HOW DO BIOLOGISTS STUDY NATURE

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What does Mendelian genetics teach us about heredity? Mendelian Genetics Part I

Learning Outcomes

By the conclusion of todays laboratory you should be able to:Distinguish between genes and alleles.

Use the correct terms to describe the genetic makeup of individuals (e.g. dominant, recessive, homozygous, heterozygous, carrier)

Describe the mechanisms behind inheritance and how genetic traits are passed from one generation to another.

Use Punnett squares in order to solve genetics problems.

Determine genotypes and phenotypes of offspring resulting from monohybrid (single heterozygous gene) and dihybrid (two different heterozygous genes) crosses.

6. Statistically test the results of genetic crosses.

introduction

A quick look around the classroom, and it should be obvious that there is variation in nature. It is this variation that drives many biological questions and in fact is why most of us are biologists. This variation may arise through unique events called mutations that change the genetic material directly or by the random recombination (shuffling) of existing genetic material contributed by the mother and father at the time of conception. This genetic variability, a consequence of sexual reproduction, assures that at least some offspring will survive even if environmental conditions, and hence selection pressures, change.

At the time Gregor Mendel began his studies of genetics in garden peas, two competing theories existed which attempted to explain variation and the patterns of inheritance in organisms. One was called blending inheritance. This theory claimed that the traits observed in the parents blend together to form an intermediate trait observed in the offspring. Another theory called inheritance of acquired characters claimed that traits present in parents are modified through use or disuse, and passed on to offspring in the modified form. We now know that both of these theories are incorrect. Gregor Mendels great contribution was that through meticulous experimentation, he demonstrated that discrete particles were passed on from parents to offspring, and depending on the specific nature of these particles, unique traits were observed in the offspring. Therefore, Mendel showed that traits were inherited as particles rather than by blending. Although Mendelian genetics does not explain every type of inheritance, its applicability is widespread enough that it has remained as one of the most important discoveries of the biological sciences, and remains in use today. It is now referred to as the particulate theory of inheritance, and it is credited to Gregor Mendel. This theory observes that traits are passed from one generation to the next through discrete particles (genes) that retain their ability to be expressed, even though they may not appear in every generation.

Gregor Mendel (1822-1884) was an Austrian monk who worked with the common garden pea, Pisum sativum. He conducted his experiments within a monastery garden in almost complete anonymity. It was only after his death that the results of his experiments became public.

His choice to use pea plants was a perfect one. These plants are easily cultivated and grow rapidly. Different varieties have clearly different characteristics that are true breeding due to the fact that they self pollinate. Pea plants self pollinate because the reproductive parts of the pants are completely enclosed within the petals. Because of this self pollinating structure, accidental cross pollination is not possible, and allows for easy manipulation.

Mendel started with 32 different varieties of pea plants. He then tracked 7 traits: seed form and color, flower position and color, pod form and color, and stem length. Mendel actually employed cross pollination in his experimental crosses. He removed the anthers from one plant (A), collected pollen containing male gametes from another plant (B), and transferred the pollen to the stigma (containing female gametes) of plant A. Mendel found in every case that the F1 (first filial) generation was not a blending of the two traits, but rather only one character was produced. However, when the F1 generation was left to self pollinate, the unexpressed trait reappeared in the second generation (F2) along with the trait of the F1 generation. Mendel termed the trait that appeared in the F1 generation Dominant, and the trait that reappeared in the F2 generation recessive. When both the dominant and recessive traits are present the dominant trait will be expressed. The results of Mendels experiments (Table 3.1) lead to the particulate theory of inheritance, which, as indicated above, states that characteristics are passed from one generation to the next through discrete particles (genes) that retain their ability to be expressed, even though they may not appear in every generation.

Mendel called the unit of inheritance a particle because at the time DNA had not yet been discovered. Today we know that these particles Mendel wrote about are genes. Genes are particular DNA sequences that code for an amino acid sequence or protein. Variability in genes allows us to see differences in the population. These variations or alternate forms of the same gene are called alleles. If one allele can mask the effect of the other, it is considered to be the dominant allele, whereas the allele whose expression is hidden is called the recessive allele. Recessive traits are only expressed if the recessive allele is present on both of the homologous chromosomes. This condition is said to be homozygous recessive. The dominant allele, however, is expressed if it is present on one or both of the homologous chromosomes. If the dominant allele is present on both homologous chromosomes, it is termed homozygous dominant. If the dominant allele is found on one homologous chromosome and the recessive allele on the other, then this situation is termed heterozygous. The allele combinations that are actually present refer to the individuals genotype.

