Mendelian Genetics

LO 3.15 The student is able to explain deviations from Mendel’s model of the inheritance of traits. [See SP 6.5]

LO 3.16 The student is able to explain how the inheritance patterns of many traits cannot be accounted for by Mendelian genetics. [See SP 6.3]

LO 3.17 The student is able to describe representations of an appropriate example of inheritance patterns that cannot be explained by Mendel’s model of the inheritance of traits. [See SP 1.2]

A brief review of Mendel(and flower parts!)

Pisum sativum

Terms you must know!Haploid Dominant MonohybridDiploid Recessive DihybridHomozygous Genotype AlleleHeterozygous Phenotype Hybrid

Check out the ratios! You have to know how to getthe expected ratios to do a chi square problem thatdeals with a genetic data.

LO 3.9 The student is able to construct an explanation, using visual representations or narratives, as to how DNA in chromosomes is transmitted to the next generation via mitosis, or meiosis followed by fertilization. [See SP 6.2]

Law of Independent Assortment-All of the parents were yellow and smooth- but those traits didn’talways stay together in the offspring.

Since we know Mendelian genetics, we don’t need the Punnett square to tell us that we get a 9:3:3:1 phenotypic ratio. There is a 1/16 chance that the offspring will be short and white.

We also need to be able to calculate percentages-

For example in a dihybrid cross between two heterozygotes,If you have 360 offspring, what are your expected values? You know that you should have a 9:3:3:1 ratio

Both dominant phenotypes 9/16 = .56 = 56% = 202One dominant; one recessive 3/16 = .19 = 19% = 68One recessive; one dominant 3/16 = .19 = 19% = 68Both recessive phenotypes 1/16 = .06 = 6% = 22

Let’s plug this into the Hardy-Weinberg equation and see if we get the same numbers. Of the 180 flowers that are tall, how many are homozygous, and how many are heterozygous?

22% is .22.22 = q2

.47 = q

What is the frequency of the dominant allele?

.53 = p

.53 x .53 = .28

.28 x 360 = 101 homozygous tall

2 (.53)(.47) = .5.5 x 360 = 179 heterozygous (tall)

So, we calculated that 180 of these flowers would be tall; now we know how many of the tall flowers are homozygous and how many are heterozygous

22% are short. What is the frequency of the recessive allele?

In a species of gecko, spots and tail size are controlled by genes that assort independently.A= spotted B= fat taila = solid b = skinny tailA scientist collected data from several gecko breeders and got the phenotypes of 135 baby geckos born within a 6 month period. Calculate the percentage of the geckos with the 4 possible phenotypes and the number of each inthis data sample if all of the parents in the trial were heterozygotes.

56% 7519% 2619% 26 6% 8

In a species of gecko, spots and tail size are controlled by genes that assort independently.A= spotted B= fat taila = solid b = skinny tail

What if the cross wasn’t between two heterozygotes? What if this was the cross? What percentages would you expect in the offspring?

aabb x AaBb

25% spotted/fat tail25% solid/fat tail25% spotted/skinny tail25% solid/skinny tail

P= smooth seeds x wrinkled seedsF1= all smooth seedsF2= 5,474 smooth seeds and 1,850 wrinkled seedsCalculate chi-square for this data and evaluate your results..

Null hypothesis- There is nosignificant difference betweenthe observed and expectedratio of smooth and wrinkledseeds.

P= smooth seeds x wrinkled seedsF1= all smooth seedsF2= 5,474 smooth seeds and 1,850 wrinkled seedsCalculate chi-square for this data and evaluate your results..

Null hypothesis- There is no significant difference betweenthe observed and expected ratio of smooth and wrinkled seeds.

I accept my null hypothesis. Mychi square value of 0.263 is lessthan the critical value of 3.84.

You don’t need Punnett squares, you can determine expected ratios

using probability

Consider a trihybrid cross-

If a plant that is heterozygous for all three characteristics is allowed to self-fertilize, what proportion of the offspring would you expect to be as follows- AaBbDd x AaBbDd

a.homozygous for the 3 dominant traits?

There is a 1/4 chance for each pair of alleles to be homozygous dominant.

1/4 x 1/4 x 1/4 = 1/64

b. homozygous for purple and tall; heterozygous for seed shape

The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring will have the following genotypes?

a. AABBCCDD

b. AaBBccDd

c. AaBBCCdd

¼ x ¼ x ¼ x ¼ = 1/256

½ x ¼ x ¼ x ½ = 1/64

½ x ¼ x ¼ x ¼ = 1/128

LO 3.15 The student is able to explain deviations from Mendel’s model of the inheritance of traits. [See SP 6.5]

LO 3.16 The student is able to explain how the inheritance patterns of many traits cannot be accounted for by Mendelian genetics. [See SP 6.3]

Incomplete Dominance

F1 hybrids have an appearance that is between that of thetwo parents. Breeding two of the hybrids produces a 1:2:1 ratio of phenotypes in the F2

Don’t just know the definition ofincomplete dominance; recognizethe ratio and know how it differsfrom a typical Mendelian cross. Youmay be given the results of a geneticcross and be asked to explain howyou would get these results.

