Chapter 2 Mendelian Inheritance
A p value of ______means that there is a 5% chance of obtaining the listed chi square value at the different degrees of freedom.
0.05
When a heterozygous plant with round seeds is mated to a plant with wrinkled seeds what fraction of the progeny are homozygous?
1/2 This question describes a cross between a plant with Rr seeds and one with rr seeds. 1/2 of the progeny will be Rr, and 1/2 of the progeny will be rr. This means that 1/2 of the progeny are homozygous.
When a heterozygous plant with round seeds is mated to a plant with wrinkled seeds what fraction of the progeny are heterozygous?
1/2 This question describes a cross between a plant with Rr seeds and one with rr seeds. 1/2 of the progeny will be Rr, and 1/2 of the progeny will be rr.
Segregation of the alleles means that a gamete has a____chance of getting a particular allele.
50%
Look at the progeny from the cross of TtYy x TtYy. If you had 1,008 progeny, how many would be expected to be short and to have green seeds?
63 Only 1/16 of the progeny will be expected to be short with green seeds. 1008/16 is 63.
The theory of pangenesis was first proposed by ________.
Hippocrates
Mendel's work was rediscovered in 1900 by which of the following individual(s)?
Hugh de Vries Erich von Tschermak Carl Correns
Which of the following characteristics made the pea plant Pisum sativum an ideal organism for Mendel's studies?
It has the ability to self-fertilize It was easy to cross-fertilize one plant with another It has easily identifiable traits
Which of the following is correct regarding the blending hypothesis of inheritance?
It was a prevailing hypothesis of inheritance prior to Mendel. It was supported by early research by Joseph Kölreuter. It was possible for the blending to change the trait from one generation to the next. It suggested that hereditary traits blended from one generation to the next.
Mendel's work with two-factor (dihybrid) crosses led directly to which of the following?
Law of independent assortment
An allele that produces an inactive enzyme would be classified as what kind of allele?
Loss-of-function
The chi square test ________________________________a hypothesis, but it tests for the probability that the difference between the observed and expected differ by random chance.
does not prove
A true breeding line of green pod pea plants is crossed with a true-breeding line of yellow pod plants. All of their offspring have green pods. From this information, it can be stated that the green color is _____ to the yellow color.
dominant
The Law of Segregation means that the number of gametes with one allele should be_____the number of gametes with the other allele.
equal to
The genetic composition of an individual is called its _____________.
genotype
If a chi square value exceeds the appropriate p value the null hypothesis is usually
rejected
An advantage of the pea plants that Mendel studied was that there were many true-breeding varieties so that he could demonstrate that the results he obtained for one trait were_____with another.
reproducible
One advantage of pea plants for the study of genetics is that they can be _____fertilized.
self
Mendel's work with monohybrid crosses provided proof of which of the following?
particulate theory of inheritance.
Cystic fibrosis is an autosomal recessive disorder that affects lung function in humans. If a couple, who are both unaffected, have an affected child, what is the probability their next child will be an affected girl?
1/8 Since both parents must be heterozygous, the likelihood they will have another affected child is 1/4 and the likelihood that the child will be a girl is 1/2. Using the multiplication rule, the likelihood of having an affected girl is 1/4 * 1/2 = 1/8.
The pedigree above shows a disease with a recessive pattern of inheritance. What is the probability individual II-4 is heterozygous for the disease allele?
100% II-4 has affected children, even though she herself does not have the disease. Since this disease exhibits a recessive pattern of inheritance, II-4 must be heterozygous to pass this allele on to her children.
The number of degrees of freedom when considering the progeny from a cross between an SsRr individual and an ssrr individual would be
3
Mendel could study a single variable at a time since he began with_____plants
true-breeding
A plant that produces wrinkled, green seeds was mated to a plant that produces phenotypically round, yellow seeds. In the F1 generation, half of the plants produced round, yellow seeds and the other half produced round, green seeds. What was the genotype of the parental plant that produced round, yellow seeds?
RRYy
The results of a study of a population is presented in the following table. The "-" indicates that the other allele is unknown. Parent 1 genotypeParent 2 genotypeRatio and genotype of offspring S - S - 989 S - 53 ss ss S - 560 S - 200 ss ss ss 700 ss Which of the conclusions listed below is correct?
The ratios of the offspring in the S - x S - matings are due to some S - parents being homozygotic and some being heterozygotic.