By convention, capital letters are used to represent the dominant allele and lower case letters represent the recessive allele. For example, the homozygous recessive genotype could be designated "aa", the homozygous dominant genotype could be "AA" and the heterozygous genotype could be "Aa". Heterozygous individuals are often referred to as carriers of the recessive condition. Although the recessive allele is not expressed, it is "carried" by the individual and is available to be passed on to the next generation. The alleles present, along with the dominance relationships involved, determine what the individual will look like. This outward manifestation or expression of an individual's genotype is called phenotype.

Table 3.1: A summary of Mendel's work with pea plants. The number of plants possessing each characteristic is indicated. TraitOriginal Cross(F2) second generation

Dominant x RecessiveDominantRecessiveTotal

Seed formSeed ColorFlower PositionFlower ColorPod FormPod ColorStem LengthRound x WrinkledYellow x GreenAxial x TerminalPurple x WhiteInflated x ConstructedGreen x YellowTall x Dwarf54746022651705882428787185020012072242991522777324802385892911815801064

These crosses or matings can be expressed in a Punnett square. The Punnett square can help you to predict the expressed traits by showing that the combinations in each square have an equal chance of occurring.

Remember that gametes (eggs and sperm) are formed by the process of meiosis. As a result of meiosis, the diploid (2n) number of chromosomes is reduced by half to the haploid (n) number, and one of each kind of chromosome is passed on to each of four daughter cells, the gametes. After fertilization of the haploid egg (n) by a haploid sperm (n), the diploid number of chromosomes (2n) is reinstated.

using a punnett square to determine outcomes for monohybrid crosses

You must first indicate your variables.R round trait r wrinkled trait

P1 (parental): Round pure breeding plant X Wrinkled pure breeding plant

RR X rr

Gametes R r

F1 generation Rr

GametesRr

RRRRr

rRrrr

F2 generation

By examining the Punnett square you can see the expected offspring genotypes are:

1RR to 2Rr to 1 rr.

This gives us the expected genotypic ratio of 1:2:1, or 1 homozygous dominant: 2 heterozygous : 1 homozygous recessive.

We can translate this genotype into a phenotype by remembering that one R is only needed for the round phenotype. When we do this we can see that the phenotype should be 3 round plants to 1 wrinkled plant or 3:1 round:wrinkled.

traits determined by single gene pairs

In humans, 22 of the 23 pairs of chromosomes are truly homologous and are collectively called autosomes. And while many inherited traits such as skin color, height, facial appearance, and fingerprints result from the interaction of many gene pairs located on autosomes (polygenic inheritance), others such as facial dimples, white forelocks, and even the length of the big toe are determined by single gene pairs.

When traits are controlled exclusively by a single gene pair then Mendel's Law of Segregation applies. This means that during meiosis and the formation of gametes, one allele of a gene pair will separate from the other such that half the gametes will receive one allele of the pair, and the other half will receive the other allele of the pair. For example, if an individual is heterozygous for a trait (Aa), then of the gametes produced by this individual half will carry the dominant allele (A), while the other half will carry the recessive allele (a).

The various alleles contained in gametes produced by meiosis and all possible combinations of alleles that could be inherited by offspring can be illustrated using a Punnett square. The different alleles that could be contributed by the female (mother) are listed above the boxes and the alleles within gametes produced by the male (father) are listed to the left of the boxes. The potential genotypes (and from these the phenotypes) of offspring are determined by filling in each box by combining alleles from gametes contributed by each parent. A cross that involves only a single trait of interest, is called a monohybrid cross.

For example, the Punnett square below describes a monohybrid cross between a male and female who are both heterozygous (Aa) for a particular trait. When either the male or female produces gametes he/she will contribute only one of each kind of chromosome to each gamete, thus half of the gametes should possess an A and the other half should contain a.