Beyond Mendel

Beyond Mendel-

Epistasis- a gene atone locus alters the effectsof a gene at another locus

B- grayb- tanC- pigment c- no pigment

If the mouse doesn’t havea capital C, regardless of thegene for coat color, nopigment will be deposited inthe fur WONKY RATIO

Two genes play a critical role in normal hearing; one is involved in the development of a normal cochlea (A). The other is responsible for a normal auditory nerve (B). Either of these genes can cause deafness if the genotype is homozygous recessive.

A man and a woman are both deaf; one as a result of a defective cochlea and the other as a result of a defective auditory nerve. Assume that they are homozygous dominant for the other gene. If they have children together (barring mutation), what is the probability that the child will be deaf?

0 AAbb x aaBB all AaBb

Beyond Mendel- Codominance & Multiple AllelesMore than two alleles for a trait, and two of those alleles show dominance

Two parents, one type A and one type B, have a type O child? Explain that!!

In the U.S. about 16% of the population is Rh negative. The allele for Rh negative is recessive to the allele for Rh positive. If the student population of a high school in the U.S. is 2000, how many students would you expect for each of the three possible genotypes?

16% = .16 = q2

q = .4p = .6

p + q = 1 p2 + 2pq + q2 = 1

Homozygous Dominant (++) 36% 720 Heterozygous (+-) 48% 960Homozygous recessive (--) 16% 320

Beyond MendelPolygenic Inheritance

Two or more geneshave an additive effecton a single character inthe phenotype

In humans, what are some other characteristicsthat are polygenic?

Beyond Mendel-

Pleiotropy

One gene =Multiple effects

Beyond Mendel-

Sex linked genes are located on the sex chromosomes (x and y). They were discovered by Thomas Morgan during his work with Drosophila. If a sex linked gene is on the X chromosome, a man who receives that X chromosome will have the trait. Since a woman has two X chromosomes, it may not be expressed in her.

Some sex linked disordersin humans-HemophiliaRed/green colorblindnessDuchennes Muscular Dystrophy

In fruit flies, the mutation for bar eyes is a sex linked dominant trait. A pure bar-eyed female is crossed with a wild (red-eyed) male. Which of the following statements gives the correct percentage of male offspring that will have bar eyes and the correct explanation for it?

a. 50%, because all the males will inherit the bar-eyed trait from the mother

b. 50%, because all the males will inherit the bar-eyed trait from the father

c. 100%, because all the males will inherit the bar-eyed trait from the mother

d. 100%, because all the males will inherit the bar-eyed trait from the father

Insert slide on how environment influences phenotype

X-inactivationX-inactivationFemales have two X chromosomes;Males only have one

Exactly what does this tell you about the disorder?

If you are heterozygous, you are a carrier of the disorder. You don’thave the disorder, but if you have children with another carrier, they havea 25% chance of having the disorder.

Examples of autosomal recessive disorders

Cystic fibrosis- primarily Caucasians, 1 in 2500 births. 1/25 is a carrierAbnormal protein is a membrane protein that transports chloride ions.Results in thick sticky mucus in lungs, digestive tract and pancreasNo cure, but can be treated

Tay Sachs Disease- Primarily Jews of European descent and Cajuns1 in 3600 birthsNonfunctional enzyme cannot breakdown lipids in brain cellsSeizures, blindness, degeneration of motor and mental functionChild dies before 5 years old

Sickle Cell AnemiaPrimarily African; 1 out of 400 African AmericansPoint mutation in hemoglobinWhen oxygen levels are low, hemoglobin crystallizes

Pittsburgh Steelers safety Ryan Clark will not play in Sunday’s playoff game against the Denver Broncos because of a blood disorder bolstered by high-altitude, low-oxygen conditions. Clark, 32, has sickle cell trait, which means he carries an abnormal version of the hemoglobin gene. Although he’s spared the severe symptoms of sickle cell disease (in which both versions of the gene — one from each parent — are abnormal), Clark could suffer life-threatening organ damage playing in Denver’s mile-high stadium.A 2007 game in Denver sent Clark into sickle cell crisis, a complication that cost him his spleen and gallbladder and ended his season.

Huntington Disease

Achondroplasia

Neurofibromatosis

Achondroplastic dwarves are heterozygous; the homozygousdominant condition is lethal. What is the probability that twoachondroplastic dwarves will have a child of normal height?

