BIO 100 LAB 7

BIO 100 LAB 7
Compose answers to the questions below and save the file as a backup copy in the event that a technical problem is encountered while attempting to submit the assignment. Make sure to run a spell check. You will be submitting your answers to the lab assignment in two parts. The first part of the lab assignment consists of the laboratory exercise questions. The second part of the lab assignment is the application question. The first textbox on the submission page corresponds to the first part of the lab. Be sure to paste the laboratory exercise questions, with your answers, into this textbox. The second textbox on the submission page will be for your response to the application question. LABORATORY EXERCISE QUESTIONS What is an allele? (1 point) Give an example of a genetic trait and two allele phenotypes for the gene that determines the trait. (3 points) Genetic Trait Allele phenotype #1 Allele phenotype #2 Compare and contrast the terms phenotype and genotype. (4 points) List the allele combinations (gamete possibilities) that can be formed by an individual with the following genotype: AABB (1 point) List the different allele combinations (gamete possibilities) that can be formed by an individual with the following genotype: AaBb. (4 points)         Given: P = purple flowers and p = white flowers and P is dominant over p. What is the phenotypic ratio of offspring from a cross between Pp x pp? (2 points) In a population with 160 individuals how many will be homozygous? (2 points) In that same population, how many will be purple? (2 points)   Which of Mendel’s laws is illustrated in a dihybrid cross? (1 point) What does this law state? (1 point) A horticulturist has a purple plant and a white plant. The horticulturist knows that purple is dominant over white. When they are bred, all of the resulting offspring are purple. What is the most likely genotype of the parent or original purple plant? (2 points) A horticulturist runs a test cross with an offspring (F1 generation) purple plant from Question 8. The phenotypic frequencies of the resulting offspring are 50% white and 50% purple. What is the true genotype of this offspring (F1 generation) purple plant? (2 points) What is the probability of a cross resulting in white offspring when two heterozygous purple pea plants (e.g. Pp x Pp) are bred? What is the genotype for this offspring? (4 points) Compare the ratios calculated in Exercises 5 and 6 (coin toss activity) to Mendel’s ratios. Were they close? Pose a possible explanation for why the ratios may not be exactly the same. (4 points) Refer to the data on the corn kernel color ratio from Part II of the lab. What was the phenotypic ratio from Step 1? (2 points) What was the phenotypic ratio from Step 2? (2 points) What was the phenotypic ratio from Step 3? (Remember there were four possible types for this part of the lab.) (2 points) Explain why your numbers did not come out exactly. (2 points) Recall from the background information that purple kernels are dominant and yellow kernels are recessive. The second ear of corn was the result of crossing two heterozygous ears of corn male purple (Pp x Pp). This is represented by the Punnett square below. Complete the Punnett square by writing the correct letters that correspond to each number indicated in the table. (4 points)         Once the Punnett square for Question 13 is complete, calculate the ratio of purple and yellow kernels (recall that if the dominant trait is present, it will be expressed). What is the ratio of purple to yellow kernels based on the Punnett square? (2 points) How did this compare to the ratio obtained from counting the corn kernels for ear number two in part II of the lab? (2 points) Recall from the background information