micah janea s. dilag bsp 1-c

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EXERCISE 40 GENETICS AND

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Page 1: Micah Janea S. Dilag BSP 1-C

EXERCISE 40

GENETICS AND HEREDITY

Page 2: Micah Janea S. Dilag BSP 1-C

IntroductionHeredity,or the inheritance of traits,falls

into the realm of genetics. Genetic information,in the form of genes found in the 23 DNA molecules that you inherit from your mother and the 23 that you inherit from your father,is passed from generation to generation.Twenty-two pairs of your DNA molecules,or chromosomes,are called autosomes.The remaining pair are called the sex chromosomes.One member of each pair is inherited from one parent,the other member of the pair from the other parent.

Page 3: Micah Janea S. Dilag BSP 1-C

Each autosomes in a pair is homologous to the other member of the pair.With this arrangement,a person has two genes for every inherited characteristic.If one gene is always expressed,whether or not its mate is the same gene,geneticists call that gene dominant.A gene that is not expressed when its mate is different is termed a recessive gene.

All of the genes in all of the chromosomes together are called the genome.The study of all genes in the human species,or the human genome,is called genomics.

Page 4: Micah Janea S. Dilag BSP 1-C

A.Human Phenotypes and Genotypes

For any inherited characteristic,a phenotype can be identified.The phenotype is the characteristics actually expressed in an individual.A person’s genotype is a statement of both genes that influence a particular trait.For example,albinism is a recessive genetic condition in which a person lacks skin pigmentation.By convention,a gene for normal skin pigmentation,which is dominant,is presented as A.The recessive gene is represented as a.A person with a normal skin piigmentation may have the genotype AA,meaning that both genes are normal.

Page 5: Micah Janea S. Dilag BSP 1-C

Geneticists may state that the person’s genotype is homozygous,meaning that both genes are the same.A person with a normal skin color could also have the Aa genotype.Because A is dominant,the abnormal a is not expressed.This person has a heterozygous genotype,meaning that the two genes are not the same.A person with albinism must have the genotype aa,because that is the only genotype that allows the recessive abnormal trait to be expressed.Thus,the normal phenotype(normal skin color)may be associated with the genotype AA or the genotype Aa.The abnormal phenotype(albinism)is always associated with the genotype aa.

Page 6: Micah Janea S. Dilag BSP 1-C

What is Albinism?>Albinism is a 

congenital disorder characterized by the complete or partial absence of pigment in the skin, hair and eyes due to absence or defect of an enzyme involved in the production of melanin.

>the congenital absence of pigmentation in the eyes and skin and hair .

Page 7: Micah Janea S. Dilag BSP 1-C

Hint…Skin color is determined by a number of factors,both environmental and genetic.A number of different A genes exist in the human gene pool,and genes at other locations in the DNA may have an impact on skin color.The example of albinism is simple when first observed,but the influence of multiple factors complicates things.This is typical of inherited characteristics.Often,many genes influence one particular trait.Also,many environmental factors can affect inherited characteristics.

Page 8: Micah Janea S. Dilag BSP 1-C

ActivityThis activity instructs you to determine

your phenotype for a variety of easily observed characteristics.Once you know your phenotype,you can determine your possible genotypes.For example,if you are an albino,you know that your genotype is aa.If you do not have albinism,your genotype is either AA or Aa.For each characteristic described,record your phenotype and possible genotypes in Lab Report 40.

Page 9: Micah Janea S. Dilag BSP 1-C

Safety First!The First step of this activity

involves tasting test papers.Be sure to use clean procedures in handling the papers before and after you taste them.Used papers are to be placed into BIOHAZARD container immediately after use.Test papers should be tasted only once to avoid ingesting a high dosage.Women who are pregnant(or maybe pregnant)should not perform the taste tests.

Page 10: Micah Janea S. Dilag BSP 1-C

Set of Characteristics

1.The first set of characteristics to be determined involves the sense of taste.It is known that the ability to taste certain compounds depends on the presence of certain genes.You will taste papers but have each been impregnated with a different compound.Before starting, put a piece of CONTROL test paper on your tongue and chew it.If you taste something, you must be sensitive to a compound in the paper itself.Ignore that taste in the tests that remain,or try to sense differences in taste.

Page 11: Micah Janea S. Dilag BSP 1-C

•Sodium benzoate test-the ability to taste something sweet,salty,or bitter in the paper is dominant.

