Genetics Lab Quizzes
When you cross two heterozygous parents (Aa), where A is a recessive lethal, what is the phenotypic ratio of the progeny?
2:1 How: If A is the recessive lethal then in this case, the allelic combination for "A" has two phenotypes!! When either the heterozygote (Aa) or the alleles coding for the homozygote recessive state (aa) are expressed in offspring, they are coding for the expression, or lack thereof, of a tail. However, if the offspring is homozygous AA, the opportunity to express a tail does not even arise as this allelic combination is LETHAL and results in death of the organism. So in fact, the homozygous AA is the recessive state solely for lethality but the A allele codes for the dominant trait of not having a tail.
In a recent study of cat food preferences between salmon and tuna, 56 cats chose tuna and 49 cats chose salmon. Use chi-square analysis to determine if this sample fits an expected ratio of 1:1. The observed chi-squared calculated from this data is_____. Chi-squared = the sum of (observed - expected)^2/expected. (2 decimal places)
0.47 How: 1. Find the expected values for tuna and salmon. (52.5 & 52.5) 2. Put the observed and expected values for tuna and salmon into the chi-squared equation. (tuna would be (56-52.5)^2/52.5; salmon would be (49-52.5)^2/52.5) 3. Calculate (0.2333 & 0.2333) 4. Add together the calculated values and calculate again. This answer is the observed chi-squared. (0.2333 + 0.2333 = 0.4666) 5. Round to specified places if needed. (0.4666 = 0.47)
In the previous chi-squared problem involving cat food preferences, how many degrees of freedom are there?
1 How: 1. Find how many categories you have in your statistical analysis. (tuna and salmon are 2 categories) 2. Subtract that number by 1. The calculated value are the degrees of freedom. (2 - 1 = 1)
In fruitflies, Wild wing shape is dominant over Crumbled wing shape. Wild eye shape is dominant over Eyeless eye shape. Using the Electronic Fly Colony you select the following Parental cross: Crumbled, Wild Male x Wild, Eyeless Female. Select two genotypes that would produce these phenotypes. aaBB male AAbb female AABB male abb female
aaBB male AAbb female
_____ ratios can be can be produced from two or more segregation ratios
assortment
The Fly Colony module used this week has a major difference from the previous weeks. Once you select a F1 offspring to use as an F2 parent, the program automatically selects the other parent to be a _________________ genotype. This choice assures that the expected mendelian assortment ratio will always be 1:1:1:1.
dihybrid double recessive
In the Fly Lab you cross a homozygous black bodied fly with a homozygous wild type fly through the F2 generation, setting the offspring value to 1000 each time. The results of your F2 cross are: 704 Wild Type flies 261 Black Bodied flies You anticipate a 3:1 ratio. In completing a chi-square calculation, what would be your expected values? (2 decimal places)
expected for Wild Type flies: 723.75 expected for Black Bodied Flies: 241.25 How: 1. Add together the total amount of observed flies. (704 + 261 = 965) 2. Multiply total amount of observed flies (965) by each part of the ratio in fraction form (3:1 would be 3/4:1/4). (965 x 3/4 = 723.75) (965 x 1/4 = 241.25) 3. Your answer for each ratio after multiplying are your expected values. 4. Round to specified places if needed.
Alignment is taking large numbers of generated DNA sequences and finding overlap between them.
false
If a male is affected with an x-linked recessive trait, his female offspring will automatically be affected.
false
Imagine you are conducting an experiment where you hypothesize that apples weigh more than bananas on average. After calculating the weight of 100 apples and 100 bananas you find the average weight of apples is 200 grams and the average weight of bananas is 100 grams. You conduct a statistical test and find this difference is significant according to the test. True or false: You have proven that apples weigh more than bananas and this is now scientific fact.
false
The GENOTYPE of the affected son and daughter at the bottom of the following pedigree is most likely _______?
homozygous recessive
For the Mendelian option, the Fly Colony program is designed so that the F1 progeny are all heterozygotes. This result stimulates exactly what happens if we were crossing real flies and occurs because one Parent is ______ dominant and the other is selected to be _____ recessive.