Aa

AAAAa

aAaaa

Complete the following series of Punnett squares representing crosses between different parental genotypes and compare your results to genotypic and phenotypic ratios summarized in Table 3.2. In reality, crosses between animals (or plants) rarely match these predictions perfectly, but they come close especially if sample sizes are large.

AAaaaaAaAaAa

AaAAaA

AaAAaa

CrossGenotypic Ratios Phenotypic Ratios

AA x AA results in100 % AA 100 % dominant

aa x aa results in100 % aa 100 % recessive

aa x AA results in100 % Aa 100 % dominant

Aa x AA results in50 % AA: 50 % Aa (1:1) 100 % dominant

Aa x aa results in50 % Aa: 50 % aa (1:1) 50 % dominant: 50 % recessive (1:1)

Aa x Aa results in25 % AA: 50 % Aa: 25 % aa (1:2:1) 75 % dominant: 25 % recessive (3:1)

Table 3.2 All possible genotypic and phenotypic ratios resulting from crosses involving a single gene pair.

ACTIVITY 1: CROSSES INVOLVING AUTOSOMES

Predictions can be made about future generations, if the genetic make up of a group is known. We can also make inferences about what the genetic composition of the parents had to be for a particular ratio of traits to appear in the offspring. To figure these things out, follow these basic steps. (1) Read the problem through and figure out from the given information what you know with certainty (this could be related to symbolism for the trait, or whether an allele is dominant or recessive). (2) Sort out the information by attempting to determine the genotypes for the individuals involved, determining all possible types of gametes that could be produced by each parent, predicting all possible combinations of genes that could occur in offspring, and predicting phenotypes from genotypes. For example, consider the following problem.

Sample Problem: The skin and hair of albinos are white. This condition, called albinism is caused by a recessive allele (n) that fails to make the enzymes necessary to manufacture pigments. A mating between two normally pigmented people results in 8 children. Of these, 6 are normally pigmented and the other 2 are albino. How many of the children would you expect to be heterozygous carriers of the albinism allele?

Solution: Look first at what you do know from the information provided. Albinism is a recessive disorder, therefore the genotypes of the two children that are albinos must be nn. One of the n alleles had to have come from the mother, and the other n allele had to come from the father. Now we know positively that both the mother and the father must possess at least one n allele.

However, the problem also tells us that the parents are normally pigmented so they must also possess a dominant allele N. The genotypes of parents are now established. The mother must be Nn and the father must be Nn.

Once genotypes of parents are known, the Punnett square may be filled in and used to predict the genotypic and phenotypic ratios of offspring. Notice that (or 75 %) of offspring should be normal and (or 25 %) should be albinos. This prediction matches the 6 normal to 2 albino children (or 3:1 ratio) in the family perfectly (it isnt always this exact!). Now you are ready to answer the last part of the question, How many children are likely to be carriers? Recall that a carrier is defined as an individual who is heterozygous for a particular trait, in this case Nn. We can see from the matrix that 50 % (or ) of the children are expected to be heterozygous or 4 ( of 8) children.

WHAT IS THE SOLUTION?

1. An albino man, whose parents are both normally pigmented, marries a normally pigmented woman. They have one child, an albino daughter. List the genotypes of all the persons mentioned.

genotype of the man ______________________genotypes of his parents ___________________genotype of the woman ____________________genotype of the child ______________________ Could this couple produce a normally pigmented offspring?__________ Why or why not?_______________________________________ _____________________________________________________________________________________________________________________

2. Huntington's Disease is a degenerative brain disorder caused by an autosomal dominant allele (H). The disease begins to manifest itself early in middle age, often after the individual has already produced children. As the disease progresses, the individual loses control over movement, speech, reason, and thinking. There is no effective treatment or cure, and eventually full time care is required. Suppose a woman who has no history of Huntingtons Disease in her family marries a man whose father died of Huntingtons Disease but whose mother is disease-free. They produce a single child. However, by the time the child is 18, the father has begun to show early signs of Huntingtons Disease. The family is devastated. What is the probability that their child will be stricken with the disease?

_________________________________________________ ______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

3. It happens that the length of the big toe is determined by a single gene pair and that big toes equal to or longer than second toes is the recessive condition (t). Presuming that your big toes are longer than your second toes and your fianc also has long big toes, what are the chances that your offspring will have short big toes, i.e. big toes that are shorter than the second toes?