It depends on who you ask!!If A = achondroplasia and a = no achondroplasia

AA= lethalAa= achondroplasiaaa= no achondroplasia

Aa x Aa

A

A

A A A

A

a

a a

a

a a

25% lethal50% heterozygous (achondroplasia)25% normal height

66% heterozygous (achondroplasia)33% normal height

LO 2.28 The student is able to use representations or models to analyze quantitatively and qualitatively the effects of disruptions to dynamic homeostasis in biological systems. [See SP 1.4]

Dihybrid Cross with Linkage

• In a normal case of Mendelian dihybrid inheritance with independent assortment of alleles, a cross between two heterozygotes produces the expected 9:3:3:1 ration in the offspring. In cases of dihybrid inheritance involving linkage, the offspring of a cross between two heterozygotes produces a greater than predicted number of parental types and a significantly smaller number of recombinants.

Linked Genes - genes that are on the same chromosome tend to be inherited together (duh)….

Unless they are separated by crossing over (recombination)

Go back and look at the data on the previous page.Calculate the RF- Answer is on the next page

There are 46 recombinants and a total of 427 offspring.46/427 = .107 or 11%

Try another one-In fruit flies, long wings (A) and gray bodies (B) are dominant to vestigial wings and black bodies. In a cross of two heterozygotes AaBb x AaBb you expect a 9:3:3:1 ratio. These are your results123 long wing, gray body21 long wing, black body27 vestigial wing, gray body129 vestigial wing, black body

Calculate the cross over value (recombination frequency) for the offspring of the test cross.

48 recombinants divided by total of 300 = .16 or 16%

There are 3 genes on a single chromosome: A, B and C. They exhibit the following crossing over frequencies:

• A-B = 35%• B-C = 10%• C-D = 15%• C-A =25%• D-B=25%• Determine the order of the genes on the

chromosome.

Start with the longest one and space itout. I marked my “chromosome” intoincrements of 5. Answer appears onthe click. Remember it could be in either order.

A D C B

Determine the sequence of genes along a chromosome based on the following recombination frequencies.A-B = 8%A-C = 28%A-D= 25%B-C = 20%B-D = 33%

B D

Do the largest first and then work fromhere. You have to put a pencil to it anddraw it out. It may seem hard at first, thenit gets easier. I filled in the first one- theanswer will appear when you click again. Map units were measured with the space bar.

C A

IMPORTANT NOTE: If this were multiple choice, these genes could be listedCBAD or DABC. Gene mapping only gives you the relative distance betweenthe genes- not their specific locus.

Another way the question may be asked-

Another way the question may be asked-

This slide combines the two skills we have just reviewed!

I don’t know about you, but this is how I feel right now…. Wonky.

P= smooth seeds crossed with wrinkled seedsF1= all smooth seedsF2= 5,474 smooth seeds and 1,850 wrinkled seedsCalculate chi-square for these results_________________What is your null hypothesis?Is the chi-square you have calculated within the boundary of “the possible”? Explain.

Phenotypes O E O-E (O-E)2 (O-E)2

E

Total XXXXXXXX XXXXXXXXX

Phenotypes O E O-E (O-E)2 (O-E)2

E

smooth 5,474 5,493 -19 361 .07

wrinkled 1,850 1,831 19 361 .20

Total XXXXXXXX XXXXXXXXX .27

Where did your expected value come from? You have to know what isthe probability of each phenotype in a genetic cross. In this cross itshould be 3:1 …..75% should be smooth and 25% should be wrinkled.There is no way you can calculate chi square if you can’t do that step.

In this case you cannot reject your null hypothesis because the chi squarevalue of .27 is less than the critical value of 3.84.

A wild type fly (heterozygous for gray body color and normal wings) is mated with a black fly with vestigial wings. The offspring have the following phenotypic distributions:

Wild type- 50Black, vestigial- 50Black, normal-5Gray, vestigial- 5

(This is the cross AaBb x aabb this info was not in the original problem)

Phenotypes O E O-E (O-E)2 (O-E)2

E

Total XXXXXXXX XXXXXXXXX

Phenotypes O E O-E (O-E)2 (O-E)2

E

Wild (gray, normal)

50 27.5 22.5 506.25 18.4

Black/vestigial 50 27.5 22.5 506.25 18.4

Black/normal 5 27.5 -22.5 506.25 18.4

Gray/vestigial 5 27.5 -22.5 506.25 18.4

Total XXXXXXXX XXXXXXXXX 73.6

In this case we will definitely reject our null hypothesis becauseour chi square value is much greater than the critical valueof 7.82. Really wonky, in fact, you should be able to explainwhat this is based on the data!!

You can also use chi square as a way to analyze your data in any type of experiment where this is an “expected” value to compare your “observed” to. See if this experiment looks familiar-

What is the probability that each of the following pairs of parents will produce the indicated offspring? (Assume independent assortment of all gene pairs.)

a. AABBCC x aabbcc------------------> AaBbCc

b. AABbCc x AaBbCc-------------------> AAbbCC

c. AaBbCc x AaBbCc--------------------> AaBbCc

d. aaBbCC x AABbcc--------------------> AaBbCc

1 x 1 x 1 = 1

½ x ¼ x ¼ = 1/32

½ x ½ x ½ = 1/8

1 x ½ x 1 = 1/2