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,; 1. GENETIC PROBLEMS 't- XI which parental pair could produce a colorblind \. .; female? \ A man heterorygous with blood type A marries blood type of their offspring could not be: 1.A 3. independent assortrnent ,t2) do*io*"" 25'/o 2. 50% 2. 4. AB O 3. segregation 4. incomplete dominance 4. S'; ts't" incomPlete dominance 3. 100% sex linkage ,:4. independent assorfinent ; An organism possessing two identical genes for l. 2. 7. with blood type AB. The A boy has brown hair and blue eyes, and his brother has brown hair and brown eyes. The fact that they have different combinations of traits is best explained by the concept known 1. multiPle alleles 6. ( ln pea plants, the trait for all tail stems is dominant over the trait for short stems. If nvo be expected to heterozygous tall plants are crossed, what percentage of the offspring would -f ' t' have the same phenotype as the tr arents? l. 5. '3. (,, \' .1"*.'..' white tulips. Onty red tulips result from a cross between homozygous red and homozygous This illustrates the PrinciPle of: [. 4. 2.8 a woman 'i ) Ho{nozygous normal-vision mother and colorblind father. 2. Colorblind mother and normal-vision father' 3. Heterozygous normal-vision mother and normal-vision father' (+.; Heterorygot t norrnal-vision mother and colorblind father- l. Z. l-*\- heterozYgous for the trait hybrid for the trait a trait is said to be: i3.: homozygous for the trait 4 incompletely dominant for the trait When two four-o'clock plants are crossed a8 pink four-o'clocks and 52 white fouro'clocks are produced. The phenotypes of the parents are: f . 2. pink and white pink and red 3. pink and pink 4. red and white as: GENETIC g. PROBLEMS polydactyly is a human characteristic in which a person has six fingers per hand. The trait for polydactyly is dominant over the trait for five fingers. If a man who is hetero4rgous for this trait manies a woman with the normal number of fingers, what are the chances that their child would be PolYdactYl? '2.'soyo I.0Y" g. 75Yo 4. t000A TI O Oo, is homozygous dominant and the other heterorygous. goth are hetero4/gous for the dominant trait. 3. 4. One is homozygous dominant and the other homozygous recessive Both are homozygous for the dominant tait. T-r tlrye O has a mother with blood type A and a father with blood brye B The parental genotypes for blood types must be: A child with blood l. 2. 11. 3. \., Which statement describes how two organisms may show the same traig yet have different genotyPes for that PhenotYPe? Z 10. 2 3. IAIB and IAIB ?-)loi and IBi IAIA and IBIB fi and IBIB ', n, agt#^' 'i?'-;, Two pea plants, hybrid for a single trait, produce 60 pea plants. Approximately how many '1-tt of the pea plants are expected to exhibit the recessive tait? L tl );., i. , 2. 4s lY 12. 1 -.', .,' (2,.-'zs"l" 'v- A colorb-lind woman marries a man who of having a colorblind daughter floz t4. 3. 30 ll-','--: 4. 60 7,-i:a ( +:._:: _l -_:_ Three brothers have blood types A" B, and O. What are the chances that the parents these three will produce a fourth child whose blood type is AB? l. 0% 13. ,' 2. 250h _r-{l_ 3.50oA has 4. to}% ' T" : in,: i- --**, li:.,.11*. vvv X hormal color vision. What are their chances 3.750A 4. 1000A A student crossed wrinkled-seeded (rr) pea plants with round-seeded (RR) pea plants. Only round seeds were produced in the resulting plants. This is caused by: l. 2. independent assortrnent segregation of 3.'dominance -4. incomplete dominance GENETIC 3 PROBLEMS end of ln peas, flowers located along the stem,(*ig) are dominant to flowers located at the th; stem (terminal.) Let A represent the allele for axial flowers and a represent the allele for terminal flowers. When plants with axial flowers are crossed with plants having terminal flowers, all of the offspring have axial flowers. In the cross the genotypes of the parent Plants are most likelY: 15. t. 16. aa x aa 2- Aa i'i.-)'a * * x Aa 4' AA x AA ln cabbage butterflies, white color flMJ is dominant and yellow color (w) is recessive. If a pure white cabbage butterlly mates with a yellow cabbage butterfly, all the resulting (F1) tutterllies are heterozygous white. Which cross represents the genotype of the parent generation? 2. WW x l.Wwxww Ww tl.)WW x ww 4. Ww x Ww 17. When a mouse with black fur is crossed with a mouse with white fir, all F1 generation results can be expected in the F2 generation? offspring - trave gray fiu. Which phenotype ',, /'Jt '.1, -.-t=l t. t00o/o gray _3. s}yoblac( 50% white '; Z. 2so/oUlack Tl%ow\ute .'",'-. ,'-4.-125Y"blac( 50%o gray,25%ow\ttte 18. There are multiple alleles for the ABO blood goup. Why are there only trvo of these alleles normally present in any one individual? l. There are not enough nucleotides in a red blood cell to produce a third allele. ,./f)tEu"h parent contributes only one allele for the ABO blood goup to the offspring. '-3. Each allele in the ABO goup must be either dominant or recessive. 