___________________________________________________________________________________________________________________________________________________________________________________________________

ACTIVITY 2: the principle of independent assortment: dihybrid crosses

In a second series of experiments, Mendel studied crosses between pea plants that were different in two characteristics. As an example lets look at a cross between one parent who has round yellow seeds and another whose peas are green and wrinkled. If we look back at Table 3.1, you will see that we are looking at the expression of two genes, one for pea shape and another for pea color. For pea shape round is dominant, and for pea color yellow is dominant. The first generation (F1) expressed only the dominant traits, all round and yellow. The second generation (F2), produced by allowing the F1 generation to self-pollinate, produced 556 seeds of which 315 showed the dominant traits of round and yellow but only 32 combined the recessive traits of wrinkled and green. The rest were unlike their parents: 108 were round and green and 108 were wrinkled and yellow.

These experiments didnt contradict Mendels earlier findings. In fact they confirmed them. In order to show this we must look at the traits independently. So first lets make a chart of the above data (Table 3.3)

Table 3.3Round, Yellow315

Round, Green108

Wrinkled, Yellow 101

Wrinkled, Green32

Now if we separate this data into two charts (Table 3.4 and 3.5) looking only at one trait at a time we can see our original hypothesis of a 3:1 ratio stands up.

Table 3.4 Table 3.5Round 423Yellow416

Wrinkled133Green110

When you look at these results you should see that the traits sorted as if they were in independent plants. This illustrates the principle of independent assortment, which states that members of each pair of genes are distributed independently when gametes are formed. Because of this principle, it is important when looking at a study of inheritance to determine how many genes are involved.

Determining Possible Gamete Combinations

In order to do a dihybrid cross we first must determine all possible gametes in that cross. Mathematically we can calculate the total gametes by the formula 2n, where n is the number of genes in our cross. In this case we have 2 genes involved, so 22 = 4; there are 4 possible gametes. If we look at the above example and indicate our variables:

R roundY- yellowr wrinkled y green

Then an individual with the genotype RrYy can produce the following gametes:

RY, Ry, rY, and ry

An individual with a genotype of RRYy can produce the following gametes:

4. Show all of the different kinds of gametes that could be produced by individuals with the following genotypes:

Aa

Bb

AA

TTRR

CcDd

AABb

Aabb

aabb

CCDdee

AaBbCc

RY and Ry

using a punnett square to determine outcomes for dihybrid crosses

Looking at the above example and the variables we have already declared, lets create a Punnett square for the dihybrid cross between the round yellow parent with the wrinkled green parent.

Parent phenotype:Round, Yellow XWrinkled, GreenGenotypeRRYYXrryy

Gametes RY ry

F1 generation:RrYy

Gametes RY, Ry, rY, ry

F2 generation:GametesRYRyrYry

RYRRYYRRYyRrYYRrYy

RyRRYyRRyyRrYyRryy

rYRrYYRrYyrrYYrrYy

ryRrYyRryyrrYyrryy

Remember that each cell has a probability of one. Here you will see that there are 16 cells. When we count the genotypes we get

1 RRYY2 RRYy1 RRyy2 RrYY4 RrYy2 Rryy1 rrYY2 rrYy1 rryy

This will give us a genotypic ratio of 1:2:1:2:4:2:1:2:1 for a dihybrid cross of heterozygous parents. To determine the phenotypes we have to relate the genotypes to the characteristics represented. We find the following:

PhenotypeAssociate Genotype.Total

Round YellowRRYY, RRYy, RrYY, and RrYy1+2+2+4=9

Round GreenRRyy, Rryy1+2=3

Wrinkled YellowrrYY, rrYy1+2=3

Wrinkled Greenrryy1

You can see that the associated phenotypic ratio for this cross is 9:3:3:1.

What is the solution?

5a. Assume fur color and eye color in cats are due to simple dominant genes, with black fur being dominant and white fur being recessive, and brown eyes being dominant and green eyes being recessive. What would the genotypes and phenotypes of both the F1 and F2 generations if we crossed a black cat with brown eyes (homozygous for both traits) with a white cat with green eyes (homozygous for both traits)?

5b. Using the same characteristics outlined in 5a, what are the expected genotypes and phenotypes for the F1 generation if we cross a black cat with brown eyes from the F1 generation from 5a with a pure breeding white cat with green eyes?