homozygous; homozygous
If a population of people had the exact same genes for a trait but only some people exhibit the trait we would say the trait has:
incomplete penetrance
If the gene for pea color is inherently separated from from the gene for pea shape, and thus the allele occurrence for one trait does not affect the other, then this demonstrates the law of _____.
independent assortment
The fragments of DNA we will be examining this week in lab will be separated based on fragment _____________.
length
You cross 1000 fruit flies with the genotype Aa and obtain the following results: AA: 208 Aa: 511 What mode of inheritance best fits these results?
lethal alleles
The branching point on a phylogenetic tree where a divergent event occurred is referred to as what?
node
An allele is _____.
one of several possible forms of a gene
What do you call the physical expression of a gene?
phenotype
The "standard" trait for a phenotype is referred to as
wild type
Performing electrophoresis in water, instead of buffer:
won't work because the buffer provides ions to allow the current to move
For the activity this week you will mostly use what gene?
tra
When each parent has one mutant allele and one wild type allele, the configuration is referred to as ______________________.
trans or replused??? Not sure but my best guess
Linkage is considered to be a violation of Mendel's Principle of Independent Assortment.
true
When using microsatellites as genetic markers, as we do in this lab exercise, it is easy to identify heterozygotes because they all show 2 bands.
true
You complete a 3-point cross and obtain the following data: Phenotype & Observed Wild, Wild, Wild 263 Wild, Singed, Wild 142 Wild, Wild, Sable 19 Wild, Singed, Sable 45 Scalloped, Wild, Wild 54 Scalloped, Singed, Wild 23 Scalloped, Wild, Sable 144 Scalloped, Singed, Sable 285 TOTAL= 975 1. Which phenotypes are the parental (P1) phenotypes? 2. Which phenotypes are the Double Cross-overs (DCO)? 3. Which phenotypes are the Single Cross-overs (SCO)? 4. Which gene is in the middle? FOR THE NEXT THREE QUESTIONS ROUND TO TWO DECIMAL PLACES 5. What is the map distance between scalloped and singed? 6. What is the map distance between singed and sable? 7. What is the map distance between scalloped and sable?
1. Wild, wild, wild & Scalloped, singed, sable. (the parentals are the two groups that have the highest observed number) 2. wild, wild, sable & scalloped, singed, wild (the DCO"s are the two groups that have the lowest observed number) 3. SCO #1= wild, singed, wild & scalloped, wild, sable SCO #2= wild, singed, sable & scalloped, wild, wild (the SCO's are usually consisted of 4 total groups. They are the groups that aren't the Parentals or the DCO's. You group them into two's by dividing them by their observed numbers. The two that are closer together is put together and the same for the other two) 4. sable (the middle gene is determined by which one is flipped by observing the parentals and DCO's. ie. you have + + + & a b c for the parentals. You have + + c & a b + for the DCO's. You can see that "c", or sable, has flipped while "a" and "b" have stayed the same) 5. 6. 7.
In fruit flies, bristle shape is found to be incompletely dominant with wild being homozygous (++), mild being heterozygous (+b) and bent bristled flies being homozygous (bb). For eye color, orange is dominant over wild in a simple Mendelian fashion. Assume that two mild bristled, oranged eyed flies (standard F1 cross) are mated, how many of their 560 progeny would you expect to be wild bristled, oranged eyed?
105 How: 1. Mild bristled and oranged eyed = +b:Rr 2. Perform a cross. +b:Rr x +b:Rr 3. Results are +b x +b = ++ (1/4); +b (1/2); bb (1/4) Rr x Rr = R_ (3/4); RR (1/4) 4. Identify genotype of progeny trying to find. Wild bristled and oranged eyed = ++:R_; (1/4):(3/4) 5. Multiply the ratios of the identified genotypes. (1/4) x (3/4) = 3/16 6. Multiply the number found in #5 by the number of progeny given. (3/16) x 560 = 105. 7. 105 is the expected progeny with wild bristled and oranged eyed
The process for creating a DNA profile using STRs is:
1: obtain sample 2: isolate DNA 3: run PCR to amplify DNA 4: determine the size of the STRs 5: determine if a match exists
Diploid organisms have _____ alleles at each locus.