Two phenotypically tall heterozygous pea plants are crossed to one another. A phenotypically tall F1 progeny plant is randomly chosen and mated to a phenotypically dwarf pea plant. If some of the F2 offspring from this mating are phenotypically dwarf, what is the genotype of the randomly chosen F1 parent plant?
Tt
Predict the probability of obtaining different progeny from the mating of SsRr x SsRr using the product rule.
When crossing Ss to Ss the genotypic ratios will be 1/4 SS to 1/2 Ss to 1/4 ss. Similarly, when crossing Rr to Rr, the genotypic ratios will be 1/4 RR to 1/2 Rr to 1/4 rr. We can use these ratios to calculate the probabilities in this question. The probability of obtaining an SsRr progeny is 1/2 x 1/2, or 1/4 = 25% The probability of obtaining an SSRr progeny is 1/4 x 1/2, or 1/8 = 12.5% The probability of obtaining an ssRR progeny is 1/4 x 1/4, or 1/16 = 6.25% The probability of obraining an SsRR progeny is 1/2 x 1/4, or 1/8 = 12.5% The probability of obtaining an ssrr progeny is 1/4 x 1/4, or 1/16 = 6.25% The probability of obtaining an SSrr progeny is 1/4 x 1/4, or 1/16 = 6.25%
According to the Law of Segregation different alleles will_______segregate into different gametes.
randomly
You crossed a plant with wrinkled, yellow seeds to a plant with round, green seeds. In the F1 generation you observed that all of the progeny had round, yellow seeds. What were the genotypes of the parental plants?
rrYY and RRyy
Suppose that we are dealing with four genes, each gene consisting of a dominant allele (capital letter) and a recessive allele (small case letter). If the cross CcMmLlPP X CCmmLlpp is made, what is the probability of obtaining an individual who is CcmmLLPp?
1/16 The Cc genotype will occur ½ the time, the mm genotype will occur ½ the time, the LL genotype will occur ¼ of the time and the Pp genotype will occur in 100% of the progeny. 1/2 * 1/2 * 1/4 = 1/16.
If two heterozygous plants with round seeds are mated, what fraction of the progeny with round seeds are genotypically heterozygous?
2/3
If 100 different ova from an SsRr individual are genotyped the number of ova that would have the SR allelic combination would be predicted to be
25
Cystic fibrosis is caused by mutations in the CF gene, and there are several different mutations that are known to result in CF disease. The CF mutations behave as recessive alleles to the WT CF allele. If two carriers that have different mutations in their CF genes have children what is the probability that one of their children will have CF disease?
25% Since we expect both mutations to inactivate the CF gene, we expect the ratios to be the same as a normal one-gene cross. Therefore, we expect that 1/4 or 25% of two carriers' children to have CF.
If two people who are heterozygous for the PKU allele have a child with PKU, what is the likelihood their next child will have PKU?
25% The existence of one affected child does not change the likelihood that they will have a second affected child. Each conception is an independent event.
A heterozygous tall pea plant is allowed to self-fertilize. The resulting F1 progeny are 25% homozygous tall, 50% heterozygous tall and 25% dwarf. If all of the F1 plants are allowed to self-fertilize, what percentage of the F2 progeny are heterozygous tall?
25% The homozygous tall and dwarf plants will only give rise to homozygous tall and dwarf plants when they self-fertilize. This will account for 50% of the F2 progeny. The progeny of the heterozygous plants will account for the other 50% of the F2 progeny. Among these progeny, 25% will be homozygous tall, 50% will be heterozygous tall, and 25% will be homozygous dwarf. Among all the F2 progeny, 50% x 50% or 25% will be heterozygous tall.
If two people who are both carriers for PKU have a child, what is the likelihood the child will be born with PKU?
25% In order to have PKU an individual must inherit two recessive "p" alleles. This will occur 25% of the time if the parents are heterozygous.
A plant that produces wrinkled, green seeds was mated to a plant that produces phenotypically round, yellow seeds. In the F1 generation, half of the plants produced round, yellow seeds and the other half produced round, green seeds. The round, yellow F1 seeds were planted and those plants were allowed to self-fertilize. In the F2 generation what fraction of the seeds will be genotypically identical to the plants of the parental generation?