5c. Using the same characteristics outlined in 5a, what are the expected genotypes and phenotypes for the F1 generation if we crossed a pure breeding black cat with green eyes with a pure breeding white cat with green eyes?

6. In peas, an allele for tall plants (T) is dominant over the allele for short plants (t). An allele of another independent gene (i.e., this gene is not located on the same chromosome as the gene for tallness) produces smooth peas (S) and is dominant over the allele for wrinkled peas (s). Calculate both the phenotypic ratios and genotypic ratios for the results of each of the following crosses:TtSs x TtSsTtss x ttssttSs x TtssTTss x ttSS

7. If the dominant allele K is necessary for hearing, and the dominant allele M of another independent gene results in deafness no matter what other genes are present, what percentage of the offspring produced by the following cross will be deaf?kkMm x Kkmm

Activity 3: Statistically comparing observed data to expectations using The Goodness-of-Fit Chi Square Test

Chi-square is a statistical test commonly used to compare observed data with data we would expect to obtain according to a specific hypothesis. For example, if, according to Mendel's laws, you expect 10 of 20 offspring from a cross to be male and the actual observed number is 8 males, then you might want to know about the "goodness of fit" between the observed and expected. Were the deviations (differences between observed and expected) the result of chance, or were they due to other factors? How much deviation can occur before the investigator must conclude that something other than chance is at work, causing the observed to differ from the expected? The chi-square test can help in making that decision. The chi-square test is always testing what scientists call the null hypothesis, which states that there is no significant difference between the expected and observed result.

The formula for calculating chi-square (2) is:

That is, chi-square is the sum of the squared difference between observed (o) and expected (e) data (or the deviation, d), divided by the expected data in all possible categories. For example, suppose that a cross between two pea plants yields a population of 880 plants, 639 with green seeds and 241 with yellow seeds. You are asked to propose the genotypes of the parents. Your hypothesis is that the allele for green is dominant to the allele for yellow and that the parent plants were both heterozygous for this trait. If your hypothesis is true, then the predicted ratio of offspring from this cross would be 3:1 (based on Mendel's laws) as predicted from the results of the Punnett square (Figure 3.1). To calculate 2, first determine the number expected in each category. If the ratio is 3:1 and the total number of observed individuals is 880, then the expected numerical values should be 660 green and 220 yellow.

Figure 3.l.

Chi-square requires that you use numerical values, not percentages or ratios.Punnett square. Predicted offspring from cross between green- and yellow-seeded plants. Green (G) is dominant (3/4 green; 1/4 yellow).

Then calculate 2 using the formula, as shown in Table B.l. Note that we get a value of 2.668 for 2. But what does this number mean? Here's how to interpret the 2 value:

1. Determine degrees of freedom (df). Degrees of freedom can be calculated as the number of categories in the problem minus 1. In our example, there are two categories (green and yellow); therefore, there is 1 degree of freedom.2. Determine a relative standard to serve as the basis for accepting or rejecting the hypothesis. The relative standard commonly used in biological research is p > 0.05. The p value is the probability that the deviation of the observed from that expected is due to chance alone (no other forces acting). In this case, using p > 0.05, you would expect any deviation to be due to chance alone 5% of the time or less.3. Refer to a chi-square distribution table (Table 3.7). Using the appropriate degrees of freedom, locate the value closest to your calculated chi-square in the table. Determine the closest p (probability) value associated with your chi-square and degrees of freedom. In this case (2 = 2.668), the p value is about 0.10, which means that there is a 10% probability that any deviation from expected results is due to chance only. Based on our standard p > 0.05, this is within the range of acceptable deviation. In terms of your hypothesis for this example, the observed chi-square is not significantly different from expected. The observed numbers are consistent with those expected under Mendel's laws.

Step-by-Step Procedure for Testing Your Hypothesis and Calculating Chi-Square

1. State the hypothesis being tested and the predicted results.2. Gather the data by conducting the relevant experiment (or, if working genetics problems, use the data provided in the problem).3. Determine the expected numbers for each observational class. Remember to use numbers, not percentages.

Chi-square should not be calculated if the expected value in any category is less than 5.