2
In the diagram below, which NUMBER represents the most recent common ancestor of Pickleball and Soccer?
2
According to the video, short match is used to search for sequences that are approximately how long?
2-30 bases
How many homologous pairs of chromosomes do Drosophila melanogaster have?
4
In fruit flies, bristle shape is found to be incompletely dominant with wild being homozygous (++), mild being heterozygous (+b) and bent bristled flies being homozygous (bb). For mandible size, another incomplete dominant trait, wild type is heterozygous (+m), short is homozygous (++) and long is homozygous (mm). Assume that two mild bristled, wild mandibled eyed flies (standard F1 cross) are mated, how many of their 480 progeny would you expect to be wild bristled, wild mandibled?
60 How: 1. Mild bristled and wild mandibled = +b:+m 2. Perform a cross. +b:+m x +b:+m 3. Results are +b x +b = ++ (1/4); +b (1/2); bb (1/4) +m x +m = ++ (1/4); +m (1/2); mm (1/4) 4. Identify genotype of progeny trying to find. Wild bristled and wild mandibled = ++:+m; (1/4):(1/2) 5. Multiply the ratios of the identified genotypes. (1/4) x (1/2) = 1/8. 6. Multiply the number found in #5 by the number of progeny given. (1/8) x 480 = 60. 7. 60 is the expected progeny with wild bristled and wild mandibled
In peas, Round seeds are dominant over Wrinkled, and Yellow is dominant over Green. A wrinkled, yellow (aaBb) plant is crossed with a round, yellow(AaBb) plant and a large number of offspring are produced. Given the results below, calculate the overall Chi-Squared (observed) value for these data. Hint: You will first need to determine the expected Phenotypic ratio for this cross. Offspring Phenotypes: 250 Wrinkled, Yellow 150 Wrinkled, Green 250 Round, Yellow 150 Round, Green
66.67
In fruit flies, bristle shape is found to be incompletely dominant with wild being homozygous (+,+), mildly bent being heterozygous (+,b) and fully bent bristles being homozygous (b,b). For mandible size, wild type is heterozygous (+,m), short is homozygous (+,+) and long is homozygous (m,m). Assume that two mildly bristled, wild mandibled flies (standard F1 cross) are mated, how many of their 736 progeny would you expect to be fully bent bristled, wild mandibled?
92 How: 1. Mild bristled and wild mandibled = +b:+m 2. Perform a cross. +b:+m x +b:+m 3. Results are +b x +b = ++ (1/4); +b (1/2); bb (1/4) +m x +m = ++ (1/4); +m (1/2); mm (1/4) 4. Identify genotype of progeny trying to find. Fully bent bristled and wild mandibled = bb:+m; (1/4):(1/2) 5. Multiply the ratios of the identified genotypes. (1/4) x (1/2) = 1/8. 6. Multiply the number found in #5 by the number of progeny given. (1/8) x 736 = 92. 7. 92 is the expected progeny with fully bent bristled and wild mandibled
In a dihybrid cross, if two heterozygotes (for both genes) are crossed once, what will be the expected phenotypic ratios? (In the following order: dominant for both traits; dominant for the first trait recessive for the second; recessive for the first trait and dominant for the second; recessive for both traits)
9:3:3:1
According to the following diagram, which two sports are most similar?
??
What are TRUE of isoforms?
??
What was the first public use of genetic fingerprinting?
??
Which of the following are true about RNA sec reads?
??
The following best describes which process? This method relies on variable number tandem repeat (VNTR) polymorphisms to distinguish various alleles, which are separated on a polyacrylamide gel using an allelic ladder. Bands can be visualized by silver staining the gel.
AFLP
Assume that allele A is dominant to a, and B is dominant to b. Match each of the following crosses with the expected phenotypic ratio. AaBb x aabb Aabb x AaBb AaBb x AaBb
AaBb x aabb = 1:1:1:1 Aabb x AaBb = 3:1:3:1 AaBb x AaBb = 9:3:3:1
Which of the following crosses is the most useful for solving a 3-point mapping problem?
AaBbCc x aabbcc
What is the most likely Mode of Inheritance in the pedigree below?