3/16 The plants of the parental generation had the genotypes rryy for wrinkled, green seeds and RRYy for round, yellow seeds because half of the F1 plants had round, green seeds and half had round, yellow seeds. The F1 round, yellow seeds had the genotype RrYy. When these plants self-fertilize to produce the F2 seeds, this will be a heteroygous two-factor cross, which can be diagrammed in a Punnett square as shown in Chapter 2. In the F2 offspring, 2/16 will have the genotype RRYy and 1/16 will have the genotype rryy, for a total of 3/16 that have genotypes of the parental generation plants.
Look at the progeny from the cross of TtYy x TtYy. How many phenotypic classes were expected in the resulting progeny?
4 Phenotypic classes include all the genotypes where the organisms are the same. There are four possible outcomes for progeny phenotypes: tall yellow; tall green; short yellow; and short green.
Two pea plants that are heterozygous for the dominant tall gene and the dominant purple flower gene are mated. The cross resulted in 9866 progeny, of which 5550 were tall with purple flowers. What are the expected ratios of the other phenotypic classes?
616 short/white flower 1850 tall/white flower 1850 short/purple flower This is a two-gene cross between heterozygotes, in which we expect a 9:3:3:1 ratio. The question tells us we have 9866 progeny, of which 5550 were tall with purple flowers. This accounts for our first category (9/16 * 9866 = 5550). Our other categories are: Tall with white flowers = 3/16 * 9866 = 1850 Short with purple flowers = 3/16 * 9866 = 1850 Short with white flowers = 1/16 * 9866 = 616
The pedigree above shows a disease with a recessive pattern of inheritance. What is the probability individual III-6 is heterozygous for the disease allele?
66% Because this is a recessive disorder, and III-6 has affected siblings, we know that both II-4 and II-5 must be carriers (heterogygous). Their progeny should have a ratio of 1 homozygous dominant: 2 heterozygous: 1 homozygous recessive. Since III-6 is not affected (symbol is not filled-in) then there is a two out of three chance he is a carrier.
Using Mendel's flower color (purple is dominant, white is recessive), if two heterozygous plants are crossed, what is the probability that the first two offspring will have purple flowers?
9/16 Here we are crossing a Pp flower with a Pp flower. We expect 1 PP: 2 Pp: 1 pp flowers. Phenotypically, 3/4 of the flowers will be purple. We can use the product rule to determine the probability that the first two offspring will have purple flowers by multiplying 3/4 times 3/4, which equals 9/16.
In tomatoes, yellow flowers (Y) are dominant to white flowers (y) and round fruit (R) is dominant to oblong fruit (r). You cross a heterozygous yellow round plant with a plant with white flowers and oblong fruit. You get the following progeny: 238 yellow round210 yellow oblong270 white round282 white oblongFormulate a null hypothesis and performing a chi square analysis on it, what chi square number would you obtain? Based upon this number, would you reject or accept your null hypothesis? Use the table listed below.
Chi square = 12.7; reject You would have expected values of 250 individuals in each category because the expected numbers is a 1:1:1:1 ratio of progeny. Your null hypothesis is that the difference between observed and expected progeny is due to chance. The degrees of freedom are 3, because with four expected categories (n = 4), n − 1 is 3. The chi square number of 12.7 lies to the right of 0.05 probability on the chart and is the basis for rejection of your null hypothesis.
The _____ indicates the probability that differences between the observed values and the expected values are due to random chance alone.
P value
The Law of Independent Assortment applies to meiosis in
both males and females
One type of error that can occur with the chi square test is the acceptance of the null hypothesis when it is
false
Mendel's Law of Independent Assortment states that two different____will be randomly assorted into the gametes.
genes
Statistical analysis determines the _______ between observed data and what was expected from the original hypothesis.
goodness of fit
An individual who has two different alleles for a trait is called ____________.
heterozygous
A cross in which a researcher investigates the patterns of inheritance of a single trait is called a __________.
monohybrid cross
n a genetic cross, the _______ represent offspring with genetic combinations that were not found in the parental lines.
nonparentals
The Law of Segregation allows for the prediction of alleleic combinations in the progeny based on
probability
Due to the Law of Independent Assortment the probability of different alleleic combinations being present in a gamete is calculated by the_____of the frequencies of those alleles.
product
The Law of Indepdent Assortment is based on Mendel's experiments with
two factors
Pea plants are useful as model organisms because they have many_______that differ in the appearance of their flowers, seeds, and pods.
varieties