4. Blood goup alleles are not segregated during meiosis. 19. Which will most probably XY be true of the children of a iolorblind man and a woman with normal color vision? \, l. Atl males will carry the gene for color blindness- 2. Nl males will be colorblind. 3. All females will be colorblind. 7|'Ntfemales will carry the gene for color \-/ 20. A father has type A blood and his child has rype O blood. Which blood goup genotype does the child's mother probably have? l. 2t. blindness. I"I" 2. The offspring of a cross between l. 2. dominance segregation a r3' J. I I I"lb Bb x 4. Iblb Bb illustrates the pattern of heredity known 3. 4. independent assorftnent sex linkage as: lr 4 GENETIC PROBLENIS On your al^wer sheet, write the best atawer to each question. for questions Use the following information 22 - 28' b, i ,- - , humans, the gene for brown eyes, B, is dominant to the gene for blue eyes, l" brownwas father whose and A bro*o-eyed man, whose mother *ui blu.-eyed eye{ and a blue-eyed womalL both of whose parents were brown-eye4 marry' I' 22. What is the woman's 23. What is tle,man's genotYPe? genotYPe? 't" ;'i. ": 24. What is the genot)?e of the man's mother? 25. What is the genotype of the woman's parents? .a l, .,. ', 7,6. If the man and woman have four children, how many would be expected to be heterozYgous for eYe color? L 27. How many of the children would be expected to have blue eyes? 28. How many of the-phildren would be expected to have brown eyes? t! t- For questions 29 & 30 refer to the follov'ing tnformation: ln pigeons, checkered pattern is dependent on a dominant gene C, and plain putt.rn is dependent on recessive allele c. Red color is controlled by a dominant gene B and brown color is controlled by recessive gene b. '', - 29. What is the result of a cross between a homozygoul checkered, redpigeon and homozygous Plain' brown Pigeon? ' 30. a t' A male pigeon who's genotype is CcBb is mated with the following female cCbB How many pigeons will be plain and brown? 31. In rabbits, black frr is dependent on dominant gene B and brown fi.r on the recessive allele b. Normal length fiu is determined b1' a dominant gene R and short fi,u on recessive gene r' :, Show the results of a cross befween rwo heterozygous black ,,orrnil'l.ngth fur rabbits. How many offspring will probably be homozygous for both traits? What is the ratio of phenotypes? , \ I rl , !t', 1,., i/'''t it' i' tt,r.. "fi-.r,ftf--ir :t,i i-. ,Lr-, i ) 5 GEI{ETIC PROBLEMS l1 32. Rolling the tongue is dominant over non-rolling and rurattached ears are dominant over attached ears. '' '' '.) f.- If a tongue rollin! woman with attaihed ears has a child who is a non-ioller with ,"r.. unattached ears, what are the probably genotypes of the woman, her husban4 and ' the child? i'.. I i ,) ,; f l"\' - ". 33. In sheep, white coud ii dominant to black. Occasionally, a black sheep appears in the flock. The farmer cannot sell black wool. How can the gene for black coat be eliminated from the flock? 34. In guinea pigs with colored hair, the hair may be black (B) or brown (b). Black is dominant to brown hair. Also, the hair may be short (S) or long (s). Short hair is dominant to long hair. ln each cross described so far, we have begun with homozygous individuals. This time we will cross animals of heterozygous genotypes: BbSs x BbSs. What is their phenotype? What will be the genotypes and phenotypes of the F1 offspring? 35. So far, we have spoken of traits as dominant or recessive. But there are other possibilities. In shorthorn cattle, for instance, when a red bull is crossed with a white cow, the heterozygous offspring are neither red nor white. lnstead, they are roan, which means they have mixed red and white hairs. In a cross between a roan bull and a roan cow, what ate are the chances it will be ioan or red? the chances that the calf will be white? *ul 36. 37 . Work out a cross of roan x ri,:i red. If a four-o'clock plant with red flowers is crossed with a white-flowered four-o'clocl<, F1 plants all have pink flowers. the From this evidence alone, what ratios would you expect to frnd in these crosses: 1. red 38. white? Suppose two newborn babies were mixed up in the hospital. From the following blood t)?es, determine which baby belongs to which parents: Baby l: Baby 2 39. 2. Piok x x pink? ' . Type Type O A Mrs. Brown Mr. Brown Type Type B AB Mrs- Smith Mr. Smith Type B Type B If a woman who is not a carrier of genes for hemophilia is married to a man who is a hemophiliac, what percentage of their male offspring could be expected to be hemophiliacs? WhY? 40. a woman whose father was colorblind, what is the probability child being a colorblind boy? If a colorblind man marries of their frst