gene disease?

A. Hemophilia A
B. Diabetes
C. Cervical Cancer

Answer 15

A. Hemophilia A

Also sickle cell disease and muscular dystrophy.

Question 16

This consists of sick children and their parents can reveal whether the child inherited two disease-causing mutations from carrier parents, or whether a dominant mutation arose anew, termed “de novo.”

Answer 16

Test of trios

Question 17

What can be true of the concept of dominance?

A. Dominant traits more often are a gain of function
B. Gain of function accounts for heterozygotes expressing the trait just as homozygotes do in dominant inheritance
C. Dominant disease whose symptoms do not appear until adulthood, or that do not severely disrupt health, remain in a population because they do not prevent a person from having a children and passing on the mutations.
D. Both A & C
E. All of the above

Answer 17

E. All of the above

Question 18

An intermediate form of disease happens in heterozygotes of single gene disease that follow dominant inheritance where they show symptoms in certain extreme conditions. An example of this is?

A. Alpha-thalassemia disease
B. Sickle cell trait
C. Huntington’s chorea
D. Juvenile onset diabetes

Answer 18

B. Sickle cell trait

Sickle cell trait is heterozygote expressing some of the symptoms in some conditions. nu dw

Question 19

What is true of recessive genes?

A. More often results in a gain of function
B. Carrier individuals who show up with the disease because the recessive gene encodes for a protein that promotes abnormal function
C. Traits from recessive genes, when it comes to disease, tend to appear later in life and are less severe
D. Disease-causing recessive alleles remain in populations because healthy homozygotes pass them to future generations
E. Consanguinity promotes the appearance of recessive genes in the population

Answer 19

E. Consanguinity promotes the appearance of recessive genes in the population.

Explanation:
A (False, should be LOSS of function)
B (Heterozygote of dominant trait, carrier individuals do not express phenotype most of the time as one functional gene is enough to show a proper trait [or wild type phenotype])
C (False, they tend to appear EARLIER in life and MORE severe)
D (False, should be HEALTHY heterozygotes)

Question 20

Dwarfism is an autosomal dominant trait. If a man with dwarfism (DD) married to a woman without the trait (dd), what are the chances of having a son that is of normal height? (heterozygotes)

A. None
B. 25%
C. 50%
D. 100%

Answer 20

A. None

Explanation:
If it is autosomal dominant, there is no heterozygosity around the X/Y chromosome. All of the children would suffer to dwarfism.

Question 21

Cystic fibrosis is an autosomal recessive trait. If the mother has cystic fibrosis and the father (homozygous dominant) does not have the gene, what is the probability that a male offspring will present signs and symptoms of the disease?

A. None
B. 25%
C. 50%
D. 75%
E.100%

Answer 21

A. None

Explanation:
The sons will be heterozygotes, X chromosome does not have influence.

Question 22

True or False

In non-Mendelian traits, the Mendelian principles and laws still operate but are hidden by environmental and gene-to-gene interactions.

Answer 22

True

Example are phenocopies and genetic heterogeneity.

Question 23

True or False

Non-Mendelian traits appear to skew the ratios provided by the Mendelian laws.

Answer 23

True

Linkage is one of the example.

Question 24

When genes are close to each other on the same chromosome, they usually do not segregate at random during meiosis and therefore their expression does not support Mendel’s predictions. This is called?

A. Sex-linked trait
B. Linkage
C. Polygenic trait
D. X-linked

Answer 24

B. Linkage

Question 25

True or False Some DNA sequences are nearly always inherited together, like two inseparable friends and causes non-random associated between DNA sequences called linkage disequilibrium (LD)

Answer 25

False ## Footnote Segregation provides the randomization, linked genes become non-random.

Question 26

What is the physical grouping of genomic variants (or polymorphisms) that tend to be inherited together?

Answer 26

Haplotype | They are NON-RANDOM, can trace them through populations.

Question 27

Which is true of sex-linkage? A. Sex linkage refers to a gene being linked to autosomal chromosome. B. Genes on the Y-chromosome have different patterns of expression in females and males because a female has two X chromosomes and a male just one. C. The X chromosome includes 4 percent of all the genes in the human genome, but accounts for about 10 percent of Mendelian (single-gene) diseases. D. The human male is considered homozygous for X-linked traits, because he has only one set of X-linked genes. E. No other Y-linked traits besides infertility are yet clearly defined, but if there are more, they will be passed from mothers to sons.

Answer 27

C. The X chromosome includes 4 percent of all the genes in the human genome, but accounts for about 10 percent of Mendelian (single-gene) diseases. ## Footnote Explanation: A (linked “ON”, potaena ewan q baket) B (False, barely find genes in Y-chromosome - should be X-chromosome) D (Human male is considered hemizygous) E (If they are more, it should be passed from FATHERS to son)

Question 28

Which is not true of sex-influence? A. A sex-limited trait that affects a structure or function of the body that is present in only males or only females. B. With a sex-influenced trait, an allele is dominant in one sex but recessive in the other. C. The difference in expression can be caused by hormonal differences between the sexes. D. Sex-limited traits are passed solely in a sex-linked manner.

Answer 28

D. Sex-limited traits are passed solely in a sex-linked manner. | Sex-influence traits are AUTOSOMAL TRAITS.

Question 29

If a female offspring has 50% chance of having Duchenne Muscular Disease (X-linked dominant trait), which of the following is true regarding the parents? (Bonus: DMD is x-linked recessive) A. The father and mother have the disease B. The father has the disease while the mother does not have the gene C. The father has the disease and the mother is a carrier D. The father is normal, mother has the disease

Answer 29

C. The father has the disease and the mother is a carrier Explanation: binonus ni sir but if it is x-linked recessive, C ang answer because DMD is an AUTOSOMAL RECESSIVE.

