1. Calculate the genotype frequencies of a population of flowers where the frequency of the CR allele is 0.8 and the CW is 0.2. There is incomplete dominance, and the population is in equilibrium. Do you understand how to calculate the following answers? CRCR 0.64 CRCW = 0.32 sangan CWCW=0.04 total = 1.00 =

1. Calculate the genotype frequencies of a population of flowers where the frequency of the CR allele is 0.8 and the CW is 0.2. There is incomplete dominance, and the population is in equilibrium. Do you understand how to calculate the following answers? CRCR 0.64 CRCW = 0.32 sangan CWCW=0.04 total = 1.00 =

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter21: Microevolution: Genetic Changes Within Populations

Section: Chapter Questions

Problem 8TYK: If the genotype frequencies in a population are 0.60 AA, 0.20 Aa, and 0.20 aa, and if the...

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EE = floppy ears Ee = floppy ears bb = brown fur ee = pointed ears BB = black fur %3D Bb = black fur %D %3D 1. A female dog with the genotype BBee is crossed with a male dog with the genotype bbEe The square is set up below. Fill it out and determine the phenotypes and proportions in the offspring. Ве Ве Ве Ве bE BbEe BbEe How many out of 16 have: black fur and floppy ears? be Bbee Bbee Bbee Bbee black fur and pointed ears? brown fur and floppy ears? bE BbEe BbEe BbEe brown fur and pointed ears? be Bbee Bbee Bbee Bbee .. ... 2. Show the cross: bbEe x BbEe How many out of 16 have: black fur and floppy ears? black fur and pointed ears? I floppy ears? own brown fur and pointed ears?
75. Under the conditions of the Hardy-Weinberg equilibrium, what term in the equilibrium equation represents the expected genotype frequency for homozygous recessive individuals? p2 2pq 2p+q p2+2pq+q2 q2
I'm confused about how I would figure out these two questions as I don't really understand the difference or where to start. What would genotypic frequencies be for Wildrose at pt-8 if it were in equilibrium? and How many individuals (to 2 decimal places) of each genotype would there be from Wildrose at pt-8 if it were in equilibrium?
26. Snow geese (Chen caerulescens) come in two color types, white “snows” and “blues” with dark bodies. A single gene controls coloration, where the dark (“blue”) allele (D) is dominant. A population of 30,012 geese includes 9236 dark individuals. Genetic testing reveals that 7636 of the 9236 dark individuals are heterozygous (Dd). What is the actual frequency of allele D? 0.819 0.181 0.435 0.347
APOE Allelic frequency Control EOAD LOAD Vascular disease e4 0.07 0.28 0.35 0.12 e3 0.88 0.70 0.63 0.87 What group is used as a control here? Based on this data, do you think APOE e4 allele increases one’s chance to develop EOAD or LOAD? Defend your answer. Your answer has to refer to the specific data in the table. Suppose that you want to improve this case control study by controlling one more variable that is a known risk factor for Alzheimer’s disease, how will you design your groups? Specify the risk factor you want to control and how you plan to control it.
4. Attached earlobes are dominant over free hanging earlobes. Complete the Punnett Square for the following individuals: Mom-Bb and Dad=bb Provide the genotype and phenotype ratios using the Punnett square given below. Genotypes 5. When two alleles are observed independently in a phenotype is called multiple alleles. Human blood group ABO is a good example of multiple alleles. The three alleles are: IA, IB, i. Mother's blood type is AB (IAI³) and father's blood type is B (I³i). Provide the genotype and phenotype ratios using the Punnett square given below. Genotypes Phenotypes Genotypes Phenotypes 6. When the allele of a trait is found on X or Y chromosome, it becomes sex-linked. The results differ from males to females. One such example is of red green colorblindness. The mother is a carrier (X© Xº ), and the father is colorblind (X© Y). Provide the genotype and phenotype ratios using the Punnett square given below. Phenotypes
1. n= 153 for BB allele. Bb, n=44, bb n=3. Is the locus in Hardy-Weinberg equilibrium, and what's the expected and observed frequency of each?
14. In class we explored how genotype frequencies can change between generations. In our experiment, we started with a population in which every individual was heterozygous for a few traits of interest. The data for the 11 offspring of these heterozygotes are below. Make sure you show at least some work for partial credit. Frecueles Eje coloN Freckles Genotype FF Eye Shape Genotype EE Eye Color Genotype RR (P 3 (P 3 Ff 5 Ee .0.3 1-0.3 3 RB 7 ff 1 ее Phenotypes Freckles No freckles Phenotypes Phenotypes Red |-0.3 FF, Ff Circle EE, Ee RR Ro.HFF) ff Star ее Purple Blue RB or BR 20.42 BB a) What were the initial allele frequencies for freckles in the starting population? FF=0.42 b) What percentage of the starting population had purple eyes? c) What are the allele frequencies for freckles after one generation? (FF)P= 0.7 (FN 2pf=0.42 (fF) = 03 d) What percentage of the population had purple eyes after one generation? e) Are the genotype frequencies for eye shape what you would expect based…
I need help calculating observed allelic frequencies. I'm in a biology lab and we did a classroom experiment where we were each given genotypes and simulated multiple generations by coinflip. This is the end data after multiple generations and we need to extrapolate p & q: AA individuals - 2 Aa individuals - 13 aa individuals - 5 My understanding is that the genotypic frequencies would simply be their %. So AA would be 0.1, Aa would be 0.65, and aa would be 0.25. This should mean that AA = 0.1 = p^2 and aa = 0.25 = q^2. Which should also mean that taking the square root of these numbers would give me p and q.... but they don't add up to 1. Squaring them gives me a p value of ~0.32 and a q value of 0.5. How is this possible? Shouldn't this method work to give me p and q since I know the exact genotypic frequencies? If anyone can clear up why this doesn't work, I would be extremely appreciative. Thank you!
This Punnett square shows allele combinations for all possible genetic crosses in the wildflower population described in Part A. CR E p= 0.7 CW q= 0.3 p= 0.7 ? CW CR CR CR CW q= 0.3 ? ? CR CW CW CW ? What are the expected genotype frequencies in the offspring generation? O 0.49 CRCR (red flowers), 0.42 CRCW (pink flowers), 0.09 CWCW (white flowers) 0.49 CRCR (red flowers), 0.21 CRCW (pink flowers), 0.30 CWCW (white flowers) O 0.09 CRCR (red flowers), 0.21 CRCW (pink flowers), 0.42 CWCW (white flowers) 0.09 CRCR (red flowers), 0.42 CRCW (pink flowers), 0.09 CWCW (white flowers)
1.: Male side-blotched lizards display throat color polymorphisms; some males have orange throats, some have yellow throat stripes, and some have blue throats. The following table gives genotypes observed in a California population in 1993. Genotype Individuals 2 10 oy ob 14 yy 32 yb bb 42 Please calculate: a) The allele frequencies. b) The observed heterozygosity. Note that you have more than one type of heterozygote at this locus. c) The expected heterozygosity.
2. Describe the phenotype of the cats with these genotypes (If allele not specified, assume wild type): CC ss OO BB dd McMc tt ww Light brown male with choco [Choose ] Light brown male with chocolate tips ctct ss OY bb TAt A uniformly colored black cat Light grey female with dark grey mackerel stripes Black female with large white patches Dark grey with black mackerel stripes Cc SS Oo BB TATA Cream male with small white patches and orange mackerel stripes CC Ss oY bb dd tt [Choose ]