4. Calculate 2 using the formula. Complete all calculations to three significant digits. Round off your answer to two significant digits.5. Use the chi-square distribution table to determine significance of the value.a. Determine degrees of freedom and locate that value in the appropriate column.

b. Locate the value closest to your calculated 2 on that degrees of freedom (df) row.

c. Move up the column to determine the p value.

Table 3.6Calculating Chi-Square

Green

Yellow

Observed (o)639241

Expected (e)660220

Deviation (o - e)-2121

Deviation2 (d2)441441

d2/e0.6682

Table 3.7Chi-Square Distribution

Degrees ofFreedomProbability (p)

(df)0.950.900.800.700.500.300.200.100.050.010.001

10.0040.020.060.150.461.071.642.713.846.6410.83

20.100.210.450.711.392.413.224.605.999.2113.82

30.350.581.011.422.373.664.646.257.8211.3416.27

40.711.061.652.203.364.885.997.789.4913.2818.47

51.141.612.343.004.356.067.299.2411.0715.0920.52

61.632.203.073.835.357.238.5610.6412.5916.8122.46

72.172.833.824.676.358.389.8012.0214.0718.4824.32

82.733.494.595.537.349.5211.0313.3615.5120.0926.12

93.324.175.386.398.3410.6612.2414.6816.9221.6727.88

103.944.866.187.279.3411.7813.4415.9918.3123.2129.59

Non-significantSignificant

Source:R. A. Fisher and F Yates, Statistical Tables for Biological Agricultural and Medical Research, 6th ed., Table IV, Longman Group UK Ltd., 1974.

6. State your conclusion in terms of your hypothesis.a. If the p value for the calculated 2 is p > 0.05, accept your hypothesis. The deviation is small enough that chance alone accounts for it. A p value of 0.6, for example, means that there is a 60% probability that any deviation from expected is due to chance only. This is within the range of acceptable deviation.b. If the p value for the calculated 2 is p 0.05, reject your hypothesis and conclude that some factor other than chance is operating for the deviation to be so great. For example, a p value of 0.01 means that there is only a 1 % chance that this deviation is due to chance alone. Therefore, other factors must be involved.

The chi-square test will be used to test for the goodness of fit between observed and expected data from several laboratory investigations.

Questions

8. Look back to Mendels results in the first section of this lab. Test the flower color, pod color, and stem length to see if they conform to the expectations Mendel described.

9. Look at question 5a. Assume we observed an F2 generation of 1002 black cats with brown eyes, 254 black cats with green eyes, 298 white cats with brown eyes, and 115 white cats with green eyes. Does this conform to our expectations?ACTIVITY 4: Simulation of gamete production and fertilization

Form two-person couples and give each person one penny, one nickel, and one dime. Lets imagine each coin is a gene on separate chromosomes, and the two sides of the coin represent alternate alleles. Self-test: Since there are three genes, how many possible gametes can form?

Designate the following:

Coin TypeGene RepresentedAllelesCoin-side

PennyHair ColorBlack (B)Heads (dominant allele)

Brown (b)Tails (recessive allele)

NickelEye ColorGreen (G)Heads (dominant allele)

Blue (g)Tails (recessive allele)

DimeHair TextureCurly (C)Heads (dominant allele)

Straight (c)Tails (recessive allele)

Simulation`When both partners are ready to have a child, each person should toss the coins gently on the bench top, and both members should greet their new child and record their genotype and phenotype for hair color, texture and eye color in Table 3.6 below. Note we are not including sex of this child for this simulation. Repeat this ten times, and then add up the number of children that show the dominant phenotype, for each trait. Does it approximate 75% (3/4) as Mendel predicts?

Also notice that using three genes in this trihybrid cross study is just a step beyond the 9:3:3:1 for the dihybrid crosses of Mr. Mendel, but each gene still behaves SEPARATELY from the others.

CHILD #Hair color: genotype/phenotypeHair texture:Genotype/phenotypeEye color:Genotype/phenotype

1

2

3

4

5

6

7

8

9

10

% Dominant phenotype

Table 3.8. Offspring phenotypes and genotypes resulting from 10 coin tosses.

Laboratory 3 Mendelian Genetics I Biology 122L 3-

kbhartney, seskandari, pjsperry 3-

kbhartney, seskandari, pjsperry

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kbhartney, seskandari, pjsperry