Question 30

Which is true of X-linked recessive traits? A. Never expressed in male B. Affected males inherit the trait exclusively from homozygous mother C. Expressed always by heterozygote female D. Affected female may inherit the trait from affected father

Answer 30

D. Affected female may inherit the trait from affected father ## Footnote Explanation: A (they CAN be expressed, because they are hemizygous) B (they can also inherit from a HETEROZYGOUS MOTHER if it carries the X-linked recessive, as it can have two different kind of alleles in two X) C (incorrect, should be HOMOZYGOTE; heterozygote female are normal meaning they are carriers [very rare])

Question 31

In familial hypercholesterolemia, the phenotypes parallel the number of receptors- individuals with two mutant alleles die in childhood of heart attacks, those with one mutant allele may suffer heart attacks in young adulthood, and those with two wild type alleles do not develop this inherited form of heart disease. The reason heterozygotes do not have heart attacks in childhood even with inheriting a dominant allele is because of? A. Codominance B. Incomplete dominance C. Epistasis D. Sex-liked inheritance

Answer 31

B. Incomplete dominance | Explanation: Blended heterozygotes.

Question 32

Homozygotes L^M L^M and L^N L^N have only M or an N antigens, respectively, on the surface of their red blood cells. However, heterozygotes L^M L^N have both types of antigens in equal numbers on the cell surface. The MN blood group then shows: A. Codominance B. Incomplete dominance C. Epistasis D. Sex-linked inheritance

Answer 32

A. Codominance ## Footnote M and N antigens are part of minor blood groups outside of ABO and RH. They are still be antigenic and cause transfusion reactions, M and N follow the dominance pattern as they have a 50/50 amount of antigen on the surface of RBCs. M is co-dominant with N.

Question 33

A certain phenotype, involving the ABO blood group system. Individuals with this phenotype lack a protein called the H antigen (genotype hh), which is used to form A and B antigens. Even though such individuals may have A or B genes, they appear to be blood group O because they lack the H antigen. Which concept is this? A. Codominance B. Incomplete dominance C. Epistasis D. Sex-linked inheritance

Answer 33

C. Epistasis | Bombay phenotype hides the A and B phenotype even if they exists.

Question 34

____ occurs when one gene influences two or more seemingly unrelated phenotypic traits. Mutations in this gene may have an effect on several traits simultaneously, due to the gene coding for a product used by a myriad of cells or different targets that have the same signaling function. A. Variable expressivity B. Pleiotropy C. Penetrance D. Phenocopies

Answer 34

B. Pleiotropy

Question 35

There is a rare congenital anomaly where the proximal aspect of an extremity is absent with the hand or foot attached directly to the trunk. A certain drug mimics the phenotype of such birth defect when taken during pregnancy. The disease caused by the drug is a ____ of the rare genetic defect. A. Variable expressivity B. Pleiotropy C. Penetrance D. Phenocopy

Answer 35

D. Phenocopy | Phocomelia is similar with the drug in those condition (»»!»!>!)

Question 36

Huntington's disease (HD) is a fatal autosomal dominant neurodegenerative disorder caused by a CAG repeat expansion in the huntingtin gene (HTT). For those with 40 or more CAG repeats, ____ is almost 100%, with all individuals with such genotype showing signs of severe disease. A. Variable expressivity B. Pleiotropy C. Penetrance D. Phenocopy

Answer 36

C. Penetrance | Explanation: Pag percentage, penetrance.

Question 37

Which of the following is an example of a disease with genetic heterogeneity? A. The different phenotypes of neurofibromatosis. B. The different genes that interact to cause Type II Diabetes Mellitus. C. The different kinds of inheritance and genes that all present as osteogenesis imperfecta. D. Some people with Marfan syndrome only develop loose joints, while some develop weakened aortas.

Answer 37

C. The different kinds of inheritance and genes that all present as osteogenesis imperfecta.

Question 38

Which of the following is an example of a disease with variable expressivity? A. The different phenotypes of neurofibromatosis. B. The different genes that interact to cause Type II Diabetes Mellitus. C. The different kinds of inheritance and genes that all present as osteogenesis imperfecta. D. None of the above.

Answer 38

A. The different phenotypes of neurofibromatosis.

Question 39

Which of the following is an example of a disease which is polygenic? A. The seemingly different phenotypes of neurofibromatosis. B. The different genes that interact to cause Type II Diabetes Mellitus. C. The different kinds of inheritance and genes that all present as osteogenesis imperfecta. D. None of the above.

Answer 39

B. The different genes that interact to cause Type II Diabetes Mellitus.

Question 40

The X-linked recessive traits of color blindness is present in 5% of males. If a mother who is a carrier and a father who is unaffected plan to have 2 children, what is the probability the children will both be male and color-blind? A. 50% B. <1% C. 25% D. 6.25%

Answer 40

C. 25% ## Footnote Explanation: The probability of having a male child is 1/2. The probability of a male child being colorblind is also 1/2. Since the child has to be a male and color blind at the same time, we apply the product rule. Since the father is color blind, he will always pass on his mutated gene to his male children. - The mother, being a carrier, can pass on either her normal or mutated gene to her children. - The probability of having a male child is 50%. - The probability of passing on the mutated gene to a male child is 50%. - Therefore, the probability of having both male children and both of them being color blind is 50% * 50% = 25%. The statement 5% does not have to do with anything.