that purple kernels are dominant and yellow kernels are recessive. Also recall that smooth kernels are dominant and wrinkled kernels are recessive. The third corn ear was the result of crossing a male ear of corn with the following gametes: PpSs, with a female ear of corn with the same gametes: PpSs. This is represented by the Punnett square, below. Complete the Punnett square by writing the correct letters that correspond to each number indicated in the table (for example, PPSS or ppss). (8 points)   PS Ps pS ps PS Ps pS 10 11 12 ps 13 14 15 16                                 Once the Punnett square for Question 15 is complete, calculate the ratio of corn kernel varieties (recall that if the dominant trait is present, it will be expressed). What is the ratio of kernels based on the Punnett square? (2 points) How does this compare to the ratio obtained from counting the corn kernels? (2 points) What are the genotypic and phenotypic ratios for kernel color and kernel texture for a dihybrid cross between PpSS x Ppss? (4 points) genotypic ratio phenotypic ratio Using the results from the Punnett Square in Question 15, answer the following question: In a population of 240 corn kernels, how many will have smooth kernel texture? (2 points) In a flower garden, a horticulturist is growing purple and white pansies. The horticulturist notices that a new pansy has sprouted. When it finally flowers, the pansy is lavender. Explain how this happened. (4 points) With a botanist’s help, an individual decides to cross the lavender pansy with the white pansy. Will this result in any purple pansies? Explain. (4 points) State an industrial concern/importance for knowing the genetic makeup of an organism. Explain why it is important in your example. (5 points) APPLICATION QUESTION (Application) How might the information gained from this lab pertaining to genetics be useful to you, or how can you apply this knowledge to your everyday life as a non-scientist? The application will be graded according to the rubric below. (20 points)
BIO 100 LAB 7
iii . T e st 3 : A AB B x a ab b ( h ead a n d t a il o f t h e n ic k e l a re A ; h ead a n d t a il o f t h e pen ny a re B ; h ead a n d t a il o f t h e d im e a re a ; h ead a n d t a il o f t h e q u arte r are b ) iv . T e st 4 : A aB b x A aB b ( h ead is d om in an t a n d t a il is r e ce ssiv e o n a ll c o in s) v. T e st 5 : A aB b x a ab b ( h ead is d om in an t a n d t a il is r e ce ssiv e o n t h e n ic k e l an d p en ny; h ead a n d t a il o n t h e q u arte r a n d d im e a re r e ce ssiv e ). B. F lip e ach c o in 2 0 t im es f o r e ach t e st a n d r e co rd r e su lt s . U se a c h eck m ark o r a n X t o in dic a te t h e o ffs p rin g’s g e n oty p es. C. D ete rm in e t h e F f r e q u en cie s a n d r a tio s b ase d o n t h e r e su lt s . N OTE : W hen r e co rd in g t h e r a tio s, r o u n d t o t h e n eare st w hole n um ber. 7. C re ate P u nnett s q u are s f o r t h e f o llo w in g d ih yb rid c ro ss: A. R RW w x R rW w i. A ssu m e w hile d oin g s o t h at it is a n in co m ple te d om in an ce s it u atio n ii. R = R ed p eta ls , r = w hit e p eta ls , R r = p in k p eta ls . iii . W = W hit e f u r, w = b la ck f u r, W w = g re y f u r. 8. C re ate a P u nnett s q u are f o r t h e f o llo w in g g e n oty p ic c ro ss X X v s. X Y. P A RT I I: C alc u la tin g t h e F re q u en cy o f C orn K ern el C hara cte ris tic s In t h is p art o f t h e la b , f r e q u en cie s o f t w o c o rn k e rn el c h ara cte ris tic s w ill b e d ete rm in ed : ke rn el c o lo r a n d k e rn el t e xtu re . I n o b se rv in g t h e e ars o f c o rn , n ote t h at t h e k e rn els a re e it h er p u rp le o r y e llo w a n d t h at t h ey a re e it h er s m ooth o r w rin kle d . W it h r e sp ect t o c o lo r, p u rp le is t h e d om in an t c o lo r, a n d y e llo w is t h e r e ce ssiv e c o lo r. W it h r e sp ect t o t e xtu re , sm ooth is t h e d om in an t t e xtu re , a n d w rin kle d is t h e r e ce ssiv e t e xtu re . U se t h e f o llo w in g desig n atio n s f o r e ach t r a it : p u rp le ( P ); y e llo w ( p ); s m ooth ( S ); w rin kle d ( s ). K eep t h is in fo rm atio n a va ila b le a s it w ill b e n eed ed la te r o n in t h e la b . 