•PTC(Phenylthiocarbamide)test-the ability to sense a bitter taste is dominant.

•Thiourea test-the ability to taste something bitter is dominant.

Page 12: Micah Janea S. Dilag BSP 1-C

2.The next set of determinations involves anatomical characteristics of your hand.

•Bent little finger-Place relaxed hand flat on the lab table.If the distal phalanx of the little(fifth)finger bends toward the fourth finger, you have the dominant trait.

•Middigital hair-Dorsal hair on the skin over the midddke phalanges of the hand is dominant.

•Hitchhiker’s thumb-If you can hyperextend the distal joint of the thumb noticeably, you have the recessive trait.

Page 13: Micah Janea S. Dilag BSP 1-C

3.Determine your phenotype for these facial features:

• Pigmented anterior of the iris-If you have pigment on the anterior and posterior of the iris, your eyes are green,brown,black,or hazel.If you lack pigment on the anterior aspect of the iris,your eyes are blue or gray.Anterior pigmentation is dominant.

• Attached earlobes-If the inferior,fatty lobe of the ear is attched rather than free,you have the recessive trait.

Page 14: Micah Janea S. Dilag BSP 1-C

• Widow’s peak-Assuming you have a hairline,if it is straight across the forehead, you have the recessive trait.If it forms a downward point near the midline, you have the dominant widow’s peak.

• Tongue roll-Try your tongue as you extend it from your mouth.If you can’t curl it, you have the recessive trait.

• Freckles-If your face has a scattering of freckles, you have the dominant form of this characteristic.If your face is free of freckles,you have the recessive condition.

Page 15: Micah Janea S. Dilag BSP 1-C

4.These are two dominant genes for ABO blood types.One is Ia, which signifies the presence of the A antigen.The other is I,signifying the B antigen.The recessive gene is i signifying neither ABO antigen.A person with the phenotype A has the genotype or Ii.A person with a phenotype type B has the genotype II or Ii.A person with a phenotype type AB has the genotype II.Phenotype type O requires the genotype ii.Record your phenotype(your ABO blood type?).What is your genotype?

Page 16: Micah Janea S. Dilag BSP 1-C

Hint…If you have not already typed your blood(as instructed in Lab exercise 25),consult your health records or perform the typing now.Refer to Lab Exercise 25 and heed the safety advice given there.

• The Rh blood type is also determined by genetics.Presence of the Rh antigen is dominant.What is your Rh blood type?What is your genotype?

Page 17: Micah Janea S. Dilag BSP 1-C

B.PROBABILITIES OF INHERITANCENow that you have a grasp of the concept of

phenotype and genotype, you can move on to the concept of probability. Probability is the likelihood of a certain outcome in a particular event. As applied to human genetics, probability refers to the likelihood that the offspring of a particular set of parents will have a certain inherited condition. Genetic counselors work with prospective parents to determine their possible genotypes for a variety of traits.Then they predict the probability of their children having those traits.In this way,parents can anticipate possible genetic conditions.

Page 18: Micah Janea S. Dilag BSP 1-C

In this exercise,you will use a simple method developed by the English geneticist Punnett. He Devised a simple grid, or Punnett square, with which one can easily predict simple ratios of genetic probability.

Page 19: Micah Janea S. Dilag BSP 1-C

Reginald Crundall Punnett

>an English geneticist who developed the simple method which is known as the PUNNETT SQUARE.

Page 20: Micah Janea S. Dilag BSP 1-C

StepsFirst, the following is an example:• The father has freckles;the mother does

not.The gene for freckles is F,and the gene for no freckles is f.The father’s genotype is either FF or Ff.The mother’s genotype must be ff.We must consider two different outcomes in offspring from this couple because the father has two possible genotypes for this trait. First,let’s assume that the father’s genotype is FF.Any sperm cell conributed by the father will have the F gene..Any oocytee contributed by the mother will have f gene. We place the possible gene for the father in the left margin of the Punnett Square and the mother’s in the top margin:

Page 21: Micah Janea S. Dilag BSP 1-C

To use the Punnett square,start in the top left square of the grid.Combine the gene from the left margin with the gene from the top margin:

f f

F

F

f f

F

F

Ff

Page 22: Micah Janea S. Dilag BSP 1-C

We now have one possible offspring genotype:Ff.We then do the same for the other three squares of the grid:

f f f f

FF

FF

Ff Ff

Ff Ff

Page 23: Micah Janea S. Dilag BSP 1-C

Our Punnett square predicts a 100% probability that any one offspring will have genotype Ff and therefore have the freckled phenotype.The second possibility is that the father’s genotype is Ff.We can set up a Punnett square for this possibility: f f

