Unit 5 AP Biology Heredity

UNIT 5 Heredity

Some text here about the topic of discussion

​Review Notes

Make sure you completed ALL Mastering Biology for 1 Completion Grade

Make sure you completed ALL Packet with questions answered/vocabulary too. 1 Completion Grade.

AP Classroom - Personal Progress Check - 1 Completion Grade.

Multiple Choice

A father with type B blood and mother with type A blood have a child. Their child has discovered that her blood group is type O, during biology experiment at her school. Does she have any concerns about her parentage?

• Yes, because she should have type AB blood if they are her true biological parents

• No, because type O is possible if her parents both had genotype AB

• No, because both of her parents could be heterozygous

• Yes, because both of her parents might be hetrozygous

• No, because blood types A and B are codominant

​Correct option is C)

ABO blood group is an example of codominance and multiple allelism wherein genotypes IA IO and IA IA gives A blood group. The possible genotypes of B blood group are IB IO and IB IB. If father is heterozygous for B blood group (IB IO) and mother is heterozygous for A blood group (IA IO), their child can have O blood group. This is because of the presence of one copy of IO allele in both parents. Thus, the correct answer is option C.

The child could have any blood type because we could receive an O allele from either parent.

Multiple Choice

Which of these is an example of polygenic inheritance?

• Freckles

• Sickle Cell Anemia

• Duchenne muscular dystrophy

• Skin Color

• Color Blindness

​Polygenetic inheritance is where multiple genes affect a single trait. Human skin color depends on three sets of alleles: Aa, Bb, and Cc. A cross between two parents with any combination of these three alleles determines skin color; there is no single skin color gene.

Duchenne Muscular Dystrophy is X-linked Recessive Inheritance.

Multiple Choice

Which occurs when a heterozygous genotype produces blended traits?

• Incomplete dominance

• Pure genotype

• Monosomy

• Codominance

• Trisomy

​Incomplete dominance involves expression of an intermediate phenotype. The heterozygotes express a phenotype that is a blend of both the dominant and recessive phenotypes. One common example is a flower with white petals and a flower with red petals sexually reproduce to create flowers with pink petals. 

Multiple Choice

Epistasis controls the color of squash, with the B locus controlling color (yellow is dominant to green), and E locus determining expression of B locus. This is a case of dominant epistasis. Which of the following is true?

• bbEe results in green squash

• Bbee results in yellow squash

• bbee results in yellow squash

• bbEe results in yellow squash

​Epistasis describes the interaction of genes, where the epistatic locus masks the effects of a gene at another locus. In this example, locus E is epistatic. As this is stated to be dominant epistasis, when the E locus is either Ee or EE this locus will mask the effect of the B locus (color). Thus, any combination of B/b with Ee or EE will result in white squash. When the E locus is homozygous recessive (ee), the effect of the B locus will not be masked. Thus, BbEe will result in yellow squash as ee will not mask the color, and B (yellow) is dominant.

Epistasis is a circumstance where the expression of one gene is modified (e.g., masked, inhibited or suppressed) by the expression of one or more other genes.

Multiple Choice

In a specific type of flower the genes coding for color display codominance. What would you expect the phenotype to be for a cross between a flower homozygous for white coloration and a flower homozygous for red coloration?

• White

• Pink

• Red

• Distinct red and white spots

​Codominance is a phenomenon in which the phenotypes associated with both alleles will be expressed in their entirety. This expression pattern results in mottled expression, creating distinct red and white spots for the flower. This is different than incomplete dominance, in which the two phenotypes appear to blend together.

Multiple Choice

In a population of flowers, red is dominant to white. A true-breeding white flower is crossed with a heterozygous flower. Determine the expected ratios of this cross. Given observed values: 63 white flowers, 37 red flowers, determine:

1) The chi squared value

2) The degrees of freedom

3) The p-value

• 1) 14.44    

  2) 2    

  3) p>0.005

• 1) 6.76    

  2) 1    

  3) p>0.025

• 1) 3.35    

  2) 4    

  3) p=0.25

• 1) 6.76    

  2) 2    

  3) p=0.05Unit

​First, determine the expected ratios using a Punnet Square. Given that red is dominant to white, the genotype of a true-breeding white flower can be denoted rr, and the genotype of a heterozygous flower can be denoted Rr. This cross will produce 50% Rr, 50% rr. Then, converting these percentages to decimals (50%= 0.50), and multiplying by the total population size of the observed population (0.50*100), gives expected values of 50 Rr (red flowers) and 50 rr (white flowers). These expected values and the observed values can then be plugged into the chi square equation (Σ(O−E)2/E). The equation will be Σ(63−50)2/50+(37−50)2/50)=6.76. Degrees of freedom is n-1, so 2-1= 1. To determine p-value, use critical values table. The chi square value of 6.76 with 1 degrees of freedom will fall between a critical value corresponding with a p-value of 0.05 and 0.025. Thus, p>0.025

Multiple Choice

In a chi squared test, the greater the difference between the observed and expected frequencies of a trait, the ___ the p-value.