1. O bse rv e t h e im age o f a c o rn e ar b elo w . N otic e t h at t h e e ar o f c o rn c o n ta in s p u rp le a n d y e llo w k e rn els . S tu dy t h e im age a n d t h en c o u nt t h re e r o w s o f k e rn els , t a lly in g th e n um ber o f p u rp le a n d t h e n um ber o f y e llo w k e rn els . R eco rd t h is in fo rm atio n o n a p ie ce o f p ap er t o r e fe r t o la te r in t h is la b ora to ry . R evie w t h e r e su lt s a n d in dic a te th e f r e q u en cy ( r a tio ) o f p u rp le t o y e llo w k e rn els . C lic k o n im age t o e n la rg e . 2. O bse rv e t h e im age o f a c o rn e ar b elo w . N otic e t h at t h e e ar o f c o rn c o n ta in s p u rp le a n d y e llo w k e rn els . S tu dy t h e im age a n d t h en c o u nt t h re e r o w s o f k e rn els , t a lly in g th e n um ber o f p u rp le a n d t h e n um ber o f y e llo w k e rn els . R eco rd t h is in fo rm atio n o n a p ie ce o f p ap er. R evie w t h e r e su lt s a n d in dic a te t h e f r e q u en cy ( r a tio ) o f p u rp le t o y e llo w k e rn els . 1 Clic k o n im age t o e n la rg e . 3. O bse rv e t h e im age o f a c o rn e ar b elo w . N otic e t h at t h e e ar o f c o rn c o n ta in s p u rp le a n d y e llo w k e rn els . A ddit io n ally , n otic e t h at s o m e o f t h e c o rn k e rn els a re s m ooth , an d s o m e o f t h e c o rn k e rn els a re w rin kle d . S tu dy t h e im age a n d t h en c o u nt t h re e ro w s o f k e rn els , t a lly in g t h e n um ber o f k e rn els . F o r t h is e xe rc is e , t h ere w ill b e f o u r dif fe re n t k e rn els t o c o u nt: p u rp le a n d s m ooth , p u rp le a n d w rin kle d , y e llo w a n d sm ooth , o r y e llo w a n d w rin kle d . R eco rd t h is in fo rm atio n o n a p ie ce o f p ap er. R evie w th e r e su lt s a n d in dic a te t h e f r e q u en cy ( r a tio ) o f c o rn k e rn els . C lic k o n im age t o e n la rg e . A sse ssin g Y ou r L earn in g ⚠ Warn in g: Y ou a re e xp ecte d to s u bm it y o ur o w n, in div id ual w ork . U sin g w ork c o m ple te d b y a nyo ne oth er th an y o urs e lf is p la gia ris m . T his in clu des re so urc e s fo und o n in te rn et s it e s. P ostin g asse ssm ents o n a n u nauth oriz e d w eb s it e , s o lic it in g asse ssm ent a nsw ers o r th e a cq uis it io n o f asse ssm ents , a sse ssm ent a nsw ers , a nd o th er aca dem ic m ate ria l is c h eatin g. C heatin g a nd/o r pla gia ris m w ill r e su lt in a fa ilin g g ra de fo r th e co urs e . D ow nlo ad t h e L a b W ork sh eet. C om pose a n sw ers t o t h e q u estio n s b elo w u sin g t h e L a b W ork sh eet a n d sa ve t h e fi le a s a b ack u p c o py in t h e e ve n t t h at a t e ch nic a l p ro b le m is e n co u nte re d w hile a tte m ptin g to s u bm it t h e a ssig n m en t. M ake s u re t o r u n a s p ell c h eck . Yo u w ill b e s u bm it t in g y o u r a n sw ers t o t h e la b a ssig n m en t in t w o p arts . T h e fi rs t p art o f t h e la b a ssig n m en t c o n sis ts o f t h e la b ora to ry e xe rc is e q u estio n s. T h e s e co n d p art o f t h e la b a ssig n m en t is t h e ap plic a tio n q u estio n . T h e fi rs t t e xtb ox o n t h e s u bm is sio n p age c o rre sp on ds t o t h e fi rs t p art o f t h e la b .