Autosomal Dominant
_______________ can compare translated nucleotides to proteins.
BLAST
You have DNA and you wish to find other DNA sequences that look like it. Which of thefive Basic Blast programs should you use?
BLASTN
What are some other names for phylogenetic trees?
Cladograms Phenograms Topologies Dendrograms
For the activity this week, which genome will you be using?
D. melanogaster
The primary cross that we will use throughout this laboratory to examine linkage is the _____________________.
Dihybrid recessive backcross
__________ is a series of techniques by which biomolecules can be separated based on size, length and/or shape.
Electrophoresis
Continuing with the chi-squared cat food preference question, what is final verdict of the chi-squared using a p value of 0.05 in the chart below.
Fail to reject How: 1. Identify your degrees of freedom (DF) and your given p-value. (1 for DF and 0.05 for your p-value) 2. Use the chart to find the statistical number associated with your DF and p-value. (3.841) 3. Determine if your calculated observed chi-squared number is more or less than the statistical number you found in the chart. (0.47 is less than 3.841) 4. If your chi-squared value is more than the number in the chart, you reject it. If your chi-squared value is less than the number in the chart, you fail to reject it. (fail to reject)
What are exons?
Genetic information coding for an amino acid sequence that will form a functional protein
What is a (are) benefit(s) of using TopHat?
It can identify splice junctions without relying on a database of known junctions Can identify splice junctions between exons. Allows finding of novel splice sites
In forensic cases what is sometimes typed because samples contain many more copies of this than what is normally typed?
Mitochondria DNA
One way to distinguish dominant and recessive traits in a pedigree is that the former will not skip generations but the latter will. But will recessive traits always skip generations?
No, if both parents are homozygous for the recessive trait it will not skip generations
The following best describes which process? The process mimics the biological process of DNA replication, but confines it to specific DNA sequences of interest. With the invention of this technique, DNA profiling took huge strides forward in both discriminating power and the ability to recover information from very small (or degraded) starting samples.
Polymerase Chain Reaction
DNA from skeletal remains was able to prove that the remains belongs to Josef Mengele because of what development(s)?
STR & PCR
Chain termination sequencing is also referred to as __________________________.
Sanger Sequencing
What is the most likely Mode of Inheritance in the following pedigree?
Sex-linked
The following best describes which process? Used to detect specific DNA sequences, this method combines transfer of electrophoresis-separated DNA fragments to a filter membrane and subsequent fragment detection by probe hybridization.
Southern Blot
After setting up a dihybrid backcross, if you fail to reject your null hypothesis, what can you infer about the linkage of loci in question?
The loci are assorting independently
Assembly is used when there is no existing reference genome and Alignment is used to compare DNA to existing sequences.
True
Complete linkage refers to when there is NO recombination between linked genes, whereas incomplete has some recombination where the distribution of offspring is heavily skewed towards the parental phenotypes.
True
These two phylogenetic trees are evolutionarily equivalent, TRUE or FALSE.
True
True or False: The majority of genetic variation in humans are found in SNPs.
True
One difference between DNA and RNA is that RNA:
contains the base Uracil.
The term HOLANDRIC refers to what type of inheritance?
Y-linked
According to Mendel's Big 6, when you cross a F1 heterozygote with a homozygous parental, you are conducting which type of cross?
back cross?
A person with the phenotype of AB for blood type has the genotype AB. What is the relationship between the A allele and B allele?
co-dominant
Which part of the DNA sequence is translated into protein?
coding sequence
What type of dominance is demonstrated by the progeny expressing the phenotype of both parents?
codominance
To test for linkage on Chromosome 1 of Drosophila, one should preform what type of cross?
reciprocal cross
When performing a ____________________ and the F2 generations yield significantly different results, it is likely that the trait is sex linked.
reciprocal cross
What was the first heritable variation detected in human DNA called?
restriction fragment-length polymorphism
De Novo sequencing is
sequencing an organism for the first time
The basic dominant/recessive relationship is also commonly referred to as __________________.
simple mendelian
Why do dideoxynucleotides cause DNA sequencing to stop?
they are lacking a 3' hydroxyl group