• More negative

• More Positive

• Greater

• Lower

​A larger difference between observed and expected values will result in a larger chi square value. Using the critical values table, for a given degrees of freedom, as chi square value increases, p-value decreases.

Multiple Choice

If the chi square value is larger than the critical value at a given level of significance, what can be stated?

• The null hypothesis is rejected

• The null hypothesis fails to be rejected

• The alternative hypothesis is rejected

• The null hypothesis is accepted

​If the chi square value is large, this indicates a large difference between the observed and expected values. This will subsequently result in a small p-value when using the critical value table. When the p-value is equal to or smaller than the significance level (alpha), the null hypothesis is rejected.

Multiple Choice

Epistasis controls the fur color of labradors, with the B locus controlling color (Black is dominant to brown), and E locus determining expression of B locus. This is a case of recessive masking epistasis. Which of the following is true?

• BBee results in black fur

• bbee results in brown fur

• BbEe results in black fur

• bbEe results in black fur

​Epistasis describes the interaction of genes, where the epistatic locus masks the effects of a gene at another locus. In this example, locus E is epistatic. As this is stated to be recessive masking epistasis, when the E locus is homozygous recessive (ee), this locus will mask the effect of the B locus (color). Thus, any combination of B/b with ee will result in yellow fur. When the E locus is heterozygous (Ee) or homozygous dominant (EE), the effect of the B locus will not be masked. Thus, BbEe will result in black fur (as black B is dominant to brown b), and the E locus does not mask the B locus here, as the E locus is heterozygous in this individual (Ee).

Multiple Choice

When two genes interact, and one gene masks the expression of a second gene, the first gene is __ to the second or ___ gene

• hypostatic, epistatic

• epistatic, hypostatic

• recessive, dominant

• dominant, recessive

​Epistasis describes the interaction of genes, where the epistatic gene masks the effects of another gene, called the hypostatic gene.

the hypostatic gene is the one being masked.

Multiple Choice

The following crossover frequencies were noted via experimentation for a set of five genes on a single chromosome:

A and B → 35%

B and C → 15%

A and C → 20%

A and D → 10%

D and B → 25%

A and E → 5%

B and E → 40%

​The crossover frequencies are an indication of the distance between the different genes on a chromosome. The farther apart they are, the greater chance there is that they will cross over during prophase I of meiosis. You are first told that A and B cross over with a frequency of 35 percent, so imagine that they are 35 units apart on a chromosome map.

We can then tell you that B and C have a frequency of 15 percent. They are 15 units apart on the map, but you cannot yet be sure what side of gene A that C is on. Gene A and C cross with 20 percent frequency. This means that gene C must be in between A and B.

	ABC	AbC	ABc	Abc	aBC	abC	aBc	abc
ABC	AABBCC	AABbCC	AABBCc	AABbCc	AaBBCC	AaBbCC	AaBBCc	AaBbCc
AbC	AABbCC	AAbbCC	AABbCc	AAbbCc	AaBbCC	AabbCC	AaBbCc	AabbCc

Gene A crosses over with D 10 percent of the time, and D crosses with B 25 percent of the time; therefore, D must also be in between A and B. It is closer to A than it is to B. You can use this knowledge to eliminate answer choices B and C.

Gene A crosses over with E with a frequency of 5 percent. You do not know which side of A gene E is on until you know its crossover frequency with B. Because the question tells you that it has a 40 percent frequency with B, you know that it must be on the left of A. This completes your map, leaving A as the correct answer.

Multiple Choice

Imagine that in squirrels, gray color (G) is dominant over black color (g). A black squirrel has the genotype gg. Crossing a gray squirrel with which of the following would let you know with the most certainty the genotype of the gray squirrel?

• GG

• Gg

• gg

• Cannot be determined

​This is a test cross. To determine the genotype of an individual showing the dominant phenotype, you cross that individual with a homozygous recessive individual for the same trait. If they have no offspring with the recessive phenotype, then the individual displaying the dominant phenotype is most likely GG. If approximately one-half of the offspring have the recessive phenotype, you know the individual has the genotype Gg.