F

f

Page 24: Micah Janea S. Dilag BSP 1-C

We can now fill in the grid with possible offspring genotypes, combining the gene from the left margin with the gene from the top margin for each square:

f f f f

FF

ff

Ff Ff

ff ff

Page 25: Micah Janea S. Dilag BSP 1-C

• Because half the squares have genotype Ff, there is a 50% probability that any offspring of this couple will have the Ff genotype (and therefore the freckled phenotype). There is also a 50% probability that the genotype will be ff, expressed as the non-freckled phenotype. If the phenotypes or genotypes of the client’s parents and grandparents are known, a genetic counselor might construct a pedigree, or family tree, to determine the father’s likely genotype. If that information is unavailable, the parents will have to deal with the two different probabilities

Page 26: Micah Janea S. Dilag BSP 1-C

Now that you have seen an example, try your hand at the following problems. Draw your Punnett squares and record your results in Lab Report 40.

• Huntington’s disease (HD), or Huntington’s chorea, is a degenerative nerve disorder with a genetic basis that becomes apparent after about the age 40. The abnormal gene that produces this disease, H, is dominant. The normal , necessive gene is h. One of Heather’s parents has HD but the other doesn’t. Can you predict the highest probability that Heather will develop HD later in her life?

Page 27: Micah Janea S. Dilag BSP 1-C

• Kevin has Rh-positive blood. His wife, Christine, has Rh-negative blood. Their first child, Andrew, has Rh-positive blood. Both of Kevin’s parents have Rh-positive blood. What is the probability that the child Kevin and Christine are expecting will be Rh-negative?

• Leo’s father has albinism, but Leo doesn’t. Cleo father has albinism, but she doesn’t. If Leo and Cleo have a child, what is the probability that it will have albinism? What is the probability that their second child will have albinism? Their third child?

• In the ABO blood typing system, Mario is type O. Ana is type AB. What ABO blood types might their children have?

Page 28: Micah Janea S. Dilag BSP 1-C

Trait Dominant Gene(s)

Recessive Gene

Your Phenotype Possible Genotypes

Sodium benzoate taste

S s

PTC taste P p

Thiourea taste T t

Bent little finger L l

Middigital hair M n

Hitchhiker’s thumb

H h

Pigmented anterior of Iris

I i

Attached earholes A a

Widow’s peak W w

Tongue roll R r

Freckles F f

ABO blood type A B

I , I

i

Rh-blood type D d

Page 29: Micah Janea S. Dilag BSP 1-C

Trait Dominant Gene(s)

Recessive Gene

Your Phenotype Possible Genotypes

Sodium benzoate taste

S s Recessive Ss,SS,ss,ss

PTC taste P p Dominant Pp,Pp,pp,ppThiourea taste T t Recessive Tt,TT,tt,ttBent little finger L l Recessive Ll,LL,ll,llMiddigital hair M n Recessive Mm,MM,mm,mmHitchhiker’s thumb

H h Recessive Hh,HH,hh,hh

Pigmented anterior of Iris

I i

Dominant Ii,II,ii,ii

Attached earholes A a Dominant Aa,AA,aa,aaWidow’s peak W w Recessive Ww,WW,ww,wwTongue roll R r Dominant Rr,Rr,rr,rrFreckles F f

Dominant Ff,Ff,ff,ffABO blood type A B

I , I

i Dominant A B A B

I I , I I , ii , iiRh-blood type D d Dominant Dd,Dd,dd,dd

Page 30: Micah Janea S. Dilag BSP 1-C

Questions in the problems:

1.What is the highest probability that Heather will develop Huntington’s disease?___%2.What is the probability that Kevin and Christine’s second child will have Rh-neagative blood.___%3.What is the probability that Leo and Cleo’s first child will be albino?___%Their second child?___%Their third child?___%4.What are the possibilities ABO blood types of Mario and Ana’s children?___%

Page 31: Micah Janea S. Dilag BSP 1-C

Dominant and Recessive Forms a.Bent fingers b.attached earlobe

c.Tongue Roll d.widow’s peak