Bio 30 Unit 3 Lesson 6 Assignment (1)

Biology 30 Unit 3 Lesson 6 Assignment Total /50 1. In the Punnett square, the filled-in squares represent: A. the distribution of the phenotypes B. heterozygotes C. homologous sets of chromosomes D. possible gametic genotypes /1 Use the following information to answer the next question. Pure-breeding long stemmed and pure-breeding short stemmed pea plants were crossed. The results of the cross are shown in the diagram. Long Stemmed Short Stemmed F 1 generation F 2 generation Cross-breeding Peas 240074 2. According to Mendel’s laws of heredity, the F 2 generation in the diagram will be

A. all long stemmed B. one half long stemmed and one half short stemmed C. three quarters short stemmed and one quarter long stemmed D. one quarter short stemmed and three quarters long stemmed /1 3. Explain Mendel’s first two laws of inheritance using green and yellow pod colour as an example. Mendel's principle number one, called the Law of Segregation, describes how different forms of a gene these are alleles separate when cells that create offspring come into being. Every living thing gets two versions for every characteristic: it receives one from its mother and another from its father. When we look at pea plants with green and yellow pod colors, if a plant has GG genes, it makes gametes that have only "G" gene inside. But if the plant's genes are gg, then every gamete will have just the "g" gene. The rule makes sure that reproductive cells have just a single version of each characteristic. Mendel's second rule, the Independent Assortment Law, means when creating reproductive cells, different genes for traits split on their own. So how one feature is passed down like the color of a pea pod (G/g) doesn't affect another one's inheritance like the color of seeds inside (Y/y). For example, the way pod color is passed on does not influence how seed color is passed on. This idea shows that different characteristics are given and sorted by themselves when making gametes, giving a structure to know how different genetic qualities go from parents to children. /4 4. The physical appearance of a particular trait in an organism is referred to as its A. genotype B. phenotype C. heterozygous D. homozygous /1 5. A certain type of dwarfism in humans is inherited as a simple monogenic trait (a trait controlled by one pair of alleles). Two dwarfs marry and have two children, one of which is dwarf and the other is normal size. Which of the following statements about this trait is correct? A. Dwarfism is produced by a recessive allele. B. Dwarfism is produced by a dominant allele. C. The father is homozygous for the normal allele. D. The mother is homozygous for the dwarfism allele.

D. ssWW /1 Use the following information to answer this question. The diagram shows possible chromosomal alignment and subsequent segregation during meiosis in pea plants. 10. The genotypes of the gametes 1, 2, 3, and 4, in order, are A. YR, yr, Yr, yR B. Yr, YR, YR, yr C. YR, YR, Yr, ry D. YR, Yr, YR, Yr /1 11. What is the best way to determine the eye-color phenotype of an organism?

A. analyze a blood sample B. do a DNA electrophoresis test C. X-ray the organism D. look at the organism's eyes /1 12.  For a given trait, the two genes of an allelic pair are alike. An individual possessing this gene is said to be A. heterozygous for that trait  B. homozygous for that trait   C. recessive for that trait D. hybrid for that trait /1 Use the information to answer the question. The allele for black coat colour ( B ) is dominant over the allele for white coat colour ( b ) in dogs. The allele for short hair ( S ) is dominant over the allele for long hair (s). The phenotypes of offspring from several crosses are given below . Parental Phenotypes Cross Phenotypes of Offspring black short black long white short white long 1 2 3 4 white, short X white, short black, short X black, long black, short X black, long black, long X black, long 16 0 6 0 15 0 5 31 0 27 3 0 0 8 2 10 50250 13. Complete the following. a. What are the genotypes for parents of each of the four crosses? Cross 1:  The black, short parent is homozygous for a black coat color and heterozygous for hair length: BB Ss.  The black, long-haired parent possesses a genotype of Bb Ss, being heterozygous for both coat color and hair length.

calculation for obtaining an heir of white long-haired progeny involves determining chances to pass on each allele ('b' and 's') from both parents: specifically, it's a 50% chance for each allele transfer resulting in overall possibility being 25%. A chance of obtaining white and long-haired offspring based on allele inheritance acknowledges the dominant traits for black hair (B) in one parent; these often override expression of the contrasting trait. If one parent carries primarily dark-haired alleles, this could significantly curtail manifestation even eliminate the expression of its light-haired counterpart within their progeny. /2 14. In corn, a dominant allele Su produces smooth, starchy kernels. The recessive allele su produces sweet, wrinkled kernels. What is the probability that an F 1 generation plant from a cross between a plant homozygous for sweet, wrinkled seeds and a plant homozygous for smooth, starchy seeds will have sweet, wrinkled seeds? A. 0%   B. 25% C. 75%   D. 100% /1 15. A plant breeder crosses two strains of plants, both of which are homozygous for two traits ( AABB × aabb ). The offspring, F 1 are then crossed to produce an F 2 generation. What is the probability that an F 2 plant will have the genotype AABB ? A. 0  B. 1/16   C. 3/16  D. 9/16 /1 16. When certain tall pea plants are crossed with short pea plants, only tall pea plants result in the offspring. This illustrates the principle of  A. incomplete dominance  B. dominance   C. independent assortment D. segregation

/1 17. Which term best describes most mutations?  A. dominant and disadvantageous to the organism B. recessive and disadvantageous to the organism   C. recessive and advantageous to the organism D. dominant and advantageous to the organism /1 18. The appearance of the recessive trait in the offspring of animals most probably indicates that A. one parent was homozygous dominant and the other parent was hybrid for the trait B. neither parent carried a recessive gene for the trait C. one parent was homozygous dominant and the other parent was homozygous recessive for the trait D. both parents carried at least one recessive gene for that trait /1 19. Albinism is a condition where pigment cells do not produce any pigment. As a result, albinos have white hair and white skin. It is caused by a recessive allele. A husband and wife, both heterozygous for the gene, plan to have a child. What is the probability that the baby will be an albino male? A. 12.5% B. 25% C. 37.5% D. 50% /1 Use the following information to answer the next 3 questions The Punnett square shown below shows the results of a dihybrid cross between pea plants. (The Punnett square does not show the gametes produced by the plants.) S represents the allele for long stem, s is the allele for short stem, W represents the allele for purple flowers and w represents the allele for white flowers.

CC Normal Pigmentation Cc Albinism Cc ½ carrier chance The Punnett square reveals the following probabilities: each baby has a 25% chance of inheriting albinism (cc); they may not carry any copies of the recessive gene; in which case their chances are eliminated altogether. Each child, whether they are in a group or individually considered, demands an individualized plan; this necessitates comprehensive understanding of their unique needs and circumstances. A one in four probability (25%) exists for the occurrence of being born with albinism (cc). The probability of both babies having albinism is determined by multiplying the individual probabilities: 0.25 times 0.25, resulting in a final value of 0.0625; this signifies that there's only a slight chance for both infants to be born with the condition.

A 6.25% chance or a ratio of 1 in 16 exists for both babies to have albinism. /4 24. A normal couple’s first child has the disease phenylketonuria. Assuming that the disease is inherited as an autosomal recessive trait, what is the probability that their next child will inherit the disorder? What are the genotypes of the parents? If two carriers procreate, each child bears a 25% chance of inheriting dual copies of the recessive allele (pp) and consequently manifests PKU. Both parents being carriers (Pp), and the autosomal recessive nature of phenylketonuria (PKU) inheritance indeed establish a 25% likelihood for their next child to inherit this condition. /3 25. In humans the gene for Rh + blood type is dominant over the gene for Rh – blood type. A husband and wife, both Rh + , have an Rh – son. What is the probability that their next child will be an Rh + male? Single gene with two alleles - Rh+ (dominant) and Rh- (recessive) determines the Rh factor. Both parents, being Rh+, should only pass on the dominant allele to their offspring; however, since they have an Rh- son, we know that both carry a recessive trait for the Rh- allele. Father's genotype is Rh+/Rh-; this indicates that he possesses a dominant trait-- specifically, Rh+. Mother's genotype is Rh+/Rh-, reflecting an identical blood type of Rh+ (RR). Each parent must carry the Rh- allele as a recessive trait (Rh-/Rh-) to potentially have an Rh- child; therefore, we consider each parent's genotype: Rh+/Rh-. Let's consider the possible genetic combinations to predict: what is the probability of their next child being an Rh+ male? The father is Rh-positive/Rh-negative, and the mother is also either Rh-positive or Rh- negative. These cross yields the following Punnett square: This demonstrates a 50% probability: specifically, out of the four potential outcomes, two points to their forthcoming child as an Rh+ male. The next child they conceive carries a 50% probability of being Rh /3 26. Co-dominance is